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3 years ago

The process by which the seafloor moves apart at mid-ocean ridges is called _____.

Physics

2 answers:

worty [1.4K]3 years ago

6 0

I think the answer would be C, because to me that's one that makes sense, I hope that I could help, Have a great Thursday!

Ira Lisetskai [31]3 years ago

5 0

Answer:

C. Sea floor spreading.

Explanation:

The geological process by which the tectonic plates separate from each other in the seabed allowing the rise of magma is known as sea floor spreading.

In this process, magma comes into contact with cold water and solidifies rapidly, generating a ridge at the bottom of the ocean, as is shown in the image.

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If the volume is held constant, what happens to the pressure of a gas as temperature is decreased? Explain.

Lady bird [3.3K]

Answer:Decreases

Explanation:

Given

Volume is held constant that is it is a isochoric process.

We know that

PV=nRT

as n,V& R are constant therefore only variables are

P & T

so $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{P_1}{P_2}=\frac{T_1}{T_2}$

As $T_1$ is decreasing therefore Pressure must also decrease so that ratio remains constant.

6 0

3 years ago

Why are the constellations in the summer sky different from those in the winter?

atroni [7]

This happens because of the earth rotating around the sun. So we see different constellations for different seasons.

4 0

3 years ago

(help please no links )

trapecia [35]

Answer:

Hello, how's your day going?

if humanity came together and made a base on the moon, it would be revolutionary. The point of a base on the moon would have multiple purposes. for example, some think that the moon contains valuable metals such as iron and titanium. a base would serve as a place for workers harvesting metals to rest. Obviously or not most of the iron harvesting would be done automatically by robots and such.

If such a base were constructed on the moon, it would be the begining of people living on other worlds and would be a great start for a base on Mars.

Hope it helped

Spiky Bob

5 0

3 years ago

A 2.1 kg steel ball strikes a massive wall at 13.2 m/s at an angle of 64.8 ◦ with the perpendicular to the plane of the wall. It

solong [7]

Answer:

112 N

Explanation:

going through the question you would notice that some detail is missing, using search engines u was able to find a similar question on "https://socratic.org/questions/a-2-3-kg-steel-ball-strikes-a-wall-with-a-speed-of-8-5-m-s-at-an-angle-of-64-wit"

and here is the question i found

"A 2.3 kg steel ball strikes a wall with a speed of 8.5 m/s at an angle of 64⁰ with the surface. It bounces off with the same speed and angle. If the ball is in contact with the wall for 0.448 s. What is the average force exerted by the ball?"

you would notice that there is a change in the values from the question posted, hence we would only take the following part to complete our question, " If the ball is in contact with the wall for 0.448 s what is the average force exerted by the ball?" while retaining all original detail.

solution

mass of ball (m) = 2.1 kg

speed of ball (v) = 13.2 m/s

angle of contact (p) = 64.8°

time of contact (t) = 0.448 s

What is the average force exerted by the ball?

The average force exerted by the ball = $\frac{change in momentum}{change in time}$

where

The momentum changes only the direction perpendicular to the wall, hence the component of momentum perpendicular to the wall = m x v x sin (p) = 2.1 x 13.2 x sin 64.8 = 25.1 kg.m/s

since the ball strikes the wall and bounces off it with the same speed

and at the same angle, the component of momentum acting

perpendicular to the wall remains the same while hitting and leaving

the wall but in opposite directions.

Hence the component of momentum acting perpendicular to the wall while hitting and leaving the wall will be 25.1 kg.m/s and -25.1 kg.m/s respectively.

change in momentum = 25.1 - (-25.1) = 25.1 + 25.1 = 50.2 kg.m/s

change in time = 0.448 s

now substituting the above into the equation we have

The average force exerted by the ball = $\frac{50.2}{0.448}$ = 112 N

6 0

4 years ago

A police officer at rest at side of highway notices speeder moving at 62 km/h along road.when speeder passes ,officer accelerate

Korolek [52]

To answer the following questions for this specific problem:

a. 11.48 secs

b. Vp = a*t*3.6 = 3*11.48*3.6 = 124.0 km/h

I am hoping that this answer has satisfied your query about and it will be able to help you.

4 0

3 years ago

Read 2 more answers