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3 years ago

From the deepest to the surface what are the parts of the earthâs interior

1 answer:

6 0

Inner core:

Outer core

Lower mantle

Transition region

Upper mantle:

Oceanic crust:

Continental crust

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The acceleration of gravity at the surface of Moon is 1.6 m/s2 . A 5.0 kg stone thrown upward on Moon reaches a height of 20 m.

kramer

Answer:

(a) 8 m/s

(b) 5 s

Explanation:

(a)

Using,

V² = U²+2gh ......................... Equation 1

Where V = final velocity, U = Initial velocity, g = acceleration due to gravity on the surface of the moon, h = height reached.

Given: V = 0 m/s ( At it's maximum height), g = -1.6 m/s² ( as its moves against gravity), h = 20 m.

Substitute into equation 1

0 = U²+[2×20×(-1.6)]

-U² = - 64

U² = 64

U = √64

U = 8 m/s.

(b)

V = U +gt.................... Equation 2

Where t = time to reach the maximum height.

Given: V = 0 m/s ( At the maximum height), g = -1.6 m/s² ( Moving against gravity), U = 8 m/s.

Substitute into equation 2

0 = 8+(-1.6t)

-8 = -1.6t

-1.6t = -8

t = -8/-1.6

t = 5 s.

5 0

3 years ago

A railroad freight car, mass 15,000 kg, is allowed to coast along a level track at a speed of 2.0 m/s. It collides and couples w

gayaneshka [121]

Answer:

The speed of the two cars after coupling is 0.46 m/s.

Explanation:

It is given that,

Mass of car 1, m₁ = 15,000 kg

Mass of car 2, m₂ = 50,000 kg

Speed of car 1, u₁ = 2 m/s

Initial speed of car 2, u₂ = 0

Let V is the speed of the two cars after coupling. It is the case of inelastic collision. Applying the conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)V$

$V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$V=\dfrac{15000\ kg\times 2\ m/s+0}{(15000\ kg+50000\ kg)}$

V = 0.46 m/s

So, the speed of the two cars after coupling is 0.46 m/s. Hence, this is the required solution.

3 0

3 years ago

When you take pictures with a camera, the distance between the lens and the film or chip has to be adjusted, depending on the di

Shalnov [3]

Answer:

Explanation:

When a camera shifts focus from a faraway object to a nearby object, the lens-to-film distance must increase. Likewise, when it shifts focus from a nearby object to a distant object, there must be an increase in the lens to film distance (that is, the image distance).

Therefore, if the picture of an object that is far away, the lens must move towards the film.

The focal length cannot be changed because it is fixed for a lens. Nevertheless, in order to focus on an object, the image distance can be changed.

8 0

4 years ago

A. what is the formula for measuring the price elasticity of supply?

daser333 [38]

Can't you look it up or something? Idk what it is but I can try and look it up for ya! If you want

7 0

3 years ago

A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)

kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

$W=Fd cos \theta$

where

F is the magnitude of the force applied

d is the displacement of the crate

$\theta$ is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied horizontally. Therefore, the work done is

$W=(230)(4.0)(cos 0^{\circ})=920 J$

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

$W=(1300)(4.0)(cos 0^{\circ})=5200 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0

3 years ago