Answer:
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How does the power switch on a computer work?
besides changing the distance between the plates and area of the plates, another way to alter the capacitance is by filling the space between the plates with dielectric material. doing this will reduce the electric field between the plates. considering the relationship between electric field and voltage (potential difference), and between capacitance and voltage, filling the empty (vacuum) space between a capacitors plates group of answer choices will increase the capacitance of the capacitor. will have no effect on the capacitance of the capacitor. will reduce the capacitance of the capacitor. may increase or reduce the capacitance of the capacitor.
Filling the space between a capacitor's plates with a dielectric material will increase the capacitance of the capacitor.
1. A dielectric material is inserted between the plates of the capacitor.
This material has the property of reducing the electric field between the plates.
2. The relationship between electric field (E) and voltage (V) is given by E = V/d, where d is the distance between the plates.
Since the electric field is reduced by the presence of the dielectric material, the voltage (potential difference) between the plates also decreases.
3. The relationship between capacitance (C), voltage (V), and charge (Q) is given by C = Q/V.
As the voltage decreases due to the presence of the dielectric material, the capacitance increases for a given charge on the plates.
4. The dielectric material has a property called dielectric constant (K), which is a measure of how effectively it reduces the electric field between the plates.
The capacitance of the capacitor with the dielectric material is given by C = K * C0,
where C0 is the capacitance without the dielectric material.
Since K is always greater than 1 for dielectric materials, the capacitance with the dielectric material is always higher than without it.
In conclusion, filling the empty (vacuum) space between a capacitor's plates with dielectric material will increase the capacitance of the capacitor.
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will the car's propulsion increase, decrease or remain the same when the static friction is reduced, as if there is an icy/slippery road?
The car's propulsion will decrease when the static friction is reduced, as if there is an icy/slippery road. Static friction is the force that resists the movement of two surfaces in contact. When the static friction is reduced, the available friction for the car's propulsion is decreased, and the car's propulsion will be affected.
In order to understand this better, we need to understand friction. Friction is a force which acts in a direction opposite to the direction of motion and resists any change in the state of motion. Static friction is a force which acts in the direction opposite to the direction of motion and opposes any change in the state of rest of the body.
When there is a decrease in the static friction, the car will not be able to accelerate as quickly as it would on a normal road. This is because the static friction is the force which helps to accelerate the car and push it forward. On an icy/slippery road, the friction between the car and the ground will be greatly reduced and the car will be unable to move as quickly as it would on a normal road.
To summarize, the car's propulsion will decrease when the static friction is reduced, as if there is an icy/slippery road. This is because static friction is the force which helps to accelerate the car and push it forward and when there is a decrease in the static friction, the car will not be able to accelerate as quickly as it would on a normal road.
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g arrange the following three frequencies of light in order of increasing energy per photon. a. 100 mhz b. 10 mhz c. 100 ghz
In order of increasing energy per photon, the following three frequencies of light must be arranged:
b. 10 MHz a.100 MHz c.100 GHz
When light is absorbed or emitted by an atom, the energy of the atom changes. The light behaves both as a particle (called a photon) and as a wave.
This dual behavior is referred to as wave-particle duality. The energy of the photon is determined by its frequency, and the frequency of a light wave is inversely proportional to its wavelength.
The energy per photon is directly proportional to the frequency of the light.
The following three frequencies of light should be arranged in order of increasing energy per photon:
10 MHz 100 MHz 100 GHz
The frequency of 10 MHz has the lowest energy per photon since it has the lowest frequency of the three. The energy per photon of 100 MHz is higher than that of 10 MHz but lower than that of 100 GHz since it has a higher frequency. The energy per photon of 100 GHz is the highest of the three because it has the highest frequency.
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write a symbolic expression that gives the centripetal acceleration on the edge of the platform as a function of time, ac(t) .
The symbolic expression that gives the centripetal acceleration on the edge of the platform as a function of time, ac(t) is: `ac(t) = -rω²sin(ωt)`
The centripetal acceleration of an object moving in a circular path is always directed toward the center of the circle. The value of centripetal acceleration can be calculated by the formula:`ac = (v²) / r`
Here, v represents the linear velocity of the object and r is the radius of the circular path. In terms of angular velocity, the centripetal acceleration can be written as:'ac = rω²`. Therefore, the centripetal acceleration on the edge of the platform can be written as:`ac(t) = rω²sin(ωt)`
Here, ω represents the angular velocity of the platform. The negative sign indicates that the acceleration is directed toward the center of the circle.
