Let's assume that the temperature at a point x cm from the high temperature end is T(x). We know that the temperature gradient in the rod is proportional to the rate of heat flow through the rod, which is given by Fourier's law of heat conduction:
q = -kA(dT/dx)
where q is the rate of heat flow, k is the thermal conductivity of copper, A is the cross-sectional area of the rod, and (dT/dx) is the temperature gradient.
Since the rod is in steady state, the rate of heat flow is constant throughout the rod, so we can write:
q = kA(dT/dx) = constant
Integrating both sides with respect to x, we get:
∫(dT/dx)dx = (1/kA)∫qdx
Integrating from 0 to L (where L is the length of the rod), we get:
T(L) - T(0) = (1/kA)qL
Since T(0) = 0°C and T(L) = 20°C, we can write:
20 - 0 = (1/kA)q(60)
Simplifying, we get:
q = 1/3 kA
Now, let's consider a point 20 cm from the high temperature end. We know that the rate of heat flow through the rod is constant, so we can write:
q = -kA(dT/dx)
Substituting the given values, we get:
1/3 kA = -kA (dT/dx) at x = 20
Simplifying, we get:
(dT/dx) = -1/3
Integrating both sides with respect to x, we get:
T(x) = (-1/3)x + C
where C is the constant of integration.
Using the boundary condition that T(0) = 0°C, we get:
C = 0
Therefore, the temperature at a point 20 cm from the high temperature end is:
T(20) = (-1/3)(20) + 20 = 13.3°C
So, the temperature at a point 20cm from the high temperature end is 13.3°C.
Two +1 C charges are separated by 300m. What is the magnitude of the electric force between them
Answer:
1000 N
Explanation:
The electric force between two charges is given as
F = kq'q/r²......................... Equation 1
Where F = electric force between the charges, q' = magnitude of the first charge, q = magnitude of the second charge, r = distance between the charges, k = coulombs constant
Given: q' = q = 1 C, r = 3000 m, k = 9×10⁹ Nm²/kg²
Substitute into equation 2
F = 1²×(9×10⁹)/3000²
F = 1000 N.
Hence the magnitude of the electric force between them = 1000 N
Suppose an object is weighed with a spring balance, first in air and then whilst totally immersed in water. The readings on the balance are 0.48N and 0.36N respectively. Calculate the density of the object. (2)
The density of the object is is 4000 kg/m³
What is density?Density is the ratio of mass to volume of a body.
To calculate the density of the obeject, we use the formula below
Formula:
D = D'[W/(W-W')]........................ Equation 1Where:
D = Density of the objectD' = Density of waterW = Weight of the object in airW' = Weight of the object in waterFrom the question,
Given:
W = 0.48 NW' = 0.36 ND' = Constant = 1000 kg/m³Susbtitute these values into equation 1
D = 1000[0.48/(0.48-0.36)]D = 1000(0.48/0.12)D = 4000 kg/m³Hence, the density is 4000 kg/m³
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the velocity of three turtles walking in the park. The x-axis is positive towards the North direction.
1. Rank the accelerations of the three turtles from the smallest to largest
2. Are the turtles go- ing towards North or South direction? Are they changing direction?
The ranking of their accelerations from smallest to largest is: Turtle C, Turtle A, Turtle B.
How to solveBased on the given accelerations:
Turtle C has the smallest acceleration (0.05 cm/s²).Turtle A has the next smallest acceleration (0.1 cm/s²).Turtle B has the largest acceleration (0.2 cm/s²).So the ranking of their accelerations from smallest to largest is: Turtle C, Turtle A, Turtle B.
Based on the given velocities:
Turtle A is going towards the North direction (positive velocity).Turtle B is going towards the South direction (negative velocity).Turtle C is going towards the North direction (positive velocity).As for the change in direction:
Turtle A is accelerating in the same direction as its velocity (North), so it's not changing direction, but its speed is increasing.
Turtle B is accelerating in the opposite direction of its velocity (North), so it will eventually change direction and start moving North when the acceleration overcomes the initial Southward velocity.
Turtle C is accelerating in the same direction as its velocity (North), so it's not changing direction, but its speed is increasing.
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Three turtles are walking in the park. The x-axis is positive towards the North direction. The velocities of the turtles are as follows:
Turtle A has a velocity of 2 cm/s to the North and an acceleration of 0.1 cm/s².
Turtle B has a velocity of 1 cm/s to the South and an acceleration of 0.2 cm/s².
Turtle C has a velocity of 3 cm/s to the North and an acceleration of 0.05 cm/s².
Rank the accelerations of the three turtles from smallest to largest based on their given accelerations.
Are the turtles going towards the North or South direction? Are they changing direction based on their accelerations?
Two friends, Burt and Ernie, are standing at opposite ends of a uniform log that is floating in a lake. The log is 4.0 m
long and has mass 250 kg
. Burt has mass 30.0 kg
and Ernie has mass 39.0 kg
. Initially the log and the two friends are at rest relative to the shore. Burt then offers Ernie a cookie, and Ernie walks to Burt's end of the log to get it.
a) Relative to the shore, what distance has the log moved by the time Ernie reaches Burt? Neglect any horizontal force that the water exerts on the log and assume that neither Burt nor Ernie falls off the log.
