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4 years ago

8

Where do most metamorphic processes take place?

1 answer:

inysia [295]4 years ago

8 0

Answer:

Most metamorphic processes take place deep underground, inside the earth's crust.

Explanation:

During metamorphism, protolith chemistry is mildly changed by increased temperature (heat), a type of pressure called confining pressure, and/or chemically reactive fluids. hope this helps you :)

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At the time of the founding of the country, most Americans worked as

STatiana [176]

Answer:

Farmers

Explanation:

At the time of founding our country most soldiers in the revolutionary war were farmers who were inexperienced.

4 0

3 years ago

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In prokaryotic and eukaryotic cells the amount of water in the cell is kept at a steady state. The organelles

Len [333]

Answer:

A

Explanation:

It would not be C because prokaryotic cells don't have a nucleus. And if you're looking for a body in the cell that controls the nutrients and water it would be A.

3 0

3 years ago

A 60kg student traveling in a 1000kg car with a constant velocity has a kinetic energy of 1.2 x 10^4 J. What is the speedometer

777dan777 [17]

Answer:

17.64 km/h

Explanation:

mass of car, m = 1000 kg

Kinetic energy of car, K = 1.2 x 10^4 J

Let the speed of car is v.

Use the formula for kinetic energy.

$K = \frac{1}{2}mv^{2}$

By substituting the values

$1.2\times 10^{4} = \frac{1}{2}\times 1000\times v^{2}$

v = 4.9 m/s

Now convert metre per second into km / h

We know that

1 km = 1000 m

1 h = 3600 second

So, $v = \left (\frac{4.9}{1000} \right )\times \left ( \frac{3600}{1} \right )$

v = 17.64 km/h

Thus, the reading of speedometer is 17.64 km/h.

4 0

3 years ago

Two spheres having masses M and 2M and radii R and 3R, respectively, are released from rest when the distance between their cent

Andrei [34K]

Answer:

$v_2 = \sqrt{\frac{GM}{3R}}$

$v_1 = 2\sqrt{\frac{GM}{3R}}$

Explanation:

As we know by energy conservation that change in gravitational potential energy of the system = change in kinetic energy of the two ball

So here we can say

$-\frac{GM(2M)}{12R} + 0 = -\frac{GM(2M)}{4R} + \frac{1}{2}Mv_1^2 + \frac{1}{2}(2M)v_2^2$

Also since there is no external force on the system of two masses so here total momentum of the two balls will remains conserved

$0 = Mv_1 + 2Mv_2$

$v_1 = -2v_2$

now we have

$\frac{GM^2}{2R} - \frac{GM^2}{6R} = \frac{1}{2}M(-2v_2)^2 + \frac{1}{2}(2M)v_2^2$

$\frac{GM^2}{3R} = Mv_2^2$

$v_2 = \sqrt{\frac{GM}{3R}}$

$v_1 = 2\sqrt{\frac{GM}{3R}}$

4 0

4 years ago

A car of mass 940.0 kg accelerates away from an intersection on a horizontal road. When the car speed is 42.5 km/hr (11.8 m/s),

MAVERICK [17]

Answer:

0.39 $m/s^2$

Explanation:

Parameters given:

Mass of car, m = 940 kg

Speed of car, v = 11.8 m/s

Power supplied by engine, P = 4300 W

To get the acceleration, we must define the relationship between Power and velocity.

Power, P, is given in terms of velocity, v, as:

P = F * v

where F = force

This is because Power is given as:

$P = \frac{E}{t} \\\\\\P = \frac{F*d}{t} \\\\\\=> P = F * v$

(where E = energy. t = time taken, d = distance moved)

Force, F, is given as:

F = m*a

Therefore, Power will be:

P = m * a * v

Acceleration, a, will then be:

$a = \frac{P}{m*v}$

$a = \frac{4300}{940 * 118} \\\\\\a = 0.39 m/s^2$

The acceleration of the car at that time is 0.39 $m/s^2$

4 0

3 years ago

Read 2 more answers