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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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irinina [24]

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2 years ago
An AC generator consists of eight turns of wire, each of area 0.0775 m2 , and total resistance of 8.53 Ω. The loop rotates in th
Bad White [126]

Answer:

44.08 Volt

Explanation:

N = 8, A = 0.0775 m^2, R = 8.53 ohm, B = 0.222 T, f = 51 Hz

e0 = N B A w

e0 = 8 x 0.222 x 0.0775 x 2 x 3.14 x 51

e0 = 44.08 Volt

3 0
3 years ago
An 8.75 kg point mass and a 14.0 kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point be
marysya [2.9K]

Answer

given,

mass of the = m₁ = 8.75 Kg

another mass of the object = m₂ = 14 Kg

distance between them = 50 cm

R₁ = 17 cm

R₂ = 50 -17 = 33 cm

a) Force applied due to the Mass 8.75  in +ve x- direction

F_1 = \dfrac{GM_1 m}{R_1^2}

F_1 = \dfrac{6.67\times 10^{-11} \times 8.75\ m}{0.17^2}

F_1 = 2.019\times 10^{8}\ m

Force applied due to mass 14 Kg in -ve x-direction

F_2 = \dfrac{GM_2 m}{R_2^2}

F_2 = \dfrac{6.67\times 10^{-11} \times 14\ m}{0.33^2}

F_2 = 0.857\times 10^{8}\ m

net force

F = F₁ + F₂

F = 2.019\times 10^{8}\ m - 0.857\times 10^{8}\ m

F = 1.162\times 10^{8}\ m

Using newton second law

a = \dfrac{F}{m}

a = \dfrac{ 1.162\times 10^{8}\ m}{m}

a =1.162\times 10^{8} \ m/s^2

b) As the acceleration of mass comes out to be  +ve hence, the direction will be toward the mass of 8.75 Kg

6 0
3 years ago
Two objects are in freefall. Object A experiences a gravitational force of
Vlad1618 [11]

<u>Answer:</u>

Both the objects A and B will have the same acceleration.

<u>Explanation :</u>

The objects will have the same acceleration as both are under free fall condition. When objects are under the free fall condition, the only force that acts on the object is its weight.

Weight is the force acting on a body of some mass, and the formula for finding the weight of a body is- Weight = mass × acceleration due to gravity(g).

Therefore, here the different weight is due to the difference masses of both bodies, and not due to the different acceleration values.

3 0
3 years ago
Examples of fixed pulley?​
Neko [114]

Answer:

One example of a fixed pulley is a Flag Pole

Explanation:

A good example of a fixed pulley is a flag pole: When you pull down on the rope, the direction of force is redirected by the pulley, and you raise the flag. Other examples of movable pulleys include construction cranes, modern elevators, and some types of weight lifting machines at the gym.

4 0
3 years ago
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