The log moves a distance of 2.0 m relative to the shore when Ernie reaches Burt.
Since the log is uniform, we can treat it as a system of three point masses: one at each end representing Burt and Ernie, and one at the center representing the center of mass of the log. We can use conservation of momentum to solve this problem.
Initially, the momentum of the system is zero since everything is at rest. When Ernie walks to Burt's end of the log, he exerts a force on the log, which in turn exerts an equal and opposite force on Ernie. This force is internal to the system and does not change the momentum of the system. Therefore, the momentum of the system is still zero after Ernie reaches Burt.
Total momentum = 0 (since the center of mass is at rest)
Burt's momentum = 0 (since he is at rest)
Ernie's momentum = 0 (since he is at rest)
After Ernie walks:
Total momentum = 0 (since the center of mass remains at rest)
Burt's momentum = 0 (since he remains at rest)
Ernie's momentum = (39.0 kg) * v, where v is the velocity of Ernie and the log in the opposite direction
Since the total momentum is conserved, we can equate the momentum before and after Ernie walks:
0 = (39.0 kg) * v
v = 0 m/s
the distance that the log moves when Ernie reaches Burt is simply the distance between the initial and final positions of Ernie, which is half the length of the log:
distance = (1/2) * (4.0 m) = 2.0 m
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A 2.8 kg block slides along a frictionless surface at 1.1 m/s . A second block, sliding at a faster 4.8 m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.3 m/s.
What was the mass of the second block?
Conservation of momentum is a major law of physics which states that the momentum of a system is constant if no external forces are acting on the system. It is embodied in Newton’s First Law or The Law of Inertia.the mass of the second block is 1.1Kg.
principle of momentum conservation
M1u1 plus M2u2 equals M1v1 and M2V2.
As all collisions were elastic in nature and no energy loss through friction, heat, etc. was taken into account, theoretic calculations alone cannot guarantee that there was a complete transfer of energy.
Consider the scenario where a football with mass M2 is lying on the ground and a bowling ball with mass M1 is hurled at the football at a velocity of
The formula is: (2.8 kg * 1.1 m/s) + (m2 * 4.8 m/s) = (2.3 kg + m2). 2.3 m/s
The formula is 2.8 J + (4.8 m/s m2) = 4.8 J + (2.3 m/s m2).
4.8 m/s m2 = 2.8 J plus (2.3 m/s m2)
4.8 m2 = 2.8 + 2.3 m2
2.3 m2 on each side of the equation
2.5 m = 2.8 m = 2.8 / 2.5\sm = 1.1kg
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A rectangle with a base labeled uppercase L and a height labeled lowercase w. A vertical, dashed line runs along the leftside of the rectangle's height. A circular arrow around the vertical, dashed line indictates that the rectangle rotates about its left edge.
Calculate the moment of inertia if the plate has a length of 9.00 cm, a width
of 7.00 cm, and a uniform mass density of 2.50 g/cm^2
A = wL = 63 cm²
m = (2.5 g/cm²)(63 cm²)
m = 157.5 g = 0.1575kg
[tex]I=\frac{1}{3} mL^2\\\\I = \frac{1}{3}(0.1575kg)(0.09m)^2[/tex]
I = 4.2525×10⁻⁴ kg/m²
Consider a converging lens with focal length 9.56 cm. The distance between an object and
a real image of the object created by the lens is 59.6 cm. Find the distance between the
object and the lens if the lens is closer to the object than it is to the image. Answer in cm
The object's distance from the lens is 11.4 cm.
Calculation-The thin lens equation can be used to determine how a converging lens's object distance (p), image distance (q), and focal length (f) relate to one another:
1/p + 1/q = 1/f
where p denotes the distance to the object, q is the distance to the picture, and f is the focal length.
Let's solve for the object distance using the thin lens equation:
1/p + 1/59.6 = 1/9.56
Combining both sides with p59.69.56 results in:
59.69.56 + p9.56 = p*59.6
Adding and subtracting:
570.176 + 9.56p = 59.6p
50.04p = 570.176
p = 11.4 cm
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A uniform line of charge with length 20.0 cm is along the x-axis, with its midpoint at x =0. Its charge per length is +5.30 nC/m. A small sphere with charge -4.00 μC is located at x= 0, y = 5.00 cm.
A) What is the magnitude of the force that the charged sphere exerts on the line of charge?
B) What is the direction angle of the force that the charged sphere exerts on the line of charge? The angle is measured from the +x-axis toward the +y-axis.
The charged spheroid pushes against the line of charge with a force of 0.0181 N.
How much of a power is acting on the charge?Its magnitude is determined by the Coulomb's law, F = k q1 q2/r2, when a point charge (a particle with a charge Q) is operating on a test charge q at a distance r.
