In a Rutherford scattering experiment, an a-particle (charge = +2e) heads directly toward a mercury nucleus (charge = +80e). The α-particle had a kinetic energy of 4.7 MeV when very far (r→ [infinity]) from the nucleus.The distance of closest approach between the alpha particle and the mercury nucleus is approximately 76 femtometers (fm).
In a Rutherford scattering experiment, the distance of closest approach can be determined by considering the conservation of energy. Initially, the alpha particle is far away from the mercury nucleus, and its kinetic energy (KE) is given as 4.7 MeV.
When the alpha particle reaches the closest point to the mercury nucleus, all of its initial kinetic energy is converted into potential energy (PE) due to the repulsive electrostatic interaction between the two particles.
Using the principle of conservation of energy, we can equate the initial kinetic energy to the final potential energy:
KE_initial = PE_final
The initial kinetic energy is given as 4.7 MeV, which can be converted to joules by using the conversion: 1 MeV = 1.6 x 10^(-13) Joules.
KE_initial = 4.7 MeV * (1.6 x 10^(-13) Joules/MeV)
= 7.52 x 10^(-13) Joules
The potential energy between the alpha particle and the mercury nucleus is given by Coulomb's law:
PE = kₑ * (|q₁| * |q₂|) / r
where kₑ is the electrostatic constant (8.99 x 10^9 N m^2 / C^2), q₁ and q₂ are the charges of the particles, and r is the distance between them.
For an alpha particle (charge = +2e) and a mercury nucleus (charge = +80e), we can substitute the values into the equation:
PE = kₑ * (2e * 80e) / r
= kₑ * (160e^2) / r
Now we can equate the initial kinetic energy to the final potential energy:
KE_initial = PE_final
7.52 x 10^(-13) Joules = kₑ * (160e^2) / r
Rearranging the equation to solve for r:
r = kₑ * (160e^2) / (KE_initial)
Substituting the known values:
r = (8.99 x 10^9 N m^2 / C^2) * (160 * (1.6 x 10^(-19) C)^2) / (7.52 x 10^(-13) Joules)
Evaluating the expression:
r ≈ 7.6 x 10^(-14) m ≈ 76 fm
Therefore, the distance of closest approach between the alpha particle and the mercury nucleus is approximately 76 femtometers (fm).
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A wire of unknown composition has a resistance of R 0
=36.5Ω when immersed in water at 26.2 ∘
C. When the wire is placed in boiling water, its resistance rises to 71.3Ω. What is the temperature when the wire has a resistance of 41.6Ω ? Number Units
Therefore, we cannot use the temperature coefficient of resistance for copper wire, which is 0.00428/°C. We would need to know the temperature coefficient of resistance for the specific wire we are using.
The temperature when the wire has a resistance of 41.6Ω is 45.7 ∘C.What is the resistance-temperature characteristic of the wire?The equation used to solve this problem isR = R0 (1 + αΔT)where R is the resistance at temperature T, R0 is the resistance at a reference temperature T0, α is the temperature coefficient of resistance, and ΔT is the difference between T and T0.Rearranging the equation givesΔT = (R - R0) / (R0α)The temperature coefficient of resistance α for a wire of unknown composition is not given. However, the resistance-temperature characteristic for most materials is known, and the temperature coefficient of resistance can be determined from it. For a copper wire, for example, α = 0.00428/°C.Substituting the given values,R0 = 36.5ΩR = 41.6ΩT0 = 26.2°CΔT = (41.6Ω - 36.5Ω) / (36.5Ω × α)For the copper wire, ΔT = (41.6Ω - 36.5Ω) / (36.5Ω × 0.00428/°C) = 28.5°C.Therefore, the temperature when the wire has a resistance of 41.6Ω is T = T0 + ΔT = 26.2°C + 28.5°C = 54.7°C.However, we were not given the material composition of the wire. Therefore, we cannot use the temperature coefficient of resistance for copper wire, which is 0.00428/°C. We would need to know the temperature coefficient of resistance for the specific wire we are using.
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A laser with wavelength 656 nm is incident on a diffraction grating with 1600 lines/mm.
(a) (15 points) Find the smallest distance from the grating that a converging lens with focal length of 20 cm be placed so that the diffracted laser light converges to a point 1.0 meter from the grating.
(b) (15 points) If a screen is placed at the location from part (a), how far apart will the two first order beams appear on the screen? (If you did not solve part (a), use a distance of 0.5 m).
(a) The converging lens should be placed at a distance of 1.95 meters from the diffraction grating to converge the diffracted laser light to a point 1.0 meter from the grating.
(b) The two first-order beams will appear approximately 0.04 meters (or 4 cm) apart on the screen.
(a) To determine the smallest distance for placing the converging lens, we can use the lens formula:
1/f = 1/v - 1/u,
where f is the focal length of the lens, v is the image distance, and u is the object distance. In this case, the lens will form an image of the diffracted laser light at a distance of 1.0 meter from the grating (v = 1.0 m). We need to find the object distance (u) that will produce this image location.
Using the diffraction grating equation:
d * sin(θ) = m * λ,
where d is the spacing between the grating lines, θ is the angle of diffraction, m is the order of the diffracted beam, and λ is the wavelength of the laser light. Rearranging the equation, we have:
sin(θ) = m * λ / d.
For the first-order beam (m = 1), we can substitute the values of λ = 656 nm (or 656 × 10^(-9) m) and d = 1/1600 mm (or 1.6 × 10^(-6) m) into the equation:
sin(θ) = (1 * 656 × 10^(-9)) / (1.6 × 10^(-6)).
Solving for θ, we find the angle of diffraction for the first-order beam. Using this angle, we can then determine the object distance u by trigonometry:
u = d / tan(θ).
Plugging in the values, we can calculate u. Finally, subtracting the object distance u from the image distance v, we get the required distance from the grating to the converging lens.
(b) Once we have the converging lens in place, we can calculate the separation between the two first-order beams on the screen. The distance between adjacent bright spots in the interference pattern can be determined by:
Δy = λ * L / d,
where Δy is the separation between the bright spots, λ is the wavelength of the laser light, L is the distance from the grating to the screen, and d is the spacing between the grating lines.
Substituting the values of λ = 656 nm (or 656 × 10^(-9) m), L = 1.95 m (the distance from the grating to the converging lens), and d = 1/1600 mm (or 1.6 × 10^(-6) m), we can calculate Δy. The resulting value will give us the distance between the two first-order beams on the screen.
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A ring of radius R = 2.20 m carries a charge q = 1.99 nC. At what distance, measured from the center of the ring does the electric field created by the ring reach its maximum value? Enter your answer rounded off to 2 decimal places. Do not type the unit. Consider only the positive distances.