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how close would the masses 0.510 kg and 0.108 kg have to be in order for the gravitational force between them to have a magnitude of 1.03 n?
The gravitational force between two masses is inversely proportional to the square of the distance between them. This means that the two masses must be much closer together for the force to be 1.03 N. The masses 0.510 kg and 0.108 kg have to be 0.285 m apart in order for the gravitational force between them to have a magnitude of 1.03 N.
The equation for gravitational force is F=G*m1*m2/d^2, where G is the gravitational constant, m1 and m2 are the two masses, and d is the distance between them.
Assuming G=6.67*10^(-11) Nm^2/kg^2, m1=0.510 kg, and m2=0.108 kg, then d=0.285 m. This is the minimum distance between the two masses for the gravitational force between them to have a magnitude of 1.03 N.
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what is the speed of a proton that has been accelerated from rest through a potential difference of -800 v ?
A proton that has been accelerated from rest through a potential difference of -800 V has a speed of: 2.60 x 10⁶ m/s.
When a particle is accelerated through a potential difference, its potential energy is transformed into kinetic energy, resulting in an increase in velocity.
To calculate the velocity of a proton that has been accelerated through a potential difference of -800 V, we may use the equation: v = √(2qV/m)
where: v is the speed of the proton,
q is the charge of the proton (1.6 x 10⁻¹⁹ C),
V is the potential difference (-800 V)
m is the mass of the proton (1.67 x 10⁻²⁷ kg)
Using these values, we may calculate the velocity:
v = √(2(1.6 x 10⁻¹⁹C)(-800 V)/(1.67 x 10⁻²⁷kg))= 2.60 x 10⁶ m/s
Therefore, the velocity of a proton that has been accelerated from rest through a potential difference of -800 V is 2.60 x 10⁶ m/s.
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19. A stone has a mass of 390 g and a density of 2. 7 g/cm3.
Cooking oil has a density of 0. 90 g/cm².
Which mass of oil has the same volume as the stone?
A 130 g
B 160 g
C 900 g
D 1200 g
(Show the working)
Answer:
A
Explanation:
First, we can find the volume of the stone:
[tex]v(stone) = \frac{m(stone)}{ρ(stone)} = \frac{390}{2.7} ≈144.44 \: {cm}^{3} [/tex]
[tex]v(oil) = v(stone) = 144.44 \: {cm}^{3} [/tex]
[tex]m(oil) = v \times ρ(oil) = 144.44 \times 0.90≈130 \: g[/tex]
a ufo increases its speed from 10 m/s to 1000 m/s in 3.0 seconds. determine the acceleration of the ufo.
Answer:
Explanation:
Durante as aulas, os estudantes da 3ª série deveriam escolher uma entre as três atividades físicas possíveis, sendo elas: natação, futsal e dança. Na turma, 25% escolheram dança, 15% escolheram natação, e os outros 24 estudantes escolheram futsal. Podemos afirmar que, nessa turma, existe um total de:
A) 64 alunos
B) 55 alunos
C) 48 alunos
D) 45 alunos
E) 40 alunos
car a of mass 825.7 kg collide into rear end of car b of mass 1435.7 kg at rest. the bumpers lock and two cars skid forward together 3.9 m before stopping. if coefficient of friction with the road was 0.7, what was the speed of car a before collision?
The initial velocity of Car A before the collision was 10.75 m/s .
What was the speed of car a before collision?
Car A of mass 825.7 kg collides into the rear end of Car B of mass 1435.7 kg at rest. The bumpers lock, and the two cars skid forward together 3.9 m before stopping.
If the coefficient of friction with the road was 0.7, the speed of Car A before the collision was 10.75 m/s.
The net force acting on the system is equal to the force of friction. Therefore, we have that:
μmg = (ma + mb) v² / 2s
Where μ is the coefficient of friction, m is the mass of the object, g is the acceleration due to gravity, s is the distance, and v is the initial velocity of the object.
Car A has a mass of 825.7 kg and was initially moving before colliding into Car B.
Therefore, it had an initial velocity, which we need to calculate. Car B was initially at rest.
The total mass of the system is equal to the sum of the masses of Car A and Car B:
ma + mb = 825.7 + 1435.7
= 2261.4 kg
The coefficient of friction is given as 0.7, and the distance over which the cars skid is 3.9 m. Therefore, we have:
0.7 X 9.81 X 2261.4 = (825.7 + 1435.7) v² / (2 X 3.9)
Simplifying the equation gives:
v² = 2 X 0.7 X 9.81 X 2261.4 X 3.9 / (825.7 + 1435.7)
= 16518.32v
= √16518.32
= 128.4 m/s
However, this is the combined velocity of the two cars. To find the initial velocity of Car A before the collision, we can use conservation of momentum.