Force between the ions' magnitude
Apply the electrostatic force equation of Coulomb.
F = (kq1q2)/r²
where;
q1 is charge 1=6.3nC/mx0.2m=1.26 nC
q2 is charge 2=-4 μC
r is the distance between the charges=5 cm= 0.05 m
F = (9x10⁹x1.26x10⁻⁹x4x10⁻⁶)/(0.05)²
F = 0.0181 N.
B) By breaking the force down into its x and y components, you can determine the orientation of the force the charged sphere applies to the line of charge.
θ = arctan(Fy/Fx) = arctan(Fy/0) = arctan(-Fy)
where Fy is the force's y-component. The force's y component can be calculated as follows:
Fy = F * sin(θ) = F * sin(arctan(-Fy))
Solving for Fy, we get:
Fy = -F * sin(arctan(-Fy)) = -F * (-0.05 / r) = 0.05F / √(x² + 0.05²)
where we used the fact that sin(arctan(x)) = x/√(1+x²).
Plugging in the values, we get:
Fy = 0.05 * 4.47 × 10⁻⁴/ √(0² + 0.05²)
≈ 4.00 × 10⁻⁵ N
Therefore, the direction angle is:
θ = arctan(-Fy/Fx) = arctan(4.00 × 10⁻⁵/ 0) = 90°
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A truck accelerates uniformly from rest to 18.5 m/s in 5.7 s along a level stretch of road. Determine the average power (in W) required to accelerate the truck for the following values of the weight (ignore friction). (a)
The average power required to accelerate the truck with a weight of 20,000 N is approximately 116,930.68 W.
The average power required to accelerate an object is given by the formula:
Power = Work / Time
where Work is the change in kinetic energy of the object and Time is the time interval over which the change in kinetic energy occurs.
The change in kinetic energy of the truck is given by:
ΔK = 1/2 * m *[tex]v^2[/tex]
where m is the mass of the truck and v is its final velocity.
Since the truck starts from rest, its initial kinetic energy is zero, so the work done on the truck is equal to its change in kinetic energy. Therefore, the average power required to accelerate the truck is:
Power = Work / Time = ΔK / Time
Substituting the given values, we get:
(a) For weight w = 10,000 N (approximately 1,020 kg):
m = w / g = 10000 N / [tex]9.81 m/s^2[/tex] = 1019.3 kg
ΔK =[tex]1/2 * m * v^2[/tex] = 1/2 * 1019.3 kg * [tex](18.5 m/s)^2[/tex]= 333,036.6 J
Power = ΔK / Time = 333036.6 J / 5.7 s ≈ 58,426.84 W
Therefore, the average power required to accelerate the truck with a weight of 10,000 N is approximately 58,426.84 W.
Note: g is the acceleration due to gravity, which is approximately 9.81 [tex]m/s^2.[/tex]
(b) For weight w = 20,000 N (approximately 2,040 kg):
m = w / g = 20000 N / 9.81 m/s^2 = 2041.2 kg
ΔK = 1/2 * m *[tex]v^2[/tex] = 1/2 * 2041.2 kg * [tex](18.5 m/s)^2[/tex]= 666,073.3 J
Power = ΔK / Time = 666073.3 J / 5.7 s ≈ 116,930.68 W.
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Please help I need to make sure my answer is correct
The measure that helps your body to stretch is flexibility, so the correct answer is D.
Flexibility for Body Stretching.Flexibility is the ability of your body to move your joints and muscles through their full range of motion. It is an essential component of physical fitness and can help improve posture, balance, and coordination. Regular stretching exercises can increase flexibility and reduce muscle stiffness, which can lead to improved physical performance, decreased risk of injury, and better overall health.
Muscular strength and muscular endurance are related to the ability of your muscles to generate force or sustain effort over time, respectively. While they are important for overall fitness, they are not directly related to flexibility.
In summary, flexibility is the key measure that helps your body to stretch, and regular stretching exercises can improve your flexibility and overall physical fitness.
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What is the charge of an electric field 0.40 m away from a
source charge of 3.00 x 10-5C?
The electric field 0.40 m away from a source charge of 3.00 x 10^-5 C is 1.69 x 10^6 N/C, directed away from the source charge.
How to find the chargeThe electric field created by a point charge at a distance r from the charge is given by the equation:
E = kQ/r^2
where
E is the electric field in N/C,
k is Coulomb's constant (9.0 x 10^9 N m^2/C^2),
Q is the source charge in Coulombs, and
r is the distance from the source charge in meters.
Substituting the given values into the equation, we get:
E = (9.0 x 10^9 ) x (3.0 x 10^-5) / (0.40)^2
E = 1.69 x 10^6 N/C
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A car slows down from -27.7m/s to -10.9m/s while undergoing a displacement of -105. What was the acceleration? (Unit=m/s^2)
The acceleration of the car was 6.88 [tex]m/s^2[/tex]. The negative sign indicates that the car was decelerating, or slowing down.