The electric field created by the ring reaches its maximum value at a distance of 1.27 meters from the center of the ring.
The electric field created by a ring with a given radius and charge reaches its maximum value at a distance from the center of the ring. To find this distance, we can use the equation for the electric field due to a ring of charge.
By differentiating this equation with respect to distance and setting it equal to zero, we can solve for the distance at which the electric field is maximum.
The equation for the electric field due to a ring of charge is given by:
E = (k * q * z) / (2 * π * ε * (z² + R²)^(3/2))
where E is the electric field, k is the Coulomb's constant (9 * [tex]10^9[/tex] N m²/C²), q is the charge of the ring, z is the distance from the center of the ring, R is the radius of the ring, and ε is the permittivity of free space (8.85 * [tex]10^{-12}[/tex] C²/N m²).
To find the distance at which the electric field is maximum, we differentiate the equation with respect to z:
dE/dz = (k * q) / (2 * π * ε) * [(3z² - R²) / (z² + R²)^(5/2)]
Setting dE/dz equal to zero and solving for z, we get:
3z² - R² = 0
z² = R²/3
z = √(R²/3)
Substituting the given values, we find:
z = √((2.20 m)² / 3) = 1.27 m
Therefore, the electric field created by the ring reaches its maximum value at a distance of 1.27 meters from the center of the ring.
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An object is located 72 cm from a thin diverging lens along the axis. If a virtual image forms at a distance of 18 cm from the lens, what is the focal length of the lens? in cm Is the image inverted or upright?
The focal length of the lens is -24 cm (negative sign indicates a diverging lens). Regarding the orientation of the image, for a diverging lens, the image formed is always virtual and upright.
To determine the focal length of the lens, we can use the lens formula:
1/f = 1/v - 1/u
where:
f is the focal length of the lens,
v is the distance of the virtual image from the lens (positive for virtual images),
u is the distance of the object from the lens (positive for objects on the same side as the incident light).
Given that the object is located 72 cm from the lens (u = -72 cm) and the virtual image forms at a distance of 18 cm from the lens (v = -18 cm), we can substitute these values into the lens formula:
1/f = 1/-18 - 1/-72
Simplifying this expression:
1/f = -1/18 + 1/72
= (-4 + 1) / 72
= -3/72
= -1/24
Now, taking the reciprocal of both sides of the equation:
f = -24 cm
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The strength of the Earth's magnetic field has an average value on the surface of about 5×10 5
T. Assume this magnetic field by taking the Earth's core to be a current loop, with a radius equal to the radius of the core. How much electric current must this current loop carry to generate the Earth's observed magnetic field? Given the Earth's core has a radius of approximately R core
=3x10 6
m. (Assume the current in the core as a single current loop).
Summary: To generate the Earth's observed magnetic field, the current loop representing the Earth's core needs to carry an electric current of approximately 1.57x10^6 Amperes.
The strength of a magnetic field generated by a current loop can be calculated using Ampere's law. According to Ampere's law, the magnetic field strength (B) at a point on the loop's axis is directly proportional to the current (I) flowing through the loop and inversely proportional to the distance (r) from the loop's center. The equation for the magnetic field strength of a current loop is given by B = (μ₀ * I * N) / (2π * r), where μ₀ is the permeability of free space, N is the number of turns in the loop (assumed to be 1 in this case), and r is the radius of the loop.
In this scenario, the Earth's core is assumed to be a single current loop with a radius (r) equal to the radius of the core, which is given as R_core = 3x10^6 meters. The average magnetic field strength on the Earth's surface is given as 5x10^-5 Tesla. Rearranging the equation for B, we can solve for I: I = (2π * B * r) / (μ₀ * N). Plugging in the given values, we get I = (2π * 5x10^-5 Tesla * 3x10^6 meters) / (4π * 10^-7 T m/A). Simplifying the expression gives us I ≈ 1.57x10^6 Amperes, which represents the electric current required for the Earth's core to generate the observed magnetic field.
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A beam of ultraviolet light with a power of 2.50 W and a wavelength of 124 nm shines on a metal surface. The maximum kinetic energy of the ejected electrons is 4.16 eV. (a) What is the work function of this metal, in eV?
(b) Assuming that each photon ejects one electron, what is the current?
(c) If the power, but not the wavelength, were reduced by half, what would be the current?
(d) If the wavelength, but not the power, were reduced by half, what would be the current?
The energy required to eject an electron from a metal surface is known as the work function. To find the work function of this metal, we can use the formula:
Work function = hυ - KEMax
Work function = hυ - KEMax
Power of ultraviolet light = 2.50 Wavelength of ultraviolet light = 124 nm Maximum kinetic energy of ejected electrons = 4.16 eV Planck's constant (h) = 6.626 × 10^-34 Js Speed of light (c) = 3 × 10^8 m/s
The energy of a photon is given by
E = hυ = hc/λ where h = Planck's constant, υ = frequency of light, c = speed of light and λ = wavelength of light.
We have to convert the wavelength of ultraviolet light from nm to m.
Therefore, λ = 124 nm × 10^-9 m/nm = 1.24 × 10^-7 m
The frequency of the ultraviolet light can be calculated by using the above equation.
υ = c/λ = (3 × 10^8 m/s)/(1.24 × 10^-7 m) = 2.42 × 10^15 Hz
Now, we can substitute these values in the formula for work function:
Work function = hυ - KEMax= 6.626 × 10^-34 Js × 2.42 × 10^15 Hz - 4.16 eV× (1.602 × 10^-19 J/eV)= 1.607 × 10^-18 J - 6.656 × 10^-20 J= 1.54 × 10^-18 J
The work function of this metal is 1.54 × 10^-18 J
The current is given by the formula:
I = nAq where I = current, n = number of electrons per second, A = area of metal surface, and q = charge on an electron
The number of photons per second can be calculated by dividing the power of ultraviolet light by the energy of one photon.
Therefore, n = P/E = (2.50 W)/(hc/λ) = (2.50 W)λ/(hc)
The area of the metal surface is not given, but we can assume it to be 1 cm^2. Therefore, A = 1 cm^2 = 10^-4 m^2.The charge on an electron is q = -1.6 × 10^-19 C. The current can now be calculated by substituting these values in the formula:
I = nAq= (2.50 W)λ/(hc) × 10^-4 m^2 × (-1.6 × 10^-19 C)= -4.03 × 10^-13 A
Current is 4.03 × 10^-13 A.
Note that the value of current is negative because electrons have a negative charge.
