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Valentin [98]
3 years ago
11

Test your prediction through calculation for the situations of the clay bob and the bouncy ball. Assume each has a mass of 100 g

rams, and each has an initial velocity of 20 m/s straight at the door. Ignore the effects of gravity. Calculate the change in momentum of
Physics
1 answer:
melamori03 [73]3 years ago
6 0

Answer:

a) Δp = -2.0 kgm / s,  b)   Δp = -4 kg m / s

Explanation:

In this exercise the change in moment of a ball is asked in two different cases

a) clay ball, in this case the ball sticks to the door and we have an inelastic collision where the final velocity of the ball is zero

         Δp = p_f - p₀

         Δp = 0 - m v₀

         Δp = - 0.100 20

         Δp = -2.0 kgm / s

b) in this case we have a bouncing ball, this is an elastic collision, as the gate is fixed it can be considered an object of infinite mass, therefore the final speed of the ball has the same modulus of the initial velocity, but address would count

         v_f = - v₀

        Δp = p_f -p₀

        Δp = m v_f - m v₀

        Δp = m (v_f -v₀)

        Δp = 0.100 (-20 - 20)

        Δp = -4 kg m / s

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Explanation:

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Which two points on the wave shown in the diagram below are in phase with each other?
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Answer:

4. B and D

Explanation:

Two points along a transverse wave (such as the one in the figure) are said to be in phase when:

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Basically, we say that two points are in phase when they are separated by a complete cycle (one complete oscillation) of the wave.

For this wave, we see that point B and C have same displacement, but they are not in phase since in B the oscillation is going down while in C is going up.

Instead, B and D are in phase, because they are separated by one complete cycle: both points have same displacement and the oscillation is going in the same way for both of them.

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slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

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where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

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the radius of the pipe is

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So the cross-sectional area is

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The flow rate along the pipe is conserved, so we can write:

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Solving for A2,

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A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m

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