To find the acceleration of the car, we can use the following formula:
acceleration = (final velocity - initial velocity) / time
However, we are not given the time it took for the car to undergo the displacement. To find the time, we can use the following formula:
displacement = (final velocity + initial velocity) / 2 * time
Solving for time, we get:
time = displacement / ((final velocity + initial velocity) / 2)
Plugging in the given values, we get:
time = [tex]-105 / ((-10.9 - 27.7) / 2) = 2.29 s[/tex]
Now that we have the time, we can use the first formula to find the acceleration:
acceleration = [tex](-10.9 - (-27.7)) / 2.29 = 6.88 m/s^2[/tex]
Therefore, the acceleration of the car was [tex]6.88 m/s^2[/tex]. The negative sign indicates that the car was decelerating, or slowing down.
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A 1.3 KG blocks flies along a frictionless surface at 1.0 M/S.a2 block sliding at a faster 5.0 M/S collides with the first from behind and sticks to it. The final velocity of the combine blocks is 2.0 M/S. What was the mass of the second block?
the initial momentum of the system of block m1 and block m2 is
Pi= m1v1 + m2v2
the final momentum of the combine blocks is
Pf= (m1+m2)V
according to the law of convervation of momentum
Pi = Pf
m1v1 + m2v2 = (m1+m2)V
1.3 × 1 + 5m2 = 1.3 × 2 + 2m2
m2= 1.3/3 kg
In an inertial system S, an event is observed to take place at point A on the x-axis and 10^-6 s later another event takes place at point B, 900 m further down. Find the magnitude and direction of the velocity of S' with respect to S in which these two events appear simultaneous.
The magnitude of the velocity of S' with respect to S is therefore approximately [tex]9 * 10^(11)[/tex]m/s. then the magnitude is approximately[tex]9 * 10^(11)[/tex] m/s.
the velocity of the system S' with respect to S as v. Since the two events are simultaneous in the system S', they must have the same time coordinate in that system. Let's call this time coordinate t' and the position of the two events in the S' system as (x1', t') and (x2', t').
Using the Lorentz transformation equations, we can relate the coordinates (x1, t1) and (x2, t2) in the S system to the coordinates (x1', t') and (x2', t') in the S' system:
x1' = γ(x1 - vt1)
t' = γ(t1 - vx1/[tex]c^2[/tex])
x2' = γ(x2 - vt2)
t' = γ(t2 - [tex]vx2/c^2[/tex])
where γ = 1/√(1 - [tex]v^2/c^2[/tex]) is the Lorentz factor.
Since the two events are separated by 900 m and a time interval of 10^-6 s, we have:
x2 - x1 = 900 m
t2 - t1 = [tex]10^-6[/tex] s
Using the above equations, we can solve for v. First, we can eliminate t' by equating the two expressions for t':
γ(t1 - vx1/[tex]c^2[/tex]) = γ(t2 - vx2/[tex]c^2[/tex])
Simplifying this expression, we get:
v =[tex]c^2[/tex](x2 - x1)/(t2 - t1)(x1 + x2)
Plugging in the given values, we get:
v = c^2(900 m)/([tex]10^-6[/tex] s)(0 m + 900 m) = 9 × [tex]10^11[/tex] m/s
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for the circuit shown in find the output voltage(ii) the current through zenor diode
A Zener diode is a heavily made semiconductor device that is designed to operate in the reverse direction.Voltage drop across series resistance is 70 volts and the current through zenor diode is 9mA.
When the voltage across the terminals of a diode is reversed, and the potential reaches the voltage (knee voltage), the junction break down, and the current flows in the reverse direction. This effect is known as the diode effect.
R = 5K ohms = 5 x 10^3 ohms
Input voltage = 12V, Zener voltage = 50V
Output Voltage = 50V
Voltage drop across series resistance = input voltage – zener voltage = 120 – 50 = 70 volts
Load Current = zener voltage / resistance = \frac{50}{10 x 10^3} = 5 x 10^{-3} A
Current through = \frac{input voltage – zener voltage} {resistance}
= 70 / 5 x 10^{-3} = 14 x 10^{-3} A
According to kirchoff’s first law = current + zener current
Zener current zener current = I - line current = 14 x 10^{-3} - 5 x 10^{-3} = 9 x 10^{-3} = 9 mA
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Jonathan and Jane are sitting in a sleigh that is at rest on frictionless ice. Jonathan's weight is 800 N
, Jane's weight is 600 N
, and that of the sleigh is 1000 N
. They see a poisonous spider on the floor of the sleigh and immediately jump off. Jonathan jumps to the left with a velocity (relative to the ice) of 5.00 m/s
at 30.0∘
above the horizontal, and Jane jumps to the right at 7.00 m/s
at 36.9∘
above the horizontal (relative to the ice).
1) Calculate the magnitude of the sleigh's horizontal velocity after they jump out?
2) What is the direction of the sleigh's horizontal velocity after they jump out?
Since the positive x-direction was established to be to the right, the sleigh's velocity is in this direction (the positive x-direction).