If the power, but not the wavelength, were reduced by half, then the number of photons per second would be halved. Therefore, the current would also be halved. The new current would be 2.02 × 10^-13 A.
If the wavelength, but not the power, were reduced by half, then the energy of each photon would be doubled. Therefore, the number of photons per second required to produce the same power would be halved. Hence, the current would also be halved. The new current would be 2.02 × 10^-13 A.
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A Find the Resistance of 100 meters of # 18 AWG Copper wire at 20° C ? B Find the Area you need to calculate the Resistance ? C Find the Resistance of 600 meters of solid Copper wire with a diameter of 5 mm ? P Find the Area you need to calculate the Resistance ? If the Resistance of some Copper wire is 80 ohms at 20° C, what is it's Resistance at 100° C ?
a. The resistance of 100 meters of #18 AWG Copper wire at 20°C is 0.2098 Ω
b. To calculate the resistance of a wire, the cross-sectional area of the wire is required.
c. The area required to calculate the resistance is 2.155 × [tex]10^{-10}[/tex] m². The resistance of copper wire at 100°C is 92.2 Ω.
a. The resistance of 100 meters of #18 AWG Copper wire at 20°C can be determined using the formula;
R = ρL/A
A = πr²ρ
where;
R = resistance
ρ = resistivity
L = length of the wire
A = area of cross-section
r = radius of the wire
Substituting the given values;
Length of wire L = 100 meters
Area of cross-section A = ?
Diameter of wire d = 0.0403 inches or 1.02462 mm
Cross-sectional area A = πd²/4 = π(1.02462 mm)²/4 = 0.8231 mm²
Resistivity ρ = 1.724 x [tex]10^{-8}[/tex] Ω-m (at 20°C for copper)
Thus;
R = ρL/A = 1.724 x [tex]10^{-8}[/tex] Ω-m x 100 meters / 0.8231 mm²R = 0.2098 Ω
a. The resistance of 100 meters of #18 AWG Copper wire at 20°C is 0.2098 Ω
b. To calculate the resistance of a wire, the cross-sectional area of the wire is required.
c. To find the resistance of 600 meters of solid Copper wire with a diameter of 5 mm, we need to know the cross-sectional area of the wire. The formula for the cross-sectional area is;
A = πr²A = π(5/2)²A = 19.63 mm²
The resistivity of copper is 1.724 × [tex]10^{-8}[/tex] Ωm. Using the formula;
R = ρL/A
where;
L = 600 mA = 19.63 mm²
ρ = 1.724 × [tex]10^{-8}[/tex] Ωm
R = 0.16 ΩP.
To find the area required to calculate the resistance, the cross-sectional area of the wire is required. If the resistance of copper wire is 80 ohms at 20°C, we can use the above formula for resistivity.
ρ = RA/L
where;
R = 80 Ω
A = ?
L = 1 m
ρ = 1.724 × [tex]10^{-8}[/tex] Ωm
A = ρL/R = 1.724 × [tex]10^{-8}[/tex] × 1/80A = 2.155 × [tex]10^{-10}[/tex] m²
The resistance of copper wire at 100°C can be determined using the formula;
Rt = R0 [1 + α(T[tex]_{t}[/tex] - T[tex]_{0}[/tex])]
where;
R0 = resistance at 20°C = 80 Ω
T0 = temperature at 20°C = 293 K (20 + 273)
Tt = temperature at 100°C = 373 K (100 + 273)
α = temperature coefficient of copper = 0.00393/°C
Rt = 80 [1 + 0.00393(373 - 293)]R[tex]_{t}[/tex] = 92.2 Ω
Answer:
Therefore area required to calculate the resistance is 2.155 × [tex]10^{-10}[/tex] m². The resistance of copper wire at 100°C is 92.2 Ω.
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A bullet of mass 10.0 g travels with a speed of 120 m/s. It impacts a block of mass 250 g which is at rest on a flat frictionless surface as shown below. The block is 20.0 m above the ground level. Assume that the bullet imbeds itself in the block. a) Find the final velocity of the bullet-block combination immediately affer the collision. (9pts) b) Calculate the horizontal range x of the bullet-block combination when it hits the ground (see figure above). (8pts) b) Calculate the horizontal range x of the bullet-block combination when it hits the ground (see figure above). ( 8 pis) c) Calculate the speed of the bullet-block combination just before it hits the ground. (8pis)
Part A, we need to find the final velocity of the bullet-block combination immediately after the collision. In part B, we are asked to calculate the horizontal range x of the bullet-block combination when it hits the ground. Part C, we need to determine the speed of the bullet-block combination just before it hits the ground.
In Part A, we can apply the principle of conservation of momentum. Since the system is isolated, the momentum before the collision is equal to the momentum after the collision. By considering the momentum of the bullet and the block separately, we can find the final velocity of the combined system.
In Part B, we can determine the time it takes for the bullet-block combination to hit the ground by using the equation of motion in the vertical direction. The displacement is the height of the block, and the initial velocity is the final velocity found in Part A. With this time, we can then calculate the horizontal range x using the equation of motion in the horizontal direction.
In Part C, the speed of the bullet-block combination just before it hits the ground can be found by considering the conservation of mechanical energy. Since the system is isolated and there is no work done due to friction or other forces, the initial mechanical energy is equal to the final mechanical energy.
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A ball of mass 113.0 g is hit by another object with a speed of 45 m/s. The ball was in contact with the object about 3.2 *10^-3 s. Find (a) the impulse imparted to the ball, (b) the average force exerted on the ball by the object.
A) The impulse imparted to the ball is 5.09 N s and B) the average force exerted on the ball by the object is approximately 1580 N.
(a) Given, Mass of the ball, m = 113.0 g
Initial velocity, u = 0
Final velocity,v = 45 m/s
Time of contact, t = 3.2 × 10⁻³ s
Here, the impulse imparted to the ball can be calculated using the above formula as,Δv = v - u = 45 - 0 = 45 m/s
Therefore, I = mΔv
I = (0.113 kg) × 45 m/sI = 5.09 N s
(b) Average force is the force that acts on an object during the time of its motion. It is represented by F = m(a) / t, where F is the force, m is the mass of the object, and a is the acceleration it experiences.
F = m(a) / t
F = m(Δv/t)
F = m[(v-u)/t]
F = m (Δv/t)
F = (0.113 kg) [(45 m/s - 0)/3.2 × 10⁻³ s]
F = 1581.5625 N ≈ 1580 N
Therefore, the average force exerted on the ball by the object is approximately 1580 N.
