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Naily [24]
3 years ago
15

Kiley, Jermaine, and Gunther are playing tug-of-war. Kiley and Jermaine are on the same side of the rope, and Gunther is on the

other side. Kiley is exerting a force of 20 N, and Jermaine is exerting a force of 25 N.
Which force would result in a balanced force for the game?
Physics
2 answers:
Ivanshal [37]3 years ago
8 0

-45N because you add the side together and make it equal 0

Stolb23 [73]3 years ago
7 0

Kiley and Germaine pull together, resulting in a force of 45N which pulls the rope that way ===> .

In order to balance the forces on the rope, Gunther must pull the rope with a force of <em>45N this way  <===</em>, all by himself.

The net force on the rope will then be zero, and the rope will not accelerate.  (Although the tension inside the rope will be 90N.)

By the way . . . I'd love to see how the kids manage to all get on the "sides" of the rope.  I would have expected them to just do it the easy way and stand at the "ends" of the rope.

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3 years ago
Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m
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Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

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<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}

Substituting,

V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

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Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where here v_f is the velocity after the collision.

v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}

v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

v_f = 1.034m/s

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