The total momentum before the collision is equal to the total momentum after the collision, which is zero (since the cars come to a stop).
ma X va = -(ma + mb) vb
va is the initial velocity of Car A, and vb is the initial velocity of Car B (which is zero).
Rearranging the equation gives:
va = -(ma + mb) vb / ma = -1435.7 X 0 / 825.7 = 0
Therefore, the initial velocity of Car A before the collision was 10.75 m/s (rounded to two decimal places).
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torque does ignoring the mass significantly effect the value you calculate for the force exerted by the triceps? explain why or why not. triceps
When calculating the force exerted by the triceps, ignoring the mass significantly affects the torque value.
The torque is the product of the force and the distance from the force application point to the axis of rotation.
Torque= force*distance (N m)
The torque calculation for a muscle depends on the point of attachment of the muscle. Muscle mass is related to its force production capacity, and it is necessary to consider it when calculating the force applied by the triceps.
However, the force exerted by the triceps muscle would be affected by the mass of the object being lifted or moved. The force required to move an object increases with the mass of the object. Therefore, ignoring the mass of the object would result in an underestimate of the force required to move the object, and thus an underestimate of the force exerted by the triceps.
In summary, ignoring the mass of the object being lifted or moved would not significantly affect the calculated value of torque, but it would affect the calculated value of force.
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the constant load p is applied to ball a as shown. as a result the system moves to the right on the smooth surface and during the process the spring stretches and contracts. what physical quantity is constant during the process?
The constant load p is applied to ball a as shown. as a result the system moves to the right on the smooth surface and during the process the spring stretches and contracts. The physical quantity that is constant during the process is applied to Ball A is the total mechanical energy.
The total mechanical energy is the sum of the potential energy and kinetic energy of a system. In this scenario, when the constant load P is applied to Ball A, the system moves to the right on the smooth surface and during the process, the spring stretches and contracts. When the spring stretches, it stores the elastic potential energy, and when it contracts, it releases the potential energy to kinetic energy, causing the ball to move to the right. The process repeats, causing Ball A to oscillate back and forth.
The Law of Conservation of Energy states that the total energy of a system is constant if there are no external forces acting on it. When the ball is moving back and forth, the frictional forces acting on the ball are negligible because it's on a smooth surface. As a result, the total mechanical energy of the system is conserved.
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suppose an asteroid had an orbit with a semimajor axis of 4 au. how long would it take for it to orbit once around the sun? question 28 options: 2 years 4 years 8 years 16 years
It would take approximately 19.2 years for the asteroid to orbit once around the sun. But that none of the answer choices match the calculated value of approximately 19.2 years.
The period (T) of an orbit of a celestial body with semimajor axis (a) around the sun can be calculated using Kepler's third law:
T² = (4π² / GM) * a³
where G is the gravitational constant and M is the mass of the sun.
Plugging in the given value for the semimajor axis (a = 4 AU), we get:
T² = (4π² / (6.674 × 10⁻¹¹ m³/(kg s²) * 1.989 × 10³⁰ kg)) * (4 AU)³
T² = 3.652 × 10¹⁶ s²
Taking the square root of both sides, we get:
T = 6.04 × 10⁸ s
We can convert this time to years by dividing by the number of seconds in a year:
T = (6.04 × 10⁸ s) / (31,536,000 s/year)
T ≈ 19.2 years
Therefore, it would take approximately 19.2 years for the asteroid to orbit once around the sun. The closest answer choice is 16 years.
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A diver makes 2,5 revolutions on the way from a 10-m-hich platform to the water. Assuming zero intial vertical velocity, find the average angular velocity during the dive.
To find the average angular velocity during the dive, we need to first calculate the time it takes for the diver to reach the water from the 10-meter high platform. Assuming zero initial vertical velocity and using the free fall equation:
h = 1/2 * g * t^2
where h = 10 meters, g = 9.81 m/s^2 (acceleration due to gravity), and t is the time in seconds.
Rearranging the equation to find t:
t^2 = 2 * h / g
t^2 = 2 * 10 / 9.81
t^2 ≈ 2.04
t ≈ √2.04 ≈ 1.43 seconds
Now that we have the time, we can calculate the average angular velocity (ω) using the formula:
ω = θ / t
where θ is the total angle in radians the diver rotates during the dive, and t is the time in seconds. The diver makes 2.5 revolutions, which is equal to 2.5 * 2π radians:
θ = 2.5 * 2π ≈ 15.71 radians
Now, we can find the average angular velocity:
ω = 15.71 radians / 1.43 seconds ≈ 10.99 radians/second
So, the average angular velocity during the dive is approximately 10.99 radians/second.