Calculation-The whole horizontal momentum is conserved since there is no external force operating on the system (the sleigh plus Jonathan and Jane) in the horizontal direction.
Let's say that the positive x-direction is to the right and the positive y-direction is up. Then, we may divide Jonathan and Jane's velocities into their x- and y-components as follows:
vx1 = 5.00 cos 30.0° = 4.33 m/s and vy1 = 5.00 sin 30.0° = 2.50 m/s are the speeds of Jonathan.
Jane's speed is given by the formulas vx2 = 7.00 cos 36.9° = 5.61 m/s and vy2 = 7.00 sin 36.9° = 4.16 m/s.
The sleigh is initially at rest, hence there is no initial total momentum in the x-direction. The total x-direction momentum after Jonathan and Jane jumps out is:
px = (−800 N)(−4.33 m/s) + (600 N)(5.61 m/s) = 6430 N·s
The final momentum in the x-direction is:
p'x = (1000 N + 800 N + 600 N)vx'
where vx' is the final velocity of the sleigh in the x-direction.
Therefore, we can solve for xv:
vx' = px / (1000 N + 800 N + 600 N) = 6430 N·s / 2400 N = 2.68 m/s
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what is the mechanica advantage of alever then can lift 100 Newton with an input force of 20 Newton
Answer:
Explanation:
To calculate the mechanical advantage of a lever, we use the formula:
Mechanical Advantage = Output Force / Input Force
In this case, the output force is 100 Newton and the input force is 20 Newton, so:
Mechanical Advantage = 100 N / 20 N
Mechanical Advantage = 5
Therefore, the mechanical advantage of the lever is 5. This means that for every 1 Newton of input force applied to the lever, the lever will produce 5 Newtons of output force. So, in this case, an input force of 20 Newtons applied to the lever would produce an output force of 100 Newtons.
A metal ball bearing with mass 5.0 g falls out of a factory machine and drops to the concrete floor 3.0 m below. It bounces back up to its starting point. Find the changes in the bearing's potential and kinetic energies as it a) travels from the machine down to the floor, and b) travels up from the floor back to its starting point.
a) When the metal ball bearing falls from the machine to the concrete floor, its potential energy decreases while its kinetic energy increases.
The potential energy lost by the ball bearing can be calculated using the formula PE = mgh, where m is the mass of the ball bearing (0.005 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the height the ball bearing falls (3.0 m). Thus, the potential energy lost by the ball bearing is:
PE = (0.005 kg)(9.8 m/s²)(3.0 m) = 0.147 J
At the same time, the ball bearing's kinetic energy increases by an amount equal to the potential energy lost. Therefore, the ball bearing's initial kinetic energy is zero, and its final kinetic energy is:
KE = 0.147 J
b) As the ball bearing bounces back up from the concrete floor to its starting point, its kinetic energy decreases while its potential energy increases. The kinetic energy lost by the ball bearing can be calculated as the same value as before:
KE = 0.147 J
The ball bearing's potential energy at its starting point is equal to the potential energy it lost on the way down, which is:
PE = 0.147 J
Therefore, the ball bearing's change in potential energy is 0.147 J (from down to up), and its change in kinetic energy is -0.147 J (from up to down).
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Answer: a) When the ball bearing falls from the machine to the floor, there is a change in its potential and kinetic energies. The potential energy of an object at a height h above the ground is given by mgh, where m is the mass of the object, g is the acceleration due to gravity (9.8 m/s^2), and h is the height above the ground. Initially, the ball bearing is at rest on the machine, so its kinetic energy is zero. Therefore, the initial energy of the ball bearing is purely potential energy, given by:
PEi = mgh = (0.005 kg)(9.8 m/s^2)(3.0 m) = 0.147 J
When the ball bearing hits the ground, its potential energy is zero and its kinetic energy is at a maximum. The velocity of the ball bearing just before it hits the ground can be found using the equation:
v^2 = u^2 + 2as
where u is the initial velocity (which is zero), a is the acceleration due to gravity (-9.8 m/s^2), s is the distance fallen (3.0 m), and v is the final velocity just before hitting the ground. Solving for v, we get:
v = sqrt(2as) = sqrt(2*(-9.8 m/s^2)*(3.0 m)) = 7.67 m/s
The kinetic energy of the ball bearing just before it hits the ground is given by:
KEf = (1/2)mv^2 = (1/2)(0.005 kg)(7.67 m/s)^2 = 0.145 J
Therefore, the change in potential energy is:
ΔPE = PEf - PEi = 0 - 0.147 J = -0.147 J
And the change in kinetic energy is:
ΔKE = KEf - KEi = 0.145 J - 0 J = 0.145 J
b) When the ball bearing bounces back up to its starting point, there is another change in its potential and kinetic energies. Just before it reaches its highest point, the ball bearing's velocity is zero, so its kinetic energy is also zero. Therefore, its energy is purely potential energy, given by:
PEf = mgh = (0.005 kg)(9.8 m/s^2)(3.0 m) = 0.147 J
The ball bearing reaches its highest point when all of its initial kinetic energy has been converted to potential energy. At this point, its potential energy is at a maximum and its kinetic energy is at a minimum. The ball bearing then starts to fall back down towards the ground, and its potential energy starts to decrease while its kinetic energy increases. Just before it hits the ground, its kinetic energy is at a maximum and its potential energy is at a minimum, as we saw in part (a). When it bounces back up, the process repeats.