Hence, the impulse imparted to the ball is 5.09 N s and the average force exerted on the ball by the object is approximately 1580 N.
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Figure 4.1 shows three charged particles located at the three corners of a rectangle. Find the electric field at the fourth vacant corner. (25 points) q 1
=3.00nC
q 2
=5.00nC
q 3
=6.00nC
x=0.600m
y=0.200m
Figure 4.1
The electric field at the fourth vacant corner is 4.05 × 10⁵ N/C.
Given,Three charged particles are located at the three corners of a rectangle.The magnitude of q1, q2 and q3 are given as 3 nC, 5 nC and 6 nC respectively.The value of x = 0.6m and the value of y = 0.2m.Figure 4.1The electric field at the fourth vacant corner can be calculated as follows:
We can make use of the formula given below to find the magnitude of the electric field,where k is the Coulomb constant and the magnitude of q1, q2 and q3 are given as 3 nC, 5 nC and 6 nC respectively, The value of x = 0.6m and the value of y = 0.2m. E = kq/r²Where k = 9 × 10⁹ N m²/C²The magnitude of q1, q2 and q3 are given as 3 nC, 5 nC and 6 nC respectively.r₁ = x² + y²r₁ = 0.6² + 0.2²r₁ = √(0.36 + 0.04)r₁ = √0.4r₁ = 0.6324 m r₂ = y²r₂ = 0.2²r₂ = 0.04 mTherefore, the electric field at the fourth vacant corner is 4.05 × 10⁵ N/C (approx).
Thus, the electric field at the fourth vacant corner is 4.05 × 10⁵ N/C.
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Which neutrino types are involved in the following decays? In your answer, please substitute the subscripts x and y that you see in the reactions below with the correct neutrino type (e, jl, or T) (i) π^+ → µ + Vx (ii) vx + p → µ^+ + n (iii) Vx + n → + p + e^-
(iv) T^- → Vx + µ^- + Vy What guiding principles do we have to follow to determine the neutrino types in the decays above?
To determine the neutrino types in the given decays, we need to follow the principles of lepton flavor conservation and charge conservation.
Lepton Flavor Conservation: According to this principle, the lepton flavor of the neutrinos involved in a decay must be conserved. In other words, the type of neutrino produced in a decay should match the type of neutrino that is present in the initial state.
Charge Conservation: Charge must also be conserved in each decay process. The sum of the charges of the particles on both sides of the reaction should be equal.
With these principles in mind, let's determine the neutrino types in each decay:
(i) π^+ → µ^+ + Vx
In this decay, a positive pion (π^+) decays into a positive muon (µ^+) and a neutrino (Vx). Since the initial state has a positive charge, the final state must also have a positive charge to conserve charge. Therefore, the neutrino type Vx must be an electron neutrino (Ve).
(ii) Vx + p → µ^+ + n
In this decay, a neutrino (Vx) interacts with a proton (p) and produces a positive muon (µ^+) and a neutron (n). Again, we need to conserve charge. Since the initial state has no charge, the final state must also have no charge. Therefore, the neutrino type Vx must be an electron neutrino (Ve).
(iii) Vx + n → p + e^- + Vy
In this decay, a neutrino (Vx) interacts with a neutron (n) and produces a proton (p), an electron (e^-), and a neutrino (Vy). Charge conservation tells us that the initial state has no charge, so the final state must also have no charge. Therefore, the neutrino type Vx must be a muon neutrino (Vμ).
(iv) T^- → Vx + µ^- + Vy
In this decay, a negative tau lepton (T^-) decays into a neutrino (Vx), a negative muon (µ^-), and a neutrino (Vy). The charge of the initial state is negative, and the final state also has a negative charge. Therefore, both neutrinos Vx and Vy must be tau neutrinos (Vτ).
By applying the principles of lepton flavor conservation and charge conservation, we can determine the appropriate neutrino types in the given decays.
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Determine the velocity required for a moving object 5.00×10 3
m above the surface of Mars to escape from Mars's gravity. The mass of Mars is 6.42×10 23
kg, and its radius is 3.40×10 3
m.
The velocity required for a moving object 5.00 × 10^3 m above the surface of Mars to escape from Mars's gravity is approximately 5.03 × 10^3 m/s.
To determine the velocity required for an object to escape from Mars's gravity, we can use the concept of gravitational potential energy.
The gravitational potential energy (PE) of an object near the surface of Mars can be given by the equation:
PE = -GMm / r
where G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2), M is the mass of Mars (6.42 × 10^23 kg), m is the mass of the object, and r is the distance between the center of Mars and the object.
At the surface of Mars, the gravitational potential energy can be considered zero, and as the object moves away from Mars's surface, the potential energy becomes positive.
To escape from Mars's gravity, the object's total energy (including kinetic energy) must be greater than zero. The kinetic energy (KE) of the object can be given by:
KE = (1/2)mv^2
where v is the velocity of the object.
At the escape point, the total energy (TE) of the object is the sum of its kinetic and potential energies:
TE = KE + PE
Since the object escapes Mars's gravity, its total energy at the escape point is zero:
0 = KE + PE
Rearranging the equation, we can solve for the velocity:
KE = -PE
(1/2)mv^2 = GMm / r
Simplifying the equation:
v^2 = (2GM) / r
Taking the square root of both sides:
v = √[(2GM) / r]
Now we can substitute the values into the equation:
v = √[(2 * 6.67430 × 10^-11 * 6.42 × 10^23) / (3.40 × 10^3 + 5.00 × 10^3)]
Calculating the value:
v ≈ 5.03 × 10^3 m/s
Therefore, the velocity required for a moving object 5.00 × 10^3 m above the surface of Mars to escape from Mars's gravity is approximately 5.03 × 10^3 m/s.
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N A siren emits a sound of frequency 1. 44 × 103 Hz when it is stationary with respect to an observer. The siren is moving away from a person and toward a cliff at a speed of 15 m/s. Both the cliff and the observer are at rest. Assume the speed of sound in air is 343 m/s. What is the frequency of the sound that the person will hear a. Coming directly from the siren and b. Reflected from the cliff?
To calculate the frequency of the sound heard by the person, we need to consider the Doppler effect, which describes the change in frequency due to the relative motion between the source of the sound and the observer.
The formula for the observed frequency due to the Doppler effect is given by:
f_observed = f_source * (v_sound + v_observer) / (v_sound + v_source)
where:
f_observed is the observed frequency,
f_source is the source frequency,
v_sound is the speed of sound in air, and
v_observer and v_source are the velocities of the observer and the source, respectively.
