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hoa [83]
3 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]3 years ago
8 0

Answer:

Explanation:

Spread the weight of the ship over a greater volume

Setler [38]3 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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Rony fills a bucket with water and whirls it in a vertical circle to demonstrate that the water will not spill from the bucket a
Romashka-Z-Leto [24]

Answer:

0 N, 3.49 m/s

Explanation:

Draw a free body diagram for the bucket at the top of the swing.  There are two forces acting on the bucket: weight and tension, both downwards.

If we take the sum of the forces in the radial direction, where towards the center is positive:

∑F = ma

W + T = m v² / r

The higher the velocity that Rony swings the bucket, the more tension there will be.  The slowest he can swing it is when the tension is 0.

W = m v² / r

mg = m v² / r

g = v² / r

v = √(gr)

Given that r = 1.24 m:

v = √(9.8 m/s² × 1.24 m)

v = 3.49 m/s

8 0
3 years ago
What is elasticity in polymer generally related to?
Andrej [43]
The elasticity of a polymer is primarily due to the structure of the molecule and the cross-linking between strands. Hydrogen bonding is a contributor to the shape of the molecule, but not a major player in terms of elasticity. We would have to answer "false".
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6 0
3 years ago
Read 2 more answers
The equation below is for potassium oxide.
ValentinkaMS [17]

Answer: The ratio of atoms of potassium to ratio of atoms of oxygen is 4:2

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed, and remains conserved. The mass of products must be same as that of the reactants.

Thus the number of atoms of each element must be same on both sides of the equation so as to keep the mass same and thus balanced chemical equations are written.

K exists as atoms and oxygen exist as molecule which consists of 2 atoms. The ratio of number of atoms on both sides of the reaction are same and thus the ratio of atoms of potassium to ratio of atoms of oxygen is 4:2.

5 0
3 years ago
Read 2 more answers
A person is filling a knee-high bucket with water using a garden hose and holding it such that water discharges from the hose at
Serggg [28]

Answer:

Yes i am agree with this suggestion

Explanation:

Given that we have to assume that there is no any frictional affects.

As we know that when height increases then the discharge level will decreases when discharge level decreases then the time of filling for the bucket will increase.So the  bucket will fill faster if the hose lowered until knee level.

Yes i am agree with this suggestion

8 0
3 years ago
What must be the length of a simple pendulum if its oscillation frequency is to be equal to that of an air-track glider of mass
Anvisha [2.4K]

Answer:

the length of the simple pendulum is 0.25 m.

Explanation:

Given;

mass of the air-track glider, m = 0.25 kg

spring constant, k = 9.75 N/m

let the length of the simple pendulum = L

let the frequency of the air-track glider which is equal to frequency of simple pendulum = F

The oscillation frequency of air-track glider is calculated as;

F = \frac{1}{2\pi } \sqrt{\frac{k}{m} } \\\\F = \frac{1}{2\pi } \sqrt{\frac{9.75}{0.25} } \\\\F = 0.994 \ Hz

The frequency of the simple pendulum is given as;

F = \frac{1}{2\pi} \sqrt{\frac{g}{l} } \\\\2\pi(F) = \sqrt{\frac{g}{l} } \\\\2\pi (0.994) = \sqrt{\frac{9.8}{l} } \\\\6.2455 = \sqrt{\frac{9.8}{l} } \\\\(6.2455)2 = \frac{9.8}{l} \\\\39.006 = \frac{9.8}{l} \\\\l = \frac{9.8}{39.006} \\\\l = 0.25 \ m

Thus, the length of the simple pendulum is 0.25 m.

8 0
3 years ago
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