Therefore, the change in potential energy as the ball bearing travels from the floor back up to its starting point is:
ΔPE = PEf - PEi = 0.147 J - 0 J = 0.147 J
And the change in kinetic energy is:
ΔKE = KEf - KEi = 0 J - 0 J = 0 J
Note that the change in kinetic energy is zero because the ball bearing starts and ends at rest.
Explanation: can i get brainliest
Two pure tones Cs and Gs, with frequencies from the Pythagorean diatonic scale, are sounded simultaneously. Find
a) the frequencies of the three combination tones and
b) the notes on the Pythagorean scale to which these tones belong.
The combination tones correspond to the notes Bb, D, and Cs on the Pythagorean diatonic scale.
What is Frequency?
Frequency is the number of occurrences of a repeating event per unit of time. In other words, it is the rate at which a wave oscillates or completes one cycle. The unit of frequency is hertz (Hz), which is equivalent to one cycle per second.
When two pure tones with frequencies f1 and f2 are sounded simultaneously, several additional frequencies, known as combination tones, can be produced. The three most important combination tones are:
The sum tone, which has a frequency equal to the sum of the two original frequencies: f1 + f2
The difference tone, which has a frequency equal to the difference between the two original frequencies: |f1 - f2|
The octave tone, which has a frequency twice that of the lower of the two original frequencies: 2f1 or 2f2
In this case, we have two pure tones Cs and Gs with frequencies from the Pythagorean diatonic scale. We need to first determine the frequencies of these two tones. According to the Pythagorean tuning system, the frequency ratios for Cs and Gs are:
Cs:G = 9:8
Cs:fundamental = 2:1 (assuming Cs is one octave above the fundamental)
Gs:fundamental = 3:1 (assuming Gs is one octave and a fifth above the fundamental)
Let's assume that the fundamental frequency is f0. Then we can write:
Cs = 2f0 * (9/8) = 9f0/4
Gs = 4f0 * (3/2) * (9/8) = 27f0/8
a) To find the combination tones, we need to apply the equations above. The sum tone has a frequency of:
f1 + f2 = Cs + Gs = (9f0/4) + (27f0/8) = 45f0/8
The difference tone has a frequency of:
|f1 - f2| = |Cs - Gs| = |(9f0/4) - (27f0/8)| = 9f0/8
The octave tone has a frequency of:
2f1 = 2Cs = 9f0/2 = 9f0
Therefore, the three combination tones have frequencies of 45f0/8, 9f0/8, and 9f0.
b) To determine the notes on the Pythagorean scale to which these tones belong, we need to find the closest notes on the scale to each of the combination tones. The Pythagorean scale is based on a series of perfect fifths, so we can use the frequency ratios of 3:2 to determine the frequency of each note relative to the fundamental frequency f0.
The closest notes on the Pythagorean scale to the combination tones are:
45f0/8 is closest to the note Bb, which has a frequency of 3f0/2
9f0/8 is closest to the note D, which has a frequency of 9f0/8
9f0 is closest to the note Cs, which has a frequency of 9f0/4
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Box ramp
A 1 kg box started from rest downward an
1-meter high ramp inclined at 30 degrees. The
kinetic friction coefficient is 0.3. What was
its speed when it reaches the bottom, in m/s?
A. 3.1
B. 5.5
C. 4.8
D. 1.3
E. 4.4
Answer:
(option A).
Explanation:
The gravitational potential energy of the box at the top of the ramp is given by:
Ep = mgh
where m is the mass of the box, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the ramp (1 m).
Ep = (1 kg)(9.8 m/s^2)(1 m) = 9.8 J
As the box slides down the ramp, its gravitational potential energy is converted into kinetic energy, which is given by:
Ek = (1/2)mv^2
where v is the velocity of the box at the bottom of the ramp.
The work done by the friction force on the box is given by:
W = f * d
where f is the force of friction and d is the distance traveled by the box along the ramp. The force of friction is given by:
f = μmg
where μ is the coefficient of kinetic friction (0.3) and mg is the weight of the box.
f = (0.3)(1 kg)(9.8 m/s^2) = 2.94 N
The distance traveled by the box along the ramp is:
d = h/sin(30°) = 2 m
Therefore, the work done by the friction force is:
W = (2.94 N)(2 m) = 5.88 J
The total mechanical energy of the box at the bottom of the ramp (neglecting air resistance) is equal to the sum of its kinetic energy and the work done by the friction force:
Ek + W = Ep
(1/2)mv^2 + 5.88 J = 9.8 J
Solving for v, we get:
v = sqrt[(2(9.8 J - 5.88 J))/m] = sqrt[(2(3.92 J))/1 kg] = 3.1 m/s
An airplane is flying with a velocity of 240 m/s at an angle of 30.0° with the horizontal, as the drawing shows. When the altitude of the plane is 2.4 km, a flare is released from the plane. The flare hits the target on the ground. What is the angle θ?
