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gregori [183]
2 years ago
8

A man is traveling from the back of a boat to the front of the boat at 2.0 m/s while the boat itself is traveling at 12.0 m/s to

ward a dock. How far will the man move across the boat in 4.0 seconds? Relative to the dock, how far will the man move in 4.0 seconds?
Physics
1 answer:
Hatshy [7]2 years ago
5 0

Based on the relative velocity of the man with respect to the boat and the dock:

  • Distance covered in 4.0 seconds relative to the boat  = 8 m
  • Distance covered in 4.0 seconds relative to the dock = 56 m

<h3>What is relative velocity?</h3>

Relative velocity is the velocity of a body relative to another body which serves as a reference point.

Relative velocity is a vector.

Considering the velocity of the man and the boat:

The relative velocity of the man with respect to the boat = 2.0 m/s

Distance covered in 4.0 seconds relative to the boat = 2.0 m/s * 4.0 s

Distance moved = 8 m

Relative velocity of the man with respect to the dock = 12 + 2 = 14 m/s

Distance covered in 4.0 seconds relative to the dock = 14.0 m/s * 4.0 s

Distance moved = 56 m

In conclusion, the relative velocity is velocity with respect to a reference point.

Learn more about relative velocity at: brainly.com/question/24337516

#SPJ1

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A ceiling fan with 90-cm-diameter blades is turning at 64 rpm . Suppose the fan coasts to a stop 28 s after being turned off. Wh
mixas84 [53]

Answer:

the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

Explanation:

Given;

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time taken for the fan to stop, t = 28 s

The distance traveled by the ceiling fan when it comes to a stop is calculated as;

d = vt = \omega r\times  t= ( \frac{64 \ rev}{\min} \times \frac{2 \pi \ rad}{rev} \times \frac{1 \min}{60 \ s} \times 0.9 \ m) \times 28 \ s\\\\d = 168.89 \ m

The speed of the tip of a blade 10 s after the fan is turned off is calculated as;

v = \frac{d}{t} \\\\v = \frac{168.89}{10} \\\\v = 16.889 \ m/s

Therefore, the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

4 0
3 years ago
You tie a cord to a pail of water and swing the pail in a vertical circle of radius 0.710 mm . What minumum speed must the pail
Blababa [14]

Answer:

The minumum speed the pail must have at its highest point if no water is to spill from it

= 2.64 m/s

Explanation:

Working with the forces acting on the water in the pail at any point.

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The normal force exerted on the water by the pail is always directed towards the centre of the circle of the circular motion.

And the centripetal force, which keeps the system in its circular motion, is the net force as a result of those two previously mentioned force.

At the highest point of the motion, the top of the vertical circle, the weight and the normal force on the water are both directed downwards.

Net force = W + (normal force)

But the speed of this motion can be lowered enough to a point where the normal force becomes zero at the moment the pail reaches the highest point of its motion. Any speed lower than this value would result in the water spilling out of the pail. The water would not be able to resist the force of gravity.

At this point of minimum velocity,

Normal force = 0

Net force = W

Net force = centripetal force = (mv²/r)

W = mg

(mv²/r) = mg

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g = 9.8 m/s²

v² = gr = 9.8 × 0.71 = 6.958

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Hope this Helps!!!

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Answer:

C.) The amount of mass used up in holding a nucleus together.

Explanation:

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