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2 years ago

A buoy is anchored tho the oceanfloor. A large wave approaches the buoy. How will the buoy move as the wave goes by?

Physics

1 answer:

nadezda [96]2 years ago

5 0

Answer:

The buoy will move upwards and downwards and well as left and right

Explanation:

As the wind is storing so had the capability to move the buoy in all directions as it is a light object despite being anchored down

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A boy starts from point A and moves 5 units toward the east, then turns back and moves 3 units toward the west. What is the disp

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The displacement of the boy is 2 units

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2 years ago

You rub a clear plastic pen with wool, and observe that a strip of invisible tape is attacted to the pen. Assuming that the pen

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Answer:

option B

Explanation:

When you rub a clear plastic pen with the wool the plastic pen gets charges this phenomenon is known as frictional charging.

Due to rubbing, the pen gets negatively charged.

We know, opposite charge attract each other and the same charge repel each other.

So, when the pen is negatively charged the tape might be positively charged or the tape might be uncharged.

Hence, the correct answer is option B

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2 years ago

Assuming a 8 kilogram bowling ball moving at 2 m/s bounces off a spring at the same speed that had before bouncing what is the a

Naya [18.7K]

a) 32 kg m/s

Assuming the spring is initially at rest, the total momentum of the system before the collision is given only by the momentum of the bowling ball:

$p_i = m u = (8 kg)(2 m/s)=16 kg m/s$

The ball bounces off at the same speed had before, but the new velocity has a negative sign (since the direction is opposite to the initial direction). So, the new momentum of the ball is:

$p_{fB}=m v_b =(8 kg)(-2 m/s)=-16 kg m/s$

The final momentum after the collision is the sum of the momenta of the ball and off the spring:

$p_f = p_{fB}+p_{fS}$

where $p_{fS}$ is the momentum of the spring. For the conservation of momentum,

$p_i = p_f\\p_i = p_{fB}+p_{fS}\\p_{fS}=p_i -p_{fB}=16 kg m/s -(-16 kg m/s)=32 kg m/s$

b) -32 kg m/s

The change in momentum of bowling ball is given by the difference between its final momentum and initial momentum:

$\Delta p=p_{fb}-p_i=-16 kg m/s - 16 kg m/s=-32 kg m/s$

c) 64 N

The change in momentum is equal to the product between the average force and the time of the interaction:

$\Delta p=F \Delta t$

Since we know $\Delta t=0.5 s$ , we can find the magnitude of the force:

$F=\frac{\Delta p}{\Delta t}=\frac{-32 kg m/s}{0.5 s}=-64 N$

The negative sign simply means that the direction of the force is opposite to the initial direction of the ball.

d) The force calculated in the previous step (64 N) is larger than the force of 32 N.

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