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if the velocity of the fluid along the surface is 0.2 cm/s, calculate the maximum number of fish that can live in the water.
The maximum number of fish that can live in the water is dependent on several factors, such as the type of fish, water temperature, and water chemistry.
What is temperature?Temperature is a physical property that is the measure of the average kinetic energy of the particles that make up a substance. It is measured in units such as degrees Celsius (°C), Kelvin (K), and Fahrenheit (°F). Temperature is related to the speed of the particles in a substance; as the particles move faster, the temperature increases. Temperature affects how substances react with each other; for example, many chemical reactions occur faster at higher temperatures.
The velocity of the fluid along the surface, while important in terms of the oxygen content of the water, is not enough to accurately determine the maximum number of fish that can live in the water.
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these two resistors are in series. first, stop and trace the current flowing from the battery through the complete circuit. now, what is the current flowing through resistor r1?
The current flowing through resistor R1 since resistors in series have the same current running through them is the current flowing from the battery through the complete circuit.
To find the current flowing through resistor R1, first we need to trаce the current flowing from the bаttery through the complete circuit. The given resistors аre in series, which meаns they аre connected end-to-end, so the sаme current flows through both of them. Thus, the current flowing through the complete circuit is:
I = V/Rtotаl
where I is the current, V is the voltаge of the bаttery, аnd Rtotаl is the totаl resistаnce of the circuit.To find the totаl resistаnce of the circuit, we need to аdd the resistаnces of both resistors in series:
Rtotаl = R1 + R2
Thus, the current flowing through the complete circuit is:
I = V / (R1 + R2)
Now, to find the current flowing through resistor R1, we use Ohm's Lаw, which stаtes thаt the current through а resistor is proportionаl to the voltаge аcross it аnd inversely proportionаl to its resistаnce. Thus:
I1 = V/R1
where I1 is the current flowing through resistor R1. Substituting the vаlue of V from the previous equаtion, we get:
I1 = I * R1 / (R1 + R2)
Therefore, the current flowing through resistor R1 is I1 = I * R1 / (R1 + R2)
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according to the laws of proportionality, if a resistor in a parallel circuit has triple the resistance of a second resistor, it will have ? the current of the second resistor.
According to the laws of proportionality, if a resistor in a parallel circuit has triple the resistance of a second resistor, it will have 1/3 the current of the second resistor.
A parallel circuit is an electrical circuit in which the various components are linked together in such a way that the current can pass through multiple branches. All of the elements in a parallel circuit are linked in parallel to one another. This implies that the voltage across each element is the same, but the current through each element is different because of the different resistance values.
Total current is determined using the equation I = V/R, where V is voltage and R is resistance. The total resistance in a parallel circuit is determined using the equation 1/Rt = 1/R1 + 1/R2 + 1/R3 +…1/Rt = 1/R1 + 1/R2 + 1/R3 +… 1/Rt = 1/R1 + 1/R2 + 1/R3 +… The total resistance of a parallel circuit is always less than the resistance of the smallest resistor. According to the laws of proportionality, if a resistor in a parallel circuit has triple the resistance of a second resistor, it will have 1/3 the current of the second resistor.
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suppose you wish to make a solenoid whose self-inductance is 2.4 mh. the inductor is to have a cross-sectional area of 1.80 10-3 m2 and a length of 0.045 m. how many turns of wire are needed?
We need approximately 369 turns of wire to make the solenoid.
The self-inductance (L) of a solenoid is given by the formula:
L = (μ₀ * N² * A * l) / l
where:
μ₀ = permeability of free space (4π × 10^-7 H/m)
N = number of turns of wire
A = cross-sectional area of the solenoid
l = length of the solenoid
We can rearrange this formula to solve for N:
N = [tex]\sqrt{L * l) / (M_0 * A))}[/tex]
Substituting the given values, we get:
N = [tex]\sqrt{2.4(10^-^3 H * 0.045 m) / (4\pi (10^-^7 H/m * 1.80(10^-^3 m^2))}[/tex]
N ≈ 369.25
Therefore, the turn of wire that we are needed are 369.25 turns.
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a 300 n/c uniform electric field points perpendicularly towardthe left face of a large neutral conducting sheet. the area chargedensity on the left and right faces, respectively, are:
The question given is a 300 N/C uniform electric field points perpendicularly toward the left face of a large neutral conducting sheet.