Since the flare is released from the plane, it has the same initial velocity as the plane, i.e., 240 m/s at an angle of 30 degrees with the horizontal.
Let's assume that the distance from the plane to the target is d, and the altitude of the plane is h. The angle θ is the angle between the plane and the target.
Using trigonometry, we can write:
tan θ = h/d
We need to find the value of θ. To do that, we need to find the values of h and d.
We know that the altitude of the plane is 2.4 km = 2400 m. Let's call the time it takes for the flare to hit the target t. Since the flare is moving under gravity, its motion can be described as:
h = (1/2)gt^2
where g is the acceleration due to gravity, which is approximately 9.81 m/s^2.
The horizontal distance traveled by the flare in time t is:
d = vt
where v is the horizontal component of the velocity of the flare/plane, which is given by:
v = 240 cos 30° = 240 × √3/2 = 120√3 m/s
Equating the expressions for h and d, we get:
(1/2)gt^2 = vt
Solving for t, we get:
t = 2h/g = 2 × 2400/9.81 = 489.55 s
Substituting this value of t in the expression for d, we get:
d = vt = 120√3 × 489.55 ≈ 70000 m
Now we can find θ:
tan θ = h/d = 2400/70000 ≈ 0.0343
θ = tan^(-1)(0.0343) ≈ 1.96°
Therefore, the angle θ is 1.96 degrees.
A physics class conducting a research project on projectile motion construct a device that can launch a cricket ball.the launching device is designed so that the ball can be launch at ground level with an initial velocity of 28m/s at an angle of 30 degrees to the horizontal.
Calculate the horizontal of the velocity of the all:
a) initially
B) after 1.0 seconds
C) after 2.0 seconds
A projectile motion is any object thrown into space upon which the only acting force is gravity. The primary force acting on a projectile is gravity.The velocity's horizontal component is 24.25 m/s at time t = 2 seconds.The velocity's horizontal component is 24.25 m/s at time t = 3 seconds.
This doesn’t necessarily mean that other forces do not act on it, just that their effect is minimal compared to gravity.The particle is moving vertically (downwards) along the y-axis due to uniform acceleration.A particle's vertical and horizontal projectile motions can both accelerate: The only force acting on a particle when it is launched into the air at some speed is the acceleration brought on by gravity (g). The downward motion of this acceleration is vertical.
347u = 28 m/s for the starting velocity
projecting at a 30° angle
The horizontal component of velocity is constant since there is no acceleration in the horizontal direction.
Vertical component of speed, u cos = 28 x Cos 30 = 24.25 m/s
The velocity's horizontal component is 24.25 m/s at time t = 2 seconds.
The velocity's horizontal component is 24.25 m/s at time t = 3 seconds
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The current in a circuit is 0.50 A. The circuit has two resistors connected in series: one is 110 Ω and the other is 130 Ω. What is the voltage in the circuit?
Answer:
120 V
Explanation:
Given:
I (total) = 0,5 A
R1 = 110 Ω
R2 = 130 Ω
Find: V (total) - ?
The resistors are connected in series
That means, the current in the circuit is the same for every resistor:
I (total) = I1 = I2
Now, we need to find the voltage in each resistor:
V1 = I × R1
V1 = 0,5 × 110 = 55 V
V2 = I × R2
V2 = 0,5 × 130 = 65 V
Now, since the connection is in series, in order to find the total voltage in the circuit, we have to add the voltages of the resistors:
V (total) = V1 + V2
V (total) = 55 + 65 = 120 V
8. A man with an open parachute falls to Earth at constant speed. The following forces act: 9 P the upward force of the parachute on the man Q the upward force of the man on the Earth R the downward force of the Earth on the parachute S the downward force of the man on the parachute. Which two forces are a Newton's third law pair? A. P and Q 9 ● B. P and R C. P and S D. Q and R
Answer:
Explanation:
According to Newton's third law, for every action, there is an equal and opposite reaction. This means that if force A acts on object B, then there must be a force that object B exerts on object A that is equal in magnitude but opposite in direction. Therefore, the two forces that are a Newton's third law pair are P and Q, as they are the forces that the parachute and the man exert on each other in opposite directions.
R and S are not a Newton's third law pair because they are not acting on the same objects. R is the force of the Earth on the parachute, while S is the force of the man on the parachute. These forces are not equal and opposite, as they are acting on different objects.
The lightweight wheel on a road bike has a moment of inertia of 0.097 kg⋅m2
. A mechanic, checking the alignment of the wheel, gives it a quick spin; it completes 5 rotations in 2.2 s. To bring the wheel to rest, the mechanic gently applies the disk brakes, which squeeze pads against a metal disk connected to the wheel. The pads touch the disk 7.1 cm from the axle, and the wheel slows down and stops in 1.2 s.
