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hoa [83]
4 years ago
15

A cruise ship is having troubles with buoyancy. What is a reasonable solution? A. Increase the weight of the ship above water B.

Increase the mass of the ship C. Decrease the volume of the ship D. Spread the weight of the ship over a greater volume
Physics
2 answers:
finlep [7]4 years ago
8 0

Answer:

Explanation:

Spread the weight of the ship over a greater volume

Setler [38]4 years ago
5 0

If a cruise ship is having troubles with buoyancy, then spread the weight of the ship over a greater volume.

Answer: Option D

<u>Explanation: </u>

Buoyancy is the upward thrusting phenomenon of water acting on any object immersed partially or fully in water body. Hence, it creates the buoyant forces that is inversely proportionate to the immersing body's density. If the immersing body's density is higher than the density of the immersing medium then the body will get completely immersed in the water.

Similarly, in case of less, the buoyant forces act on the body will prevent it from complete immersion and allow it to float on water. Mostly cruise ships and other navy vessels use this phenomenon to keep on floating on surface of water.

In the present condition, the solution for buoyancy problem faced by a cruise ship can be solved by decreasing the density of the ship. And the ship's density can be decreased by increasing the ship's volume or by spreading the ship's weight over a greater volume.

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3 years ago
A 10 kg ball strikes a wall with a velocity of 3 m/s to the left. The ball bounces off with a velocity of 3 m/s to the right. If
lord [1]

Answer:

The force is 272.73 newtons

Explanation:

We're going to use impulse-momentum theorem that states impulse is the change on the linear momentum this is:

\overrightarrow{J}=\overrightarrow{p}_{f}-\overrightarrow{p}_{i} (1)

Impulse is also defined as average force times the time the force is applied:

\overrightarrow{J}=\overrightarrow{F}_{avg}(\varDelta t) (2)

By (2) on (1):

\overrightarrow{F}_{avg}(\varDelta t)= \overrightarrow{p}_{f}-\overrightarrow{p}_{i}

solving for \overrightarrow{F}_{avg}:

\overrightarrow{F}_{avg}=\frac{\overrightarrow{p}_{f}-\overrightarrow{p}_{i}}{\varDelta t} (3)

We already know Δt is equal to 0.22 s, all we should do now is to find \overrightarrow{p}_{f}-\overrightarrow{p}_{i} and put on (3) (\overrightarrow{p_{i}} the initial momentum and \overrightarrow{p_{f}} the final momentum). Linear momentum is defined as \overrightarrow{p}=m\overrightarrow{v} , using that on (3):

\varDelta\overrightarrow{p}=m \overrightarrow{v_{f}}-m \overrightarrow{v_{i}} (4)

Velocity (v) are vectors so direction matters, if positive direction is the right direction and negative direction left \overrightarrow{v_{i}}=+3\, \frac{m}{s} and \overrightarrow{v_{f}}=-3\, \frac{m}{s} so (4) becomes:

\varDelta\overrightarrow{p}=m(-3\frac{m}{s}- (+3\frac{m}{s}))=-(10kg)(6\frac{m}{s})

\varDelta\overrightarrow{p}=-60\, \frac{mkg}{s} (5)

Using (5) on (3):

\overrightarrow{F}_{avg}=\frac{-60\, \frac{mkg}{s}}{0.22s}

F_{avg}=272.73N

8 0
4 years ago
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