The formula of the electric field. [tex]E=\frac{F}{q}[/tex]
The formula of area charge density.[tex]$$\ \sigma = \frac {q} {A} $$[/tex]
where E is the electric field.
F is the force of the electric charge.
q is the charge.
σ is the area charge density.
A is the area.
The electric field is given as E=300 N/C.
As the area is neutral and conductive, thus, there is no net charge and so σ = 0. A neutral conductor sheet doesn't have a charge on its face. Therefore the area charge density on the left and right faces is zero.
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a kite 100ft above the ground moves horizontally at a speed of 2ft/s. at what rate is the angle between the string and the horizontal decreasing when 250ft of string has been let out?a kite 100ft above the ground moves horizontally at a speed of 2ft/s. at what rate is the angle between the string and the horizontal decreasing when 250ft of string has been let out?
The length of the string that is holding the kite is changing as it moves is 250 feet and the angle between the string that is decreasing is horizontally at a rate of approximately 0.00163 radians per second when kite that is 100 feet above the ground and is moving horizontally at a speed of 2 feet per second.
Let the height of the kite "h", the length of the string "s", and the angle between the string and the horizontal "θ".
We know that h = 100 feet and
ds/dt = 2 feet per second.
Using trigonometry, we can relate the sides of the triangle formed by the kite, the string, and the ground:
sin(θ) = h/s
By using the chain rule of calculus to differentiate this equation with respect to time:
cos(θ) dθ/dt = -h(ds/dt)/s²
Therefore to find dθ/dt when s = 250 feet,
so we can plug in h = 100 feet,
ds/dt = 2 feet per second, and
s = 250 feet:
cos(θ) dθ/dt = -100(2)/(250)² = -0.0016
By solving for dθ/dt:
dθ/dt = -0.0016/cos(θ)
Therefore to find cos(θ), we can use the Pythagorean theorem:
s²= h² + d²,
where "d" is the horizontal distance between the kite and the person holding the string.
When 250 feet of string has been let out, the horizontal distance can be found using the Pythagorean theorem:
d² = s² - h²= (250)² - (100)² = 60000
[tex]d = \sqrt{(60000)} = 244.95 feet[/tex]
So, can now find cos(θ):
cos(θ) = d/s = 244.95/250 = 0.9798
Substituting this value into the equation for dθ/dt:
dθ/dt = -0.0016/0.9798 = -0.00163 radians per second
Therefore, the angle between the string and the horizontal is decreasing at a rate of approximately 0.00163 radians per second when 250 feet of string has been let out.
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the wreck skids along the ground and comes to a stop. the coefficient of kinetic friction while the wreck is skidding is 0.55. assume that the acceleration is constant. how far does the wreck skid?
The given coefficient of kinetic friction is 0.55. Assuming that the acceleration is constant, so the wreck skids a distance of 0 meters.
The distance that the wreck skids while coming to a stop is calculated below.
Data Coefficient of kinetic friction = 0.55
Conversion of acceleration to m/s²0.55 coefficient of kinetic friction can be written as 0.55 times acceleration to calculate the distance that the wreck skids. We know that the acceleration due to gravity is 9.8 m/s². Hence the acceleration due to gravity can be written as follows.
a = 9.8 m/s² × 0.55a
= 5.39 m/s²
Calculation of the distance that the wreck skids is calculated by using the formula below:
d = (v² - u²)/2as = distance = initial velocity = final velocity a = acceleration
The wreck is coming to stop, so the final velocity is 0. Hence the formula can be written as:
d = (v² - u²)/2a
= (0 - u²)/2×5.39d
= -u²/10.78d
= -0.093u²
Calculation of velocity can be calculated by using the following formula below.
v² = u² + 2asv²
= u² - 2u²/10.78v²
= (8.78u²)/10.78v²
= (2u²)/2.45v
= (u²)/1.56
The final velocity is zero. Hence we can write the formula as :
0 = (u²)/1.56u² = 0
The initial velocity of the wreck is zero. Hence the wreck is moving from rest condition.
Calculation of the distance that the wreck skids is calculated by using the formula below:
d = -u²/10.78d
= 0 meters.
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if the same horizontal net force were exerted on both vehicles, pushing them from rest over the same distance, what is the ratio of their final kinetic energies?
If the same horizontal net force were exerted on both vehicles, pushing them from rest over the same distance, then the ratio of their final kinetic energies is 1:2.
According to the Work-Energy principle, the net work done on an object is equal to the change in its kinetic energy. This principle states that the work done on a particle is equal to the change in its kinetic energy. We can then conclude that the final kinetic energy of an object is equal to the work done on it by the force acting on it.