Given:
Source frequency (f_source) = 1.44 × 10^3 Hz
Speed of sound in air (v_sound) = 343 m/s
Velocity of the siren (v_source) = 15 m/s
Velocity of the observer (v_observer) = 0 m/s (since the observer is at rest)
(a) Frequency of the sound directly from the siren:
For this scenario, the observer and the siren are moving away from each other. Substituting the given values into the Doppler effect formula:
f_observed = 1.44 × 10^3 * (343 + 0) / (343 + 15)
(b) Frequency of the sound reflected from the cliff:
In this case, the sound waves are reflected by the cliff, resulting in a change in direction. The relative motion between the observer and the reflected sound is the sum of their individual velocities. Thus, we consider the observer's velocity as -15 m/s (since it's moving towards the observer).
f_observed = 1.44 × 10^3 * (343 + 0) / (343 - 15)
By performing the calculations, we can determine the frequencies of the sound heard by the person directly from the siren and reflected from the cliff.
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A charged capacitor with a capacitance of C=5.00×10 −3
F, has an initial potential of 5.00 V. The capacitor is discharged by connecting a resistance R between its terminals. The graph below shows the potential across the capacitor as a funtion of the time elapsed since the connection. C.alculate the value of R. Note that the curve passes through an intersection point. Tries 1/20 Previous Tries
The value of resistance R is 3.48 kΩ.
The capacitance of a charged capacitor is C=5.00×10−3F, and its initial voltage is 5.00V. When a resistor R is connected between its terminals, it is discharged. The potential across the capacitor versus time since the connection is plotted in the graph shown.The capacitor's voltage and current change as it charges and discharges. The voltage across the capacitor as a function of time elapsed since the connection is shown in the graph.
The voltage of the capacitor decreases exponentially and eventually approaches zero as it discharges.The capacitor discharge is given by the following equation:q = Q × e−t/RCWhere R is the resistance, C is the capacitance, t is the time elapsed, and q is the charge stored in the capacitor at time t. The voltage across the capacitor can be determined using the following formula:V = q/C = Q/C × e−t/RC.
The voltage across the capacitor is plotted in the graph, and the intersection point is located at t = 5.0ms and V = 2.5V. As a result, the charge stored on the capacitor at that moment is Q = CV = 5.00×10−3F × 2.50V = 12.5×10−3C.The value of R can now be calculated using the formula:R = t/ln(V0/V) × C = 5.0×10−3s/ln(5.00V/2.50V) × 5.00×10−3F ≈ 3.48kΩTherefore, the value of resistance R is 3.48 kΩ.
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A cow (200 g) is accidentally accelerated to 0.6 c. Determine the kinetic energy of the cow. (Use Special Relativity).
To determine the kinetic energy of a cow accelerated to 0.6 times the speed of light (c) using special relativity, we can utilize the relativistic kinetic energy equation.
In special relativity, the relativistic kinetic energy equation takes into account the effects of high velocities. It is given by the equation:
K = (γ - 1) * mc^2,
where K is the kinetic energy, γ is the Lorentz factor, m is the mass of the object, and c is the speed of light.
The Lorentz factor, γ, is defined as:
γ = 1 / √(1 - v^2/c^2),
where v is the velocity of the object
To calculate the kinetic energy of the cow, we first need to convert the mass from grams to kilograms (200 g = 0.2 kg). The speed of light, c, is approximately 3.0 x 10^8 m/s.
Next, we calculate the Lorentz factor, γ, using the given velocity:
γ = 1 / √(1 - (0.6c)^2/c^2).
Using the Lorentz factor, we can plug it into the relativistic kinetic energy equation along with the mass and the speed of light to find the kinetic energy of the cow:
K = (γ - 1) * mc^2.
By substituting the values into these equations, we can determine the kinetic energy of the cow accelerated to 0.6 times the speed of light using special relativity.
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Complete each statement with the correct term. A collision in which some kinetic energy is lost is a(n)_____collision. A collision in which the objects become one and move together is a(n)_____inelastic collision.
A radio transmitter broadcasts at a frequency of 96,600 Hz. What is the wavelength of the wave in meters? What is the wavelength (in nanometers) of the peak of the blackbody radiation curve for something at 1,600 kelvins?
The wavelength of a radio wave with a frequency of 96,600 Hz is approximately 3.10 meters. The peak wavelength of blackbody radiation for an object at 1,600 kelvins is around 1,810 nanometers.
To calculate the wavelength of a radio wave, we can use the formula: wavelength = speed of light / frequency. The speed of light is approximately 299,792,458 meters per second. Therefore, for a radio wave with a frequency of 96,600 Hz, the calculation would be: wavelength = 299,792,458 m/s / 96,600 Hz ≈ 3.10 meters.
Blackbody radiation refers to the electromagnetic radiation emitted by an object due to its temperature. The peak wavelength of this radiation can be determined using Wien's displacement law, which states that the peak wavelength is inversely proportional to the temperature of the object. The formula for calculating the peak wavelength is: peak wavelength = constant / temperature. The constant in this equation is approximately 2.898 × 10^6 nanometers * kelvins.
Plugging in the temperature of 1,600 kelvins, the calculation would be: peak wavelength = 2.898 × 10^6 nm*K / 1,600 K ≈ 1,810 nanometers. Thus, for an object at 1,600 kelvins, the peak wavelength of its blackbody radiation curve would be around 1,810 nanometers.
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Find the attached image illustrates the thermal resistance model for two devices mounded on single heatsink : Tj1 1 kQ 1 kQ www ww Rjc1 Device Ta 1 KQ 1 www Rsa Tj2 1kQ 1 ΚΩ www www Rcs2 Device Rjc2 2 Where, * Tj1 - Device 1 junction temperature = 180°C * Tj2 - Device 2 junction temperature = 180°C * Rjc1 - Device 1 junction to case thermal resistance = 4 K/W * Rjc2 - Device 2 junction to case thermal resistance = 2 K/W * Rcs1,Rcs2 - Device 1 and device 2 case to heatsink thermal resistance (heatsink grease) = 0.038 K/W * Rsa - heat sink thermal resistance ( need to be find). * Ta - ambient temperature = 40°C * The formula for heatsink (as specifically available based on its thermal resistance, Rsa) is * Rsa = Tj1 - Ta - Pd1 (Rjc1 + Rcs1)/(Pd1 + Pd2) Where, * Pd1 - power dissipated by device 1 * Pd2 - power dissipated by device 2 * Then, * Rsa = 180 - 40 - 16(4+0.038) / (16+24) * Rsa = 1.88 K/W * The heatsink thermal resistance (Rsa) = 1.88 K/W. Rcs1
Two MOSFETS are used to control the brightness of a high power spotlight. Under maximum power both MOSFETS in the circuit as shown are conducting. M1 dissipates a maximum of 16 W and has a junction to case thermal resistance of 4 K/W. M2 dissipates a maximum of 24 W and has a junction to case thermal resistance of 2 K/W. Both MOSFETs are mounted on a common heatsink (with isolation). The maximum junction temperature of the MOSFETs is 180 °C and the circuit must operate in an ambient temperature of 40 °C. Please assist with getting the required heatsink. A thermal circuit will aid my understanding so please draw the thermal circuit first.
