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Nata [24]
3 years ago
9

At the same moment, one rock is thrown upward at 4.5 m/s and another thrown downward at 6.2 m/s. What is the relative velocity o

f the second rock from the perspective of the first rock? Assume that up is positive.
A) 10.7 m/s
B) -1.7 m/s
C) -10.7 m/s
D) 1.7 m/s
Physics
1 answer:
erastova [34]3 years ago
8 0
The correct answer is 
<span>C) -10.7 m/s 

In fact, the first rock is moving upward with velocity +4.5 m/s, while the second rock is moving downward with velocity -6.2 m/s, with respect to a fixed reference frame. In the reference frame of the first rock, instead, the second rock is moving with velocity equal to its velocity in the fixed frame minus the velocity of the reference frame of the first rock:
</span>v=-6.2 m/s -(+4.5 m/s) = -10.7 m/s<span>
</span>
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What is 64 nanometers to m?
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\sf Hello!

\sf We\: know \:that,
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\sf Distance\: in \:m = \sf Distance\: in\: nm × \dfrac{\sf 1}{\sf 10^{9}}\: \sf m

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3 years ago
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A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T
miv72 [106K]

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C.   In the first collision has twice the momentum as when it stays still ( second colllions)

Explanation:

To see which statement is correct, it is best to solve the problem, the momentum is equal to the variation of the moment

     I = Δp = m vf - m v₀

     I = m (vf -v₀)

Case 1. In car bounces, the initial speed is 0.3 m / s, say that this direction is positive, when the magnitude of the speed bounces it remains constant, but its direction is reversed (vf = -0.3 m / s)

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Case 2. The expensive one that still after the crash so its speed is zero (vf = 0)

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Let's calculate the relationship between the two impulses

     I₁ / I₂ = -0.6m / -0.3m

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When it bounces it has twice the momentum as when it stays still

Now let's analyze the answers:

A.   False The momentum changes

B. False. The momentum is less in the second collision

C. True.  The momentum is double in this collision

D. False. Can be calculated, because the mass is the same throughout the exercise and is eliminated in the equations

E. False.  When they say bounces it implies the same speed with the opposite direction

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