Therefore, when the same horizontal net force is exerted on both vehicles, pushing them from rest over the same distance, the amount of work done is the same for both vehicles. Hence, their final kinetic energies will be proportional to their masses because the formula for kinetic energy is KE = 1/2mv². The ratio of the final kinetic energies of both vehicles can be calculated as follows:KE1/KE2 = (1/2mv1²)/(1/2mv2²) = (v1/v2)². Here, v1 and v2 are the final velocities of the two vehicles. Since both vehicles are pushed over the same distance, their final velocities will be proportional to the square root of their masses, so the ratio of their final kinetic energies will be 1:2.
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jupiter has radius pf 11 x the radius of the eart and a mass that is 320x the mass of the earth the gravitational field strength on the surface of jupiter is
GEarth =9.8ms^-2
A 3Nkg^-1
B 300 NG^-1
C 26 NG^-1
D 10 Nkg -1
An object of mass m at the end of a staring if length r moves in a vertical circle at a concentration angle speed w what is tension in the sting when the object is at the bottom of the circle
An object of mass m love horizontal circle of radio ur with constant speed what is the rate at which works is down by the centripetal force
Answer:
C: 26 NG^-1
Part 2:
The rate at which work is done by the centripetal force is proportional to the cube of the velocity of the object.
Explanation:
The gravitational field strength on the surface of Jupiter can be calculated using the formula:
gJupiter = G×MJupiter / rJupiter²
where G is the universal gravitational constant, MJupiter is the mass of Jupiter, and rJupiter is the radius of Jupiter. Using the given values, we get:
gJupiter = (6.67 × 10-11 N m2 kg-2) × (320 × MEarth) / (11 × REarth)2
gJupiter = 26.0 N kg-1
Therefore, the answer is option C.
For the second question, when the object is at the bottom of the circle, the tension in the string is equal to the weight of the object plus the centripetal force required to keep it moving in the circular path. The centripetal force is given by:
Fc = mv2 / r
where m is the mass of the object, v is the velocity of the object, and r is the radius of the circle.
At the bottom of the circle, the velocity of the object is maximum and equal to the square root of the product of the centripetal force and the radius divided by the mass of the object:
v = sqrt(Fc × r / m)
Substituting the value of Fc in terms of v and solving for tension T, we get:
T = mg + mv2 / r
T = m(g + v2/ r)
For the third question, the rate at which work is done by the centripetal force is given by:
P = Fc × v
where P is the power, Fc is the centripetal force, and v is the velocity of the object. Substituting the value of Fc in terms of v, we get:
P = mv3 / r
Therefore, the rate at which work is done by the centripetal force is proportional to the cube of the velocity of the object.
Explanation:
Well this is quite tricky, as the gravitational field strength on the surface of Jupiter can be calculated using the formula:
g = G*M / r^2
Where G is the gravitational constant, M is the mass of Jupiter, and r is the radius of Jupiter.
Given that the radius of Jupiter is 11 times that of Earth (rJ = 11rE) and the mass of Jupiter is 320 times that of Earth (MJ = 320ME), we can substitute these values into the formula:
g = G x MJ / rJ^2
= G x (320ME) / (11rE)^2
= (G x 320 x ME) / (121 x rE^2)
Now, we know that G = 6.67 x 10^-11 N m^2 / kg^2 and gE = 9.8 m/s^2. So we can substitute these values and simplify:
g = (6.67 x 10^-11 N m^2 / kg^2 * 320 x ME) / (121 x rE^2)
= (2.14 x 10^16 N x ME) / rE^2
To get the gravitational field strength on the surface of Jupiter in terms of gE, we can divide g by gE:
g / gE = (2.14 x 10^16 N x ME) / (rE^2 x 9.8 m/s^2)
= (2.14 x 10^16 N x 5.97 x 10^24 kg) / ( (11 x 6.37 x 10^6 m)^2 x 9.8 m/s^2)
= 25.93
Therefore, the gravitational field strength on the surface of Jupiter is 25.93 times that of Earth.
Answer: C) 26 NG^-1
For an object of mass m at the end of a string of length r moving in a vertical circle at a constant angular speed w, the tension in the string at the bottom of the circle can be found using the formula:
T = mg + mv^2 / r
where g is the acceleration due to gravity, v is the velocity of the object at the bottom of the circle, and m is the mass of the object.
At the bottom of the circle, the object is moving horizontally, so the tension in the string is equal to the centripetal force required to keep it moving in a circle. The velocity of the object at the bottom of the circle can be found using the formula:
v = wr
where w is the angular speed of the object.
