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Tomtit [17]
2 years ago
14

What area of earth contains semi-solid rock and lava?

Physics
1 answer:
ioda2 years ago
4 0
Below the crust is the mantle, a dense, hot layer of semi-solid rock approximately 2,900 km thick. The mantle, which contains more iron, magnesium, and calcium than the crust, is hotter and denser because temperature and pressure inside the Earth increase with depth.

Hope this helped! :)
Can I be brainliest please?
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Aye, I'm b0red so here's an easy one:
asambeis [7]

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8 0
2 years ago
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There are 152 steps between the ground floor and the sixth floor in a building. Each step is 16.2 cm tall