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Ann [662]
2 years ago
14

A 72.8-kg swimmer is standing on a stationary 265-kg floating raft. The swimmer then runs off the raft horizontally with a veloc

ity of 5.21 m/s relative to the shore. Find the recoil velocity that the raft would have (again relative to the shore) if there were no friction and resistance due to the water.
Physics
1 answer:
nalin [4]2 years ago
7 0

Answer:

-1.43 m/s relative to the shore

Explanation:

Total momentum must be conserved before and after the run. Since they were both stationary before, their total speed, and momentum, is 0, so is the total momentum after the run off:

m_sv_s + m_rv_r = 0

where m_s = 72.8, m_r = 265 are the mass of the swimmer and raft, respectively. v_s = 5.21 m/s, v_r are the velocities of the swimmer and the raft after the run, respectively. We can solve for v_r

265v_r + 72.8*5.21 = 0

v_b = -72.8*5.21/265 = -1.43 m/s

So the recoil velocity that the raft would have is -1.43 m/s after the swimmer runs off, relative to the shore

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Answer:

Part a)

Reading = 2.00 kg

Part b)

Reading = 2.00 kg

Part c)

Reading = 4.04 kg

Part d)

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

Explanation:

Part a)

When elevator is ascending with constant speed then we will have

F_{net} = 0

T - mg = 0

T = mg

So it will read same as that of the mass

Reading = 2.00 kg

Part b)

When elevator is decending with constant speed then we will have

F_{net} = 0

T - mg = 0

T = mg

So it will read same as that of the mass

Reading = 2.00 kg

Part c)

When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have

F_{net} = ma

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Reading = 2.00\frac{9.81 + 10}{9.81}

Reading = 4.04 kg

Part d)

Here the speed of the elevator is constant initially

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

3 0
3 years ago
From the top of the engineering building, you throw a ball vertically upward. the ball strikes the ground 4.00 s later. the engi
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Equations of the vertical launch:

Vf = Vo - gt

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Answer: 11.25 m/s


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Answer:

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A 0.10 newton spring toy with a spring constant of 160 newtons per meter is compressed 0.05 meter before it is launched. When re
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Answer:

(1) V = 0.2 J (2) 0.05J

Explanation:

Solution

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(2)Now, we solve for how much of the internal energy is produced as the toy springs up to its maximum height.

By using the energy conversion, we have the following

ΔV = mgh

=(0.1/9.8) * 9.8 * 1.5 = 0.15J

The internal energy = 0.2 -0.15

=0.05J

8 0
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