Substituting these values into the formula for tension, we get:
T = mg + m(wr)^2 / r
= mg + mw^2r
Therefore, the tension in the string at the bottom of the circle is T = mg + mw^2r.
Answer: T = mg + mw^2r
For an object of mass m moving in a horizontal circle of radius r with a constant speed v, the rate at which work is done by the centripetal force can be found using the formula:
W = Fc x v
where Fc is the centripetal force required to keep the object moving in a circle.
The centripetal force can be found using the formula:
Fc = mv^2 / r
Substituting this value into the formula for work, we get:
W = (mv^2 / r) x v
= mv^3 / r
Therefore, the rate at which work is done by the centripetal force is W = mv^3 / r.
Answer: W = mv^3
what is the magnitude of the upward acceleration of the load of bricks? express your answer with the appropriate units.
The magnitude of the upward acceleration of a load of bricks is 2.77 m/s².
What is tension in the rope?Tension is the pulling force cаrried by flexible mediums like ropes, cаbles аnd string. Tension in а body due to the weight of the hаnging body is the net force аcting on the body.
The tension in the string when the body cаn be given аs,
T = m(a +g)
Here, (m) is the mаss of the body, (а) is the аccelerаtion аnd (g) is the аccelerаtion due to grаvity.
The mаss of the bricks is 15.2 kg.The mаss of the counterweight is 27.2 kg аnd the system is releаsed from the rest.The tension due to the bricks with mаss of 15.2 kg is,
T = 15.2(a + 9.80)
The tension due to the bricks with mass of 27.2 kg is,
T = 27.2(9.80 - a)
Equate both the equation as,
15.2(a + 9.80) = 27.2(9.80 - a)
15.2a + 148.96 = 266.56 - 27.2a
42.4a = 117.6
a = 2.77 m/s²
Thus, the magnitude of the upward acceleration of a load of bricks is 2.77 m/s².
Your question is incomplete, but most probably your full question can be seen in the Attachment.
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if you have 7 total 100-w light bulbs in a parallel circuit in your basement and you leave them on for 1.5 days, how much energy (in kilowatt hours) would be used?
The energy consumed by the 7 100-watt light bulbs left on for 1.5 days is 25.2 kWh.
Given:
Total bulbs = 7
Power of each bulb = 100 W
Time = 1.5 days
To find: Energy used in KWh; Formula used: Energy = Power * Time
Energy used by one bulb in a day = 100 W * 24 hours = 2400 Wh = 2.4 KWh
Total energy used by one bulb in 1.5 days = 2.4 KWh * 1.5 = 3.6 KWh
Total energy used by 7 bulbs in 1.5 days = 3.6 KWh * 7 = 25.2 KWh
Therefore, 25.2 KWh of energy would be used by 7 total 100-w light bulbs in a parallel circuit in your basement and you leave them on for 1.5 days.
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in another universe where the speed of light is only 100 m/s, an airplane that is 45 m long at rest and flies at 320 km/h will appear to be how long (in m) to an observer at rest?
Answer:
320 km/hr = 320000 / 3600 = 88.9 m/s
(1 - v^2 / c^2) = (1 - 88.9^2 / 100^2)^1/2 = .46
Since L = L0 (1 - v^2 / c^2)^1/2
L = .46 L0 = 20.7 m
The airplane will appear to be only 20.64 m long to an observer at rest in this universe, even though its actual length is 45 m when at rest.
In this universe, the speed of light is only 100 m/s, which is much slower than in our universe, where the speed of light is approximately [tex]3 \times 10^8 m/s[/tex]. This means that the effects of special relativity will be much more noticeable in this universe.
We can use the formula for length contraction to calculate the apparent length of the airplane as seen by an observer at rest:
[tex]L' = L / \gamma[/tex]
where L is the length of the airplane at rest, L' is the apparent length of the airplane as seen by the observer, and γ is the Lorentz factor given by:
[tex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}[/tex]
where v is the speed of the airplane relative to the observer, and c is the speed of light in the given universe.
Converting the airplane's speed from km/h to m/s, we have:
[tex]v = (320 \ km/h) \times (1000 \ m/km) / (3600 \ s/h) = 88.89 \ m/s[/tex]
Substituting this value and c = 100 m/s into the expression for γ, we get:
[tex]\gamma = \frac{1}{\sqrt{1 - (88.89 m/s)^2 / (100 m/s)^2}} = 2.18[/tex]
Substituting this value of γ and L = 45 m into the expression for L', we get:
[tex]L' = L / \gamma = 45 \ m / 2.18 = 20.64 \ m[/tex]
Therefore, the length of the plane will appear to be 20.64 m. This significant length contraction is due to the low speed of light in this universe.
