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RSB [31]
2 years ago
7

What are the two sources of earth's internal heat energy

Physics
1 answer:
RUDIKE [14]2 years ago
6 0

Answer:

Hey

Your answer would be Radioactive Decay and Heat of formation.

When earth first formed it was very violant. Some planetary scientists believe that the moon was created from a collision involving earth and Thea (a theoretical dwarf planet). these collisions that have formed earth heated it so much that it is still hot from them.

Radioactivce decay is another major fource of internal heat for earth.

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Why is it important for scientists to replicate each other’s experiments?
Vilka [71]

Explanation:

to determine if important scientific results are repeatable

6 0
2 years ago
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To determine an epicentral distance scientists consider the arrival times of what wave types
Rudiy27
 The answer is P-waves and S-waves
3 0
3 years ago
In an incompressible three-dimensional flow field, the velocity components are given by u = ax + byz; υ = cy + dxz. Determine th
Ksivusya [100]

An incompressible flow field F in a 3D cartesian grid with components u,v,w:

F = u + v + w

where u,v,w are functions of x,y,z

Must satisfy:

∇·F = du/dx + dv/dy + dw/dz = 0

We have a field F defined:

F = u+v+w, u = ax+byz, v = cy+dxz

du/dx = a, dv/dy = c

Recall ∇·F = 0:

∇·F = du/dx + dv/dy + dw/dz = 0

a + c + dw/dz = 0

dw/dz = -a-c

Solve for w by separation of variables:

w = ∫(-a-c)dz

w = -az - cz + f(x,y)

f(x,y) is some undetermined function of x and y

The question states that w is not a function of x and y, therefore f(x,y) = 0...

w = -az - cz

8 0
2 years ago
A 18.0-kg rock is sliding on a rough, horizontal surface at 7.10 m/s and eventually stops due to friction. the coefficient of ki
Bond [772]
A = .3*g = 2.94 m/s² 

<span>t = v/a = 9/2.94 = 3.061 sec </span>

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4 0
3 years ago
Two boxes are at rest on a smooth, horizontal surface. The boxes are in contact with one another. If box 1 is pushed with a forc
Maslowich

Answer:

mass of box 1 = 2.20 kg

mass of box 2 = 5.93 kg

Explanation:

Let the mass of box 1 and box 2 is respectively

m_1 and m_2

so we will have

Force applied on box 1 then acceleration

a = \frac{F}{m_1 + m_2}

1.50 = \frac{12.2}{m_1 + m_2}

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Now we know that contact force between them in above case is given as

F_n = m_2a

8.90 = m_2(1.5)

m_2 = 5.93 kg

now we have

m_1 = 2.20 kg

8 0
3 years ago
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