The problem involves two MOSFETs mounted on a common heatsink, and the goal is to determine the required thermal resistance of the heatsink.
Given the power dissipation and thermal resistance values of the MOSFETs, along with the maximum junction temperature and ambient temperature, the thermal circuit needs to be analyzed to find the required heatsink thermal resistance.
To analyze the thermal circuit and determine the required heatsink thermal resistance, we can start by visualizing the circuit as a thermal network. The key components in the circuit are the MOSFETs (M1 and M2), their junction-to-case thermal resistances (Rjc1 and Rjc2), the case-to-heatsink thermal resistances (Rcs1 and Rcs2), and the unknown heatsink thermal resistance (Rsa). We also have the maximum junction temperature (Tj1 = Tj2 = 180°C) and the ambient temperature (Ta = 40°C).By applying the thermal circuit equations, we can write the following expression to calculate Rsa:
Rsa = (Tj1 - Ta - Pd1 * (Rjc1 + Rcs1)) / Pd1
where Pd1 is the power dissipated by device M1 (16 W) and Rjc1 is the junction-to-case thermal resistance of M1 (4 K/W). We can substitute these values into the equation and solve for Rsa.
Similarly, for M2, we have:
Rsa = (Tj2 - Ta - Pd2 * (Rjc2 + Rcs2)) / Pd2
where Pd2 is the power dissipated by device M2 (24 W) and Rjc2 is the junction-to-case thermal resistance of M2 (2 K/W).
Once we have the values of Rsa from both equations, we can compare them and choose the larger value as the required heatsink thermal resistance to ensure proper heat dissipation and keep the MOSFETs within their maximum temperature limits.
In conclusion, by constructing the thermal circuit and applying the thermal equations, we can determine the required heatsink thermal resistance (Rsa) to keep the MOSFETs within their temperature limits. This ensures the reliable operation of the circuit under the given power dissipation and ambient temperature conditions. The thermal circuit analysis helps in understanding the heat flow and designing effective cooling solutions to maintain the components at safe operating temperatures.
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A parallel-plate capacitor has a capacitance of 21μF when filled with air and it can withstand a potential difference of 49 V before it suffers electric breakdown. (a) What is the maximum amount of charge we can place on this air-filled capacitor? The dielectric strength of 3.00×106 V/m. c (b) If we fill this capacitor with polyethylene, what will be its new capacitance? F (c) What will be the maximum potential difference that this new capacitor can withstand? V (d) What will be the corresponding maximum amount of charge we can place on this capacitore is 1.80×107 V/m. C
a) The formula for capacitance is given as:
C=Q/V
Where Q is the charge on the capacitor and
V is the voltage across the capacitor.
Rearranging the formula gives the charge on the capacitor, Q=CV
The maximum amount of charge we can place on this air-filled capacitor is:
Q = CV = 21 × 10⁻⁶ × 49 = 1.029 × 10⁻³ C
b) The new capacitance of the capacitor if we fill this capacitor with polyethylene is given by:
Cnew = εrε0A/d
Where εr is the relative permittivity of the polyethylene, ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.
Cnew = εrε0A/d
= 2.3 × ε0 × A/d
c) The maximum potential difference that this new capacitor can withstand is:
Vmax = Ed
Where E is the dielectric strength of the polyethylene, and d is the distance between the plates.
Vmax = Ed = 1.8 × 10⁷ V/md)
The corresponding maximum amount of charge we can place on this capacitor is given by:
Q= CVmax
The value of Vmax has been obtained in the previous part.
Hence,Q = Cnew
Vmax = 2.3 × ε0 × A/d × 1.8 × 10⁷ V/m
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The magnetic field is 1.50uT at a distance 42.6 cm away from a long, straight wire. At what distance is it 0.150mT ? 4.26×10 2
cm Previous Tries the middle of the straight cord, in the plane of the two wires. Tries 2/10 Previous Tries
The magnetic field strength of [tex]0.150 \mu T[/tex] is achieved at a distance of approximately 13.48 cm from the long, straight wire.
The magnetic field generated by a long, straight wire decreases with distance according to the inverse square law. This means that as the distance from the wire increases, the magnetic field strength decreases.
For calculating distance at which the magnetic field strength is [tex]0.150 \mu T[/tex], a proportion is set using the given information. Denote the distance from the wire where the field strength is[tex]0.150 \mu T[/tex] as x.
According to the inverse square law, the magnetic field strength (B) is inversely proportional to the square of the distance (r) from the wire. Therefore, following proportion can be set as:
[tex](B_1/B_2) = (r_2^2/r_1^2)[/tex]
Plugging in the given values,
[tex](1.50 \mu T/0.150 \mu T) = (42.6 cm)^2/x^2[/tex]
Simplifying the proportion:
[tex]10 = (42.6 cm)^2/x^2[/tex]
For finding x, rearrange the equation:
[tex]x^2 = (42.6 cm)^2/10\\x^2 = 181.476 cm^2[/tex]
Taking the square root of both sides,
x ≈ 13.48 cm
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Find the speed (in terms of c) of a particle (for example, an electron) whose relativistic kinetic energy KE is 5 times its rest energy E in - -
. For example, if the speed is 0.500c, enter only 0.500. Keep 3 digits after the decimal point.
The speed of the particle is approximately 0.993c.
According to Einstein's theory of relativity, the relativistic kinetic energy (KE) of a particle can be expressed as KE = (γ - 1)[tex]mc^2[/tex], where γ is the Lorentz factor and m is the rest mass of the particle.
We are given that the kinetic energy is 5 times the rest energy, which can be expressed as KE = 5[tex]mc^2[/tex].Setting these two equations equal to each other, we have (γ - 1)[tex]mc^2[/tex] = 5[tex]mc^2[/tex]. Simplifying, we get γ - 1 = 5, which leads to γ = 6.
The Lorentz factor γ is defined as γ = 1/√[tex](1 - v^2/c^2)[/tex], where v is the velocity of the particle. We can rearrange this equation to solve for v: v = c√(1 - 1/γ^2).
