What causes a submarine to rise and descend?

Physics

1 answer:

photoshop1234 [79]3 years ago

6 0

Answer:

When the ballast tanks are filled with air, the submarine rises to the surface because it has positive buoyancy. With water inside the tanks, the sub has negative buoyancy so it sinks deeper into the ocean. The ballast tanks can also be used to help a submarine surface very quickly in an emergency

Explanation:

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A stretched string is observed to have four equal segments in a standing wave driven at a frequency of 480 hz. what driving freq

Korvikt [17]

600Hz is the driving frequency needed to create a standing wave with five equal segments.

To find the answer, we have to know about the fundamental frequency.

<h3>How to find the driving frequency?</h3>

The following expression can be used to relate the fundamental frequency to the driving frequency;

f(n) = n * f (1)

where, f(1) denotes the fundamental frequency and the driving frequency f(n).

The standing wave has four equal segments, hence with n=4 and f(n)=4, we may calculate the fundamental frequency.

f(4) = 4× f (1)

480 = 4× f(1)

f(1) = 480/4 =120Hz.

So, 120Hz is the fundamental frequency.

To determine the driving frequency necessary to create a standing wave with five equally spaced peaks?
For, n = 5,

f(n) = n 120Hz,

f(5) = 5×120Hz=600Hz.

Consequently, 600Hz is the driving frequency needed to create a standing wave with five equal segments.

Learn more about the fundamental frequency here:

brainly.com/question/2288944

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A plane leaves Hartsfield Airport and flies around the world returning to Hartsfield airport What is the planes displacement?

rewona [7]

The way around the world

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A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit

Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

$M=8.7\cdot 10^{13}kg$ is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

$F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N$

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

$m\frac{v^2}{r}=G\frac{Mm}{r^2}$