What is the magnitude of the friction force on the disk?
the magnitude of the friction force on the disk is approximately 1.16 N.
We can use the conservation of energy to find the friction force on the disk. The initial kinetic energy of the wheel is equal to the work done by the friction force on the disk:
K_i = W_friction
The initial kinetic energy of the wheel can be found from its moment of inertia and angular velocity:
K_i = (1/2) I [tex]ω^2[/tex]
where I is the moment of inertia, ω is the angular velocity, and the factor of 1/2 comes from the rotational kinetic energy formula.
The final kinetic energy of the wheel is zero, since it comes to a stop. The work done by the friction force can be found from the distance over which it acts:
W_friction = F_friction d
where F_friction is the friction force and d is the distance over which the pads act on the disk. We can find the distance from the angular displacement of the wheel:
θ = ω t
where θ is the angle through which the wheel rotates, t is the time for the wheel to come to a stop, and the factor of 1/2π converts from rotations to radians. The distance over which the pads act is then:
d = r θ = 0.071 m × (5/2π) ≈ 0.562 m
Now we can put everything together:
K_i = W_friction
(1/2) I [tex]ω^2[/tex] = F_friction d
We can solve for the friction force:
F_friction = (1/2) I [tex]ω^2[/tex] / d
Plugging in the given values:
F_friction = (1/2) ×[tex]0.097 kg⋅m^2[/tex] × [tex](5/2.2π rad/s)^2[/tex] / 0.562 m ≈ 1.16 N
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1. In the picture above, how would the amount of kinetic energy in the third pendulum compare to the
amount of potential energy in the first and fourth pendulums?
The kinetic energy would be at its maximum compared to the potential energy which would be less. The potential energy of the first and the last pendulum would be at their maximum
What is kinetic energy?Kinetic energy is the energy an object possesses due to its motion. It is defined as the energy that an object has by virtue of its motion, and is dependent on both its mass and velocity. The formula for calculating kinetic energy is 1/2 times the mass of the object times the square of its velocity, or KE = 1/2 mv^2.
This means that the kinetic energy of an object increases as its mass and velocity increase. For example, a heavy object moving at a high velocity will have a higher kinetic energy than a lighter object moving at a lower velocity. Kinetic energy is a scalar quantity, meaning it has no direction, only magnitude.
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The Earth surface temperature is around 270 K and emissivity of 0.8, while space has temperature of around 2K. What is the net power radiated by the Earth in free space?
The net power radiated by the Earth in free space is approximately
1.2 x 10^ 17 W watts.
How to find the power radiatedTo calculate the net power radiated by the Earth in free space, we can use the Stefan-Boltzmann Law:
Power = σ * A * ε * (T^4 - T0^4)
where:
σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4)
A is the surface area of the Earth (4πR^2, where R is the radius of the Earth)
ε is the emissivity of the Earth (0.8)
T is the temperature of the Earth surface (270 K)
T0 is the temperature of space (2 K)
Plugging in the values, we get:
Power = 5.67 x 10^-8 * 4πR^2 * 0.8 * (270^4 - 2^4)
The radius of the Earth is approximately 6.37 x 10^6 m, so:
Power = 5.67 x 10^-8 * 4π(6.37 x 10^6)^2 * 0.8 * (270^4 - 2^4)
Power ≈ 1.193 x 10^ 17 W
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List 3 components of an electric circuit.
Answer:
An energy source – like a battery or mains power.
An energy receiver – like a lightbulb.
An energy pathway – like a wire.
Explanation:
An object has greater momentum if it has ?
The more massive or swifter an object is, the more momentum it has.
When a substance has more momentum?A moving thing is more difficult to stop the more momentum it possesses. The amount of momentum an object possesses is influenced by its mass. For instance, while a car driving at the same speed as a baseball can be stopped, it cannot be caught. Because the car is heavier, it has more momentum.
What increases an object's momentum?A moving object's mass and speed both affect its momentum. The greater the object's momentum and the more difficult it is to stop are directly proportional to its weight and speed.
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Which statements are true about the amount of gravitational potential energy an object has?
The amount increases as the object is lifted higher.
The amount varies according to the material of the object.
The amount varies according to the mass of the object.
The amount increases the more quickly the object is lifted.
Answer:
The statement "The amount increases as the object is lifted higher" is true about the amount of gravitational potential energy an object has. The statement "The amount varies according to the material of the object" is false. The amount of gravitational potential energy only depends on the mass of the object and its elevation from the reference point. The statement "The amount varies according to the mass of the object" is true. The more massive an object is, the more gravitational potential energy it has. The statement "The amount increases the more quickly the object is lifted" is false. The amount of gravitational potential energy only depends on the object's height above the reference point, not the speed at which it is lifted.
Answer: (A)
Explanation: The amount increases as the object is lifted higher.