Plugging in γ = 6, we find v ≈ 0.993c. Therefore, the speed of the particle is approximately 0.993c.
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A sound source is detected at a level of 54 dB Intensity of 2512-07 W/m?) when there is no background noise. How much will the sound level increase if there were 53,5 dB (Intensity of 2.239-07 W/m?) b
If the sound level increases from 54 dB (intensity of 2.512×10⁻⁷ W/m²) to 53.5 dB (intensity of 2.239×10⁻⁷ W/m²), the sound level will increase by approximately 0.5 dB.
Sound level is measured in decibels (dB), which is a logarithmic scale used to express the intensity or power of sound. The formula to calculate the change in sound level in decibels is ΔL = 10 × log₁₀(I/I₀), where ΔL is the change in sound level, I am the final intensity, and I₀ is the reference intensity.
Given that the initial sound level is 54 dB, we can calculate the initial intensity using the formula I₀ = 10^(L₀/10). Similarly, we can calculate the final intensity using the given sound level of 53.5 dB.
Using the formulas, we find that the initial intensity is 2.512×10⁻⁷ W/m² and the final intensity is 2.239×10⁻⁷ W/m².
Substituting these values into the formula to calculate the change in sound level, we get ΔL = 10 × log₁₀(2.239×10⁻⁷ / 2.512×10⁻⁷) ≈ 0.5 dB.
Therefore, the sound level will increase by approximately 0.5 dB when the intensity changes from 2.512×10⁻⁷ W/m² to 2.239×10⁻⁷ W/m².
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Suppose 2000 J of heat are added to 3.4 mol of argon gas at a constant pressure of 140 kPa. Part A Find the change in internal energy. Part B Find the change in temperature for this gas. Express your answer using two significant figures
Part C Calculate the change in volume of the gas.
The resulting change in temperature of the argon gas is approximately 34.62 Kelvin.
To determine the change in temperature of the argon gas, we can use the formula:
ΔQ = nCpΔT
where:
ΔQ is the heat added to the gas (in joules),
n is the number of moles of the gas,
Cp is the molar specific heat capacity of the gas at constant pressure (in joules per mole per kelvin),
ΔT is the change in temperature (in kelvin).
In this case, we have:
ΔQ = 2000 J
n = 3.4 mol
Cp (specific heat capacity of argon at constant pressure) = 20.8 J/(mol·K) (approximately)
We need to rearrange the formula to solve for ΔT:
ΔT = ΔQ / (nCp)
Substituting the given values into the equation, we have:
ΔT = 2000 J / (3.4 mol * 20.8 J/(mol·K))
Calculating the result:
ΔT ≈ 34.62 K
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--The complete Question is, Suppose 2000 J of heat are added to 3.4 mol of argon gas at a constant pressure of 140 kPa. What will be the resulting change in temperature of the gas? Assume the argon gas behaves ideally.--
Jupiter, Saturn, Uranus, and Neptune are larger than the terrestrial planets because They formed in cooler parts of the solar nebula where the most abundant elements could condense They formed before the Sun formed whereas the rocky planets formed from leftover material They formed in a different solar system and were captured by the Sun's gravity They formed close to the Sun but have been gradually moving away from the Sun for the past 4.6 billion years
Jupiter, Saturn, Uranus, and Neptune are larger than the terrestrial planets because they formed in cooler parts of the solar nebula where the most abundant elements could condense.
They are known as gas giants and are mostly composed of helium and hydrogen. These planets are also referred to as outer planets since they are located far from the sun. It is said that these planets are colder than the rocky planets.
Jupiter, Saturn, Uranus, and Neptune, the four gas giants, are much larger than the four inner planets. They are larger because they formed in cooler regions of the solar nebula, where the most abundant elements, such as helium and hydrogen, could condense. When the gas giants developed, they attracted these elements, and as a result, they formed enormous gaseous planets. These gas giants have a more complex structure than the inner planets. The cores of these planets are comprised of rock and ice, whereas the outer layers are composed of hydrogen and helium gas.
The gas giants are far from the sun and are referred to as outer planets. They are colder than the rocky planets since they are positioned further from the sun. Additionally, the outer planets rotate faster than the inner planets. Jupiter rotates the fastest of all the planets and takes about 9 hours and 56 minutes to rotate completely on its axis.
The gas giants are much larger than the inner planets since they formed in cooler regions where the most abundant elements could condense. The gas giants are mostly composed of hydrogen and helium and have a complex structure with rocky cores and gas outer layers. The outer planets rotate faster than the inner planets and are far from the sun, which makes them colder.
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Suppose that two stars in a binary star system are separated by a distance of 90 million kilometers and are located at a distance of 110 light-years from Earth. What is the angular separation of the two stars? Give your answer in degrees. Express your answer using two significant figures. Part B What is the angular separation of the two stars? Give your answer in arcseconds. Express your answer using two significant figures.
Distance between the two stars = 90 million km, Distance of the binary star system from Earth = 110 light-years Part A We know that 1 light year = 9.461 × 10¹² km
Therefore, Distance of binary star system from Earth = 110 × 9.461 × 10¹² km Distance of binary star system from Earth = 1.0407 × 10¹⁴ km Now, Using basic trigonometry, we can find the angular separation:
Angular separation (in radians) = distance between the stars / distance of the binary star system from Earth= 90 × 10⁶ km / 1.0407 × 10¹⁴ km Angular separation (in radians) = 8.65 × 10⁻⁹ radians
Now, We know that 2π radians = 360 degrees. Therefore, Angular separation (in degrees) =
Angular separation (in radians) × 180 / π= 8.65 × 10⁻⁹ radians × 180 / π
Angular separation (in degrees) = 0.00000156 degrees Angular separation (in degrees) = 1.6 × 10⁻⁶ degrees Part B We know that 1 degree = 3600 arcseconds. Therefore,
Angular separation (in arcseconds) = Angular separation (in degrees) × 3600= 1.6 × 10⁻⁶ degrees × 3600
Angular separation (in arcseconds) = 0.0056 arcseconds Angular separation (in arcseconds) = 0.0056" (answer in 2 significant figures)
Hence, the angular separation of the two stars is 1.6 × 10⁻⁶ degrees and 0.0056".
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A student wears eyeglasses that are positioned 120 cm from his eyes. The prescription for the eyeglasses should be open Wut the case he can see clearly without vision correction State answer in centers with 1 digit right of decimal Do not include
A student wears eyeglasses that are positioned 120 cm from his eyes..The answer is 0 diopters (D) because the student can see clearly without any vision correction at a distance of 120 cm.
