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Lana71 [14]
3 years ago
10

What are the cracks between the plates of the lithosphere called

Physics
1 answer:
Eddi Din [679]3 years ago
4 0
They're called plate boundaries
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The answer is two (4 -3)
Zero is not a SF
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a 4kg box is resting on top of a shelf that is 2m high, what is the boxs gravitational potential relative to the floor​
Aloiza [94]

Answer:

80 J

Explanation:

PE = mgh

PE = (4 kg)(9.8 m/s^2)(2 m)

PE = 78.4 J and with sig figs, it would be 80 J

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2 years ago
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a 2.0-mole sample of an ideal gas is gently heated at constant temperature 330 k. it expands from initial volume 19 l to final v
shutvik [7]
Isothermal Work =  PVln(v₂/v₁)

PV = nRT =  2 mole * 8.314 J/ (k.mol) * 330 k = 5487.24 J

Isothermal Work =  PVln(v₂/v₁)            v₂ = ? v₁ = 19L, 

1.7 kJ = (5487.24)In(v₂/19)

1700 = (5487.24)In(v₂/19)

In(v₂/19) = (1700/5487.24) = 0.3098

In(v₂/19) = 0.3098

(v₂/19) = e^{0.3098}


v₂  =  19* e^{0.3098}

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3 years ago
Last one I swear, 50 points again! plz help this is schoolwork
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It is a stretch of the atmosphere ranging from the upper mesosphere to the lower parts of the thermosphere. It’s useful to us in radio communication.
8 0
3 years ago
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You are holding a positive charge and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east. Wh
lara [203]

If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.

Reasoning:

It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.

Let the magnitude of the charge in my hand is Q, and the magnitude of the other charges is q.

Thus the electric force applied on the charge in my hand due to each other is,

F=\frac{kQq}{r^2}

Here k is the Coulomb constant, and r is the distance between the charges.

It is also known that the force on a positive charge due to another positive charge is acted outwards.

Thus, the force on the charge due to the charge on the east is,

\vec{F_1}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{i}

And the force on the charge due to the charge on the north is,

\vec{F_2}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{j}

As the forces are equal in magnitude and one is perpendicular to the other, thus the net force will be acted at an angle of 45^\circ from the north or from the north direction.

Thus the net force is acting in the north-east direction.

Learn more about the direction of the force here,

brainly.com/question/2037071

#SPJ4

3 0
2 years ago
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