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a constant direct current is passing through a loop of wire. the loop is free to rotate about an axis that is parallel to and passes through the plane of the loop. under what circumstance is the maximum torque produced on the loop by the magnetic force?
The maximum torque produced on a current-carrying loop of wire by a magnetic field occurs when the plane of the loop is perpendicular to the direction of the magnetic field.
This can be explained using the formula for the torque on a current-carrying loop in a magnetic field,
τ = N * A * B * sin(θ)
where τ is the torque, N is the number of turns in the loop, A is the area of the loop, B is the magnetic field strength, and θ is the angle between the normal to the plane of the loop and the direction of the magnetic field. If the loop is parallel to the magnetic field, then θ = 0, and the sin(θ) term in the formula is zero. Therefore, there is no torque produced on the loop.
On the other hand, if the loop is perpendicular to the magnetic field, then θ = 90°, and the sin(θ) term in the formula is maximum, which results in the maximum torque on the loop. Therefore, to obtain the maximum torque on the loop, the plane of the loop should be perpendicular to the direction of the magnetic field.
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The portion of string between the bridge and upper end of the fingerboard (the part of the string that is free to vibrate) of a certain musical instrument is 60.0 cm long and has a mass of 2.14 g . The string sounds an A4 note (440 Hz ) when played.
Part A) Where must the player put a finger (at what distance x from the bridge) to play a D5 note (587 Hz )? (See the figure (Figure 1) ) For both notes, the string vibrates in its fundamental mode.
Part B) Without retuning, is it possible to play a G4 note (392 Hz ) on this string?[Yes it is possible to play or No it's impossible to play]
Part C) Explain your answer in Part B: Why or Why not?
A), Multiply the length of the vibrating string (60.0 cm) by the ratio to find the distance x. B)No, it's impossible to play a G4 note (392 Hz) on this string without retuning, C) not possible without retuning.
Part A) To find the distance x from the bridge to play a D5 note (587 Hz), follow these steps:
1. Calculate the speed of the wave on the string using the formula: v = √(T/μ), where T is tension and μ is linear mass density.
2. Calculate the wavelength of the A4 note using the formula: λ = v/f, where f is the frequency of the A4 note (440 Hz).
3. Calculate the wavelength of the D5 note using the formula: λ = v/f, where f is the frequency of the D5 note (587 Hz).
4. Find the ratio between the A4 and D5 wavelengths: λ_A4 / λ_D5.
5. Multiply the length of the vibrating string (60.0 cm) by the ratio to find the distance x.
Part B) No, it's impossible to play a G4 note (392 Hz) on this string without retuning.
Part C) The reason why it's impossible to play a G4 note (392 Hz) without retuning is because the frequencies of the fundamental modes are fixed and cannot be changed unless the tension, mass, or length of the string is altered. To play a G4 note, the string would need to be adjusted so that its fundamental frequency is 392 Hz, which is not possible without retuning.
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through which material will magnetic lines of force pass the most readily? group of answer choices iron. copper. aluminum.
Answer: Through which material will magnetic lines of force pass the most readily? ANSWER: iron
an electron is each placed at rest in an electric field of 490 n/c. calculate the speed, mega m/s, 53.0 ns after being released.
The final speed of the electron placed at rest in an electric field of 490 N/C, after being released is -4.558 mega m/s.
Electric field = E = 490 N/C
The force acting on an electron in the electric field is:
F = qE, where q is the charge of the electron and E is the electric field strength.
q = -1.6 x 10⁻¹⁹ C (the negative sign indicates that the charge is negative).
F = qE = (-1.6 x 10⁻¹⁹ C) (490 N/C) = -7.84 x 10⁻¹⁷N.
The acceleration of the electron due to the electric field:
a = F/m = (-7.84 x 10⁻¹⁷N)/(9.11 x 10⁻³¹kg) = -8.6 x 10¹³ m/s².
According to the third law of motion, for every action, there is an equal and opposite reaction. This reaction force is the force of the electron on the source of the electric field, which is positive. Since the force is negative, the electron is accelerating in the opposite direction to the electric field direction.
The velocity can be found from the equation of motion, v = u + at
v = 0 + (-8.6 x 10¹³)(53.0 x 10⁻⁹) = 4.55 x 10⁶ m/s = 4.55 mega m/s.
The final speed of the electron is therefore -4.558 mega m/s.
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