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Irina18 [472]
3 years ago
8

Which of the following is likely to contribute to geological events that take place on Earth?

Physics
2 answers:
Dennis_Churaev [7]3 years ago
5 0
Crust sitting on top of Milton rock of the mantle
MakcuM [25]3 years ago
4 0

Answer:

Crust sitting on top of molten rock of the mantle

Explanation:

Earth's crust is the geological name for what we usually call the <em>surface</em> of the planet, and it is immediately above the molten rock of the mantle, the next layer of the planet.

The crust happens to be fragmented in large pieces or tectonic plates, that can be seen as "drifting" on the mantle's molten rock, also known as magma.

In some areas tectonic plates are drifting away, allowing magma to reach the surface (volcanoes for example), or the seabed (forming mid-oceans crests).

In other areas tectonic plates are crashing each other, causing earthquakes, tidal waves, mountain formation and other relevant geological events.

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Which of the following is an example of projectile
jonny [76]

Answer:

The answer is choice A.

Explanation:

Assuming you are in a situation with a gravitational field. You can divide the motion of the bullet into two components. One horizontal and the other in the vertical.

7 0
3 years ago
Suppose the force of gravity on the electron was comparable to the electric force created by the deflection plates. how would th
Alexandra [31]
"F=Vector Sum Of The Two Forces" Is the answer.
3 0
3 years ago
Suppose a person pushes thumbtack that is 1/5 centimeter long into a bulletin board, and the force (in dynes) exerted when the d
mario62 [17]

Answer:

W = 290.7 dynes*cm

Explanation:

d = 1/5 cm = 0.2 cm

The force is in function of the depth x:

F(x) = 1000 * (1 + 2*x)^2

We can expand that as:

F(x) = 1000 * (1 + 4*x + 4x^2)

F(x) = 1000 + 4000*x + 4000*x^2

Work is defined as

W = F * d

Since we have non constant force we integrate

W = \int\limits^{0.2}_{0} {(1000 + 4000*x + 4000*x^2)} \, dx

W = [1000*x + 2000*x^2 + 1333*X^3] evaluated between 0 and 0.2

W = 1000*0.2 + 2000*0.2^2 + 1333*0.2^3 - 1000*0 - 2000*0^2 - 1333*0^3

W = 200 + 80 + 10.7 = 290.7 dynes*cm

3 0
3 years ago
A small but measurable current of 5.8 × 10-10 A exists in a copper wire whose diameter is 3.0 mm. The number of charge carriers
swat32

Answer:

a) The current density ,J = 2.05×10^-5

b) The drift velocity Vd= 1.51×10^-15

Explanation:

The equation for the current density and drift velocity is given by:

J = i/A = (ne)×Vd

Where i= current

A = Are

Vd = drift velocity

e = charge ,q= 1.602 ×10^-19C

n = volume

Given: i = 5.8×10^-10A

Raduis,r = 3mm= 3.0×10^-3m

n = 8.49×10^28m^3

a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]

J = (5.8×10^-10) /(2.83×10^-5)

J = 2.05 ×10^-5

b) Drift velocity, Vd = J/ (ne)

Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)

Vd = (2.05×10^-5)/(1.36 ×10^10)

Vd = 1.51× 10^-5

8 0
3 years ago
Read 2 more answers
Convert 1erg into joule by dimensional method​
Valentin [98]

Answer:

1 * 10^-7 [J]

Explanation:

To solve this problem we must use dimensional analysis.

1 ergos [erg] is equal to 1 * 10^-7 Joules [J]

1[erg]*\frac{1*10^{-7} }{1}*[\frac{J}{erg} ] \\= 1*10^{-7}[J]

4 0
3 years ago
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