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love history [14]
3 years ago
10

A ball is thrown downward; and its speed is v just before it strikes the ground. After the collision with the earth, it heads up

ward with a speed v. Which one of the following statements concerning this situation is truea) Only the magnitude of momentum of the earth is changed by collision (the correct answer, an explanation of why this is true would be very helpful)
b) the momentum of the earth is not changed by the collision
c) only the momentum of the ball is changed by the collision
d) the momentum of the ball is not changed by the collision
e) the collision causes both the momentum and the kinetic energy of the ball to change
Physics
1 answer:
d1i1m1o1n [39]3 years ago
4 0

For this case it is necessary to consider the assessments made between collisions of 'immovable' objects. In this case the earth is the immovable object or at least, it is approached by its mass. Considering the case of an inflatable ball of mass m that travels at a speed v towards the ground, it hits the floor and bounces with speed -v (Negative). The earth does not move, however, the momentum of the ball has changed by 2mv since the speed went from positive to negative. Applying the conservation of the momentum we know that the change of the momentum on the fly would be given under the function

p = m\Delta V

Considering the direction of the velocities this expression can be rewritten as

p = m (v-(-v))

p = 2mv

Therefore the correct answer would be

C. Only the momentum of the ball is changed by the collision

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Answer:

1920Joules

Explanation:

The formula for calculating the kinetic energy of a body is expressed as;

KE = 1/2 mv²

m isthe mass

V is the speed

For the two masses, the combined KE is expressed as;

KE  = 1/2(m1+m2)v²

KE = 1/2(45+15)(8)²

KE  = 1/2 * 60 * 64

KE  = 30 * 64

KE  = 1920J

Hence the combined kinetic energy of the boy and the bicycle is 1920Joules

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What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1
notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is  F_{net}=  3.22*10^{-5} \ J

Explanation:

From the question we are told that

    The third charge is  q_3 =  55 nC =  55 *10^{-9} C

    The position of the third charge is  x = -1.220 \ m

     The first charge is q_1 =  -16 nC  =  -16 *10^{-9} \ C

     The position of the first charge is x_1 =  -1.650m

      The second charge is  q_2 =  32 nC  =  32 *10^{-9} C

      The position of the second charge is  x_2 =   0  \ m  

The distance between the first and the third charge is

      d_{1-3} =  -1.650 -(-1.220)

     d_{1-3} = -0.43 \ m

The force exerted on the third charge by the first is  

     F_{1-3} =  \frac{k  q_1 q_3}{d_{1-3}^2}

Where k is the coulomb's constant with a value  9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

      F_{1-3} =  \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}

       F_{1-3} = 4.28 *10^{-5} \ N

 The distance between the second and the third charge is      

  d_{2-3} =  0- (-1.22)

   d_{2-3} =1.220 \ m

The force exerted on the third charge by the first is mathematically evaluated as

       F_{2-3} =  \frac{k  q_2 q_3}{d_{2-3}^2}

substituting values

       F_{2-3} =  \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}

       F_{2-3} =  1.06*10^{-5} N

The net force is

      F_{net} =  F_{1-3} -F_{2-3}

substituting values

    F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}

    F_{net}=  3.22*10^{-5} \ J

6 0
3 years ago
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