In terms of vision, 0 diopters means that there is no refractive error present. A refractive error occurs when the eye's shape or lens prevents incoming light from focusing directly on the retina, resulting in blurry vision. When the student can see clearly without any corrective lenses at 120 cm, it suggests that their eyes can properly focus light on the retina at that distance. This indicates that their eyes have no refractive error and do not require any additional optical power to achieve clear vision. Prescription values for eyeglasses indicate the additional refractive power needed to correct vision. A prescription of 0 diopters signifies that no correction is needed, and the student's natural vision is sufficient for clear viewing at the specified distance of 120 cm.
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A Force of F= (4.20i +3.60j) N is applied to a rigid body of mass 1.50 kg rotating around a fixed axis . Determine the torque experienced by the particle when the force is applied at the position of r= (1.50i+ 2.20j)
Which direction is the Torque oriented?
The torque experienced by the particle is 10.38 N·m, and its direction is perpendicular to the plane formed by the position vector and the force vector.
To determine the torque experienced by the particle, we need to calculate the cross product of the position vector and the force vector. The formula for torque is given by:
τ = r × F
where τ represents the torque, r is the position vector, and F is the force vector. In this case, the position vector r is (1.50i + 2.20j) and the force vector F is (4.20i + 3.60j).
Taking the cross product of these vectors, we have:
τ = (1.50i + 2.20j) × (4.20i + 3.60j)
Expanding the cross product, we get:
τ = (1.50 * 3.60 - 2.20 * 4.20)k
Simplifying the equation, we have:
τ = (5.40 - 9.24)k
τ = -3.84k
Therefore, the torque experienced by the particle is -3.84 N·m. The negative sign indicates that the torque is oriented in the opposite direction to the positive z-axis.
Since torque is a vector quantity, it has both magnitude and direction. The direction of the torque is determined by the right-hand rule. In this case, the torque is oriented along the negative z-axis, which means it is pointing into the plane formed by the position vector and the force vector.
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An airplane starts from west on the runway. The engines exorta constant force of 78.0 KN on the body of the plane (mass 9 20 104 KO) during takeofc How far down the runway does the plane reach its takeoff speed of 46.1m/s?
An airplane starts from west on the runway. The engines extort constant force of 78.0 KN on the body of the plane (mass 9 20 104 Kg) during takeoff . The plane reaches its takeoff speed after traveling approximately 1135.17 meters down the runway.
To find the distance the plane travels down the runway to reach its takeoff speed, we can use the equations of motion.
The force exerted by the engines is given as 78.0 kN, which can be converted to Newtons:
Force = 78.0 kN = 78.0 × 10^3 N
The mass of the plane is given as 9.20 × 10^4 kg.
The acceleration of the plane can be determined using Newton's second law:
Force = mass × acceleration
Rearranging the equation, we have:
acceleration = Force / mass
Substituting the given values, we find:
acceleration = (78.0 × 10^3 N) / (9.20 × 10^4 kg)
Now, we can use the equations of motion to find the distance traveled.
The equation that relates distance, initial velocity, final velocity, and acceleration is
v^2 = u^2 + 2as
where:
v = final velocity = 46.1 m/s (takeoff speed)
u = initial velocity = 0 m/s (plane starts from rest)
a = acceleration (calculated above)
s = distance traveled
Plugging in the values, we have:
(46.1 m/s)^2 = (0 m/s)^2 + 2 × acceleration × s
Simplifying the equation, we can solve for 's':
s = (46.1 m/s)^2 / (2 × acceleration)
Calculating this, we find:
s ≈ 1135.17 m
Therefore, the plane reaches its takeoff speed after traveling approximately 1135.17 meters down the runway.
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Objects Cooling in Air Animal Size and Heat Transfer Room temperature T 2
= The miope of yroph in (T− 7
1
T. vs t is oqual to - . Computer Graph: thang Excel to Plos in (T. Ty vs f for (1 in; 2 in and 3 in Spbares). From each 3reph, deternaine the values of f, the conling rates. 3 plets (conviant flots Analyals: if f - D, where r is the cocling rate and D is the diameter ef the sphere, then 10gr=n 69
D. The slope of log rvs
log D
is the power n. r=4−int d=x−int facwill itek of iclationilf. lefoes the slope aid. collanigrate: Computer Graph: Using Excel to Plot log r vs
log D
. Slope = How does the cooling rate, r, depend on the diameter, D, of the sphere? Circle the equation best describes this dependence. r=1/D 3
r=1/D 2
r=1/Dr−Dr=D 2
r=D 3
The cooling rate, r, depends on the diameter, D, of the sphere such that r=D2.
The given slope of log r vs log D is -2. The equation which best describes the dependence of the cooling rate, r, on the diameter, D, of the sphere is given by:r = D2. Explanation: The cooling rate, r, for a given sphere depends on its diameter, D.
The cooling rate can be expressed as: r = k Dn, where k is a proportionality constant and n is the power to which D is raised. We need to find how the cooling rate depends on the diameter of the sphere. The slope of log r vs log D is the power n. Given: Slope of log r vs log D is -2. Therefore, n = -2.The relation between r and D is given as:r = k Dnr = k D-2r = k / D2From the above equation, we can see that the cooling rate is inversely proportional to the square of the diameter. Therefore, the cooling rate, r, depends on the diameter, D, of the sphere such that r = D2.
Thus, the equation which best describes the dependence of the cooling rate, r, on the diameter, D, of the sphere is given by:r = D2.
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At what absolute temperature do the Celsius and Fahrenheit temperature scales give the same numerical value? What is the value? (include a minus sign if required.) The Celsius and Fahrenheit temperature scales give the same numerical value at an absolute temperature of The Celsius temperature is ∘C. The Fahrenheit temperature is
The Celsius temperature is -40 degrees Celsius, and the Fahrenheit temperature is also -40 degrees Fahrenheit at this absolute temperature.
To find the absolute temperature at which the Celsius and Fahrenheit scales give the same numerical value, we can set up an equation and solve for the unknown temperature. The relationship between Celsius (C) and Fahrenheit (F) temperatures is given by the equation:
F = (9/5)C + 32
Since we want the Celsius and Fahrenheit temperatures to be equal, we can set up the equation:
C = (9/5)C + 32
To solve for C, we can simplify the equation:
C - (9/5)C = 32
(5/5)C - (9/5)C = 32
(-4/5)C = 32
Now we can solve for C:
C = 32 × (-5/4)
C = -40
Therefore, the Celsius temperature is -40 degrees Celsius, and the Fahrenheit temperature is also -40 degrees Fahrenheit at this absolute temperature.
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