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Kisachek [45]
3 years ago
14

Which sphere of Earth is associated with plate tectonics?

Physics
2 answers:
SOVA2 [1]3 years ago
8 0

Answer:

D. Lithosphere

Explanation:

Hope this helps!

Marrrta [24]3 years ago
5 0

When looking at this question, we can easily start by eliminating certain answers. In the selections you've provided, you've shown atmosphere. We can easily eliminate letter A, as that makes absolutely no sense. Moving on, you also eliminate letter B, as that deals with ecosystems and whatnot. And finally, you can eliminate hydrosphere, letter C - as that's not the same. That deals with water, like oceans or rivers.

That leaves you with D) Lithosphere for your answer. The Lithosphere is the rigid part of the earth, the outermost layer, I would say. The crust / mantle. That's why it would be letter D - plate tectonics seem to have relations with the Lithosphere. The lithosphere is affected.

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The force exerted on the tires of a car that directly accelerate it along a road is exerted by the
azamat

The force exerted on the tires of a car that directly accelerate it along a road is exerted by the road friction.

<h3>What is force?</h3>

Force is defined as the product of mass and acceleration of an object.

Friction is defined as the force that resists the movement of an object over another.

Therefore, the force exerted on the tires of a car that directly accelerate it along a road is exerted by the road friction.

Learn more about force here:

brainly.com/question/12970081

#SPJ12

7 0
2 years ago
While water skiing behind her father’s boat, Letty is pulled at constant speed by a force of 164 N from the tow rope that makes
kap26 [50]

Answer:

0.265

Explanation:

Draw a free body diagram.  There are four forces:

Normal force Fn pushing up.

Weight force mg pulling down.

Tension force T at an angle θ.

Friction force Fn μ pushing left.

Sum the forces in the y direction:

∑F = ma

Fn + T sin θ − mg = 0

Fn = mg − T sin θ

Sum the forces in the x direction:

∑F = ma

T cos θ − Fn μ = 0

Fn μ = T cos θ

μ = T cos θ / Fn

μ = T cos θ / (mg − T sin θ)

Given T = 164 N, θ = 10.0°, m = 65.0 kg, and g = 9.8 m/s²:

μ = (164 N cos 10.0°) / (65.0 kg × 9.8 m/s² − 164 N sin 10.0°)

μ = 0.265

7 0
3 years ago
Buttery popcorn contained in a large 1__bowl has a mass of about 50 __<br>and about 650 calories.​
Rasek [7]

Answer:

Buttery popcorn contained in a large 1 liter bowl has a mass of about 50 mg and about 650 calories.

Explanation:

Liter is the most appropriate unit to measure a bowl. Usually 1 liter of liquid has a mass of 1000 gram. Since popcorn is something lightweight and only a few can fill the bowl quickly so 50 mg makes perfect sense with 1 liter of bowl and 650 calories in buttery popcorn.

4 0
3 years ago
Read 2 more answers
A 25.0-g sample of copper at 363 K is placed in I 00.0 g of water at 293 K. The copper and water quickly come to the sa me tempe
Simora [160]

Answer:

Final temperature is 295K

Explanation:

Where the sample of copper is placed in the water, the heat transferred from the copper is equal that the heat absorbed by the water.

The heat transferred from the copper is:

C×\frac{1mol}{63,546g}×mass×ΔT

Where C is molar heat capacity of copper (24,5J/molK)

Mass is 25,0g

And ΔT is final temperature - initial temperature (X-363K)

Also, the heat absorbed by the water is:

-C×\frac{1mol}{18,02g}×mass×ΔT

Where C is molar heat capacity of water (75,2J/molK)

Mass is 100,0g

And ΔT is final temperature - initial temperature (X-293K)

As heat transferred is equal to heat absorbed:

24,5J/molK×\frac{1mol}{63,546g}×25,0g×(X-363K) = -75,2J/molK×\frac{1mol}{18,02g}×100,0g× (X-293K)

9,64X J/K - 3499J = - 417X J/K + 122273J

426,64X J/K = 125772 J

<em>X = 295K</em>

<em></em>

Final temperature is 295K

I hope it helps!

6 0
4 years ago
A 4-kg toy car with a speed of 5 m/s collides head-on with a stationary 1-kg car. After the collision, the cars are locked toget
mihalych1998 [28]

Kinetic energy lost in collision is 10 J.

<u>Explanation:</u>

Given,

Mass, m_{1} = 4 kg

Speed, v_{1} = 5 m/s

m_{2} = 1 kg

v_{2} = 0

Speed after collision = 4 m/s

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

Before collision, the kinetic energy is

\frac{1}{2} m1 (v1)^2 + \frac{1}{2} m2(v2)^2

By plugging in the values we get,

KE = \frac{1}{2} * 4 * (5)^2 + \frac{1}{2} * 1 * (0)^2\\\\KE = \frac{1}{2} * 4 * 25 + 0\\\\

K×E = 50 J

Therefore, kinetic energy before collision is 50 J

Kinetic energy after collision:

KE = \frac{1}{2} (4 + 1) * (4)^2 + KE(lost)

KE = 40J + KE(lost)

Since,

Initial Kinetic energy = Final kinetic energy

50 J = 40 J + K×E(lost)

K×E(lost) = 50 J - 40 J

K×E(lost) = 10 J

Therefore, kinetic energy lost in collision is 10 J.

4 0
3 years ago
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