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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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The average intensity of light emerging from a polarizing sheet is 0.708 W/m2, and that of the horizontally polarized light inci
Pachacha [2.7K]

Answer:

Angle θ = 30.82°

Explanation:

From Malus’s law, since the intensity of a wave is proportional to its amplitude squared, the intensity I of the transmitted wave is related to the incident wave by; I = I_o cos²θ

where;

I_o is the intensity of the polarized wave before passing through the filter.

In this question,

I is 0.708 W/m²

While I_o is 0.960 W/m²

Thus, plugging in these values into the equation, we have;

0.708 W/m² = 0.960 W/m² •cos²θ

Thus, cos²θ = 0.708 W/m²/0.960 W/m²

cos²θ = 0.7375

Cos θ = √0.7375

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θ = Cos^(-1)0.8588

θ = 30.82°

4 0
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5 0
2 years ago
Express 4,560 m in Km
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7 0
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SP 15The magnetic field in a region of space is measured to be:This field is known to be caused by a cluster of long-straight wi
tino4ka555 [31]

Answer:

 i = 0.477 10⁴ B

the current flows in the  counterclockwise

Explanation:

For this exercise let's use the Ampere law

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Where the path is closed

Let's start by locating the current vines that are parallel to the z-axis, so it must be exterminated along the x-axis and as the specific direction is not indicated, suppose it extends along the y-axis.

From BiotSavart's law, the field must be perpendicular to the direction of the current, so the magnetic field must go in the x direction.

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Segment on the y axis

        L₀ = (y2-y1)

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Segment on the point x = 2 cm

        L₁ = 3-0

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       B L = μ₀ I

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The current is perpendicular to the magnetic field whereby the current flows in the  counterclockwise

8 0
3 years ago
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