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Furkat [3]
3 years ago
9

Two speedboats are traveling at the same speed relative to the water in opposite directions in a moving river. An observer on th

e riverbank sees the boats moving at 4.9 m/s and 7.0 m/s.(a) What is the speed of the boats relative to the river? (b) How fast is the river moving relative to the shore?
Physics
1 answer:
DENIUS [597]3 years ago
3 0

Answer:

a) vboat = 5.95 m/s  b) vriver= 1.05 m/s

Explanation:

a) As observed from the shore, the speed of the boats can be expressed as the vector sum, of the boat speed relative to the water and the river speed relative to the shore, as follows:

vb₁s = vb₁w + vrs

In one case, the boat is moving in the same direction as the water:

vb₁s = vb₁w + vrs = 7.0 m/s (1)

For the other boat, it is clear that is moving in an opposite direction:

vb₂s = vb₂w - vrs = 4.9 m/s (2)

As  we know that vb₁w = vb₂w, adding both sides, we can remove the river speed from the equation, as follows:

vb₁w = vb₂w =  \frac{7.0 m/s + 4.9 m/s}{2} =5.95 m/s

b) Replacing this value in (1) and solving for vriver, we have:

vriver = 7.0 m/s - 5.95 m/s = 1.05 m/s

(we could have arrived to the same result subtracting both sides in (1), and (2))

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Answer:

The bicyclist's acceleration is 2.2m/s^2

Explanation:

Given

u = 2.5m/s ---- Initial Velocity

v = 12.5m/s ---- Final Velocity

t = 4.5s ---- Time

Required

Determine the acceleration

This will be solved using the first equation of motion

v = u + at

Substitute values for v, u and a

12.5 = 2.5 + a * 4.5

12.5 = 2.5 + 4.5a

Collect Like Terms

4.5a = 12.5 - 2.5

4.5a = 10.0

Solve for a

a = 10.0/4.5

a = 2.2m/s^2 ---- (approximated)

<em>Hence, the bicyclist's acceleration is 2.2m/s^2</em>

3 0
3 years ago
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Matter that organisms require for their life processes are
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A body is thrown vertically upward. Its velocity keep on decreasing. What happens to its kinetic energy when it reaches the maxi
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Answer:

Explanation:

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3 years ago
It has been suggested that rotating cylinders about 10 mi long and 5.9 mi in diameter be placed in space and used as colonies. T
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Answer:

ω = 0.05 rad/s

Explanation:

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\omega = \sqrt{\frac{9.8\ m/s^{2}}{4023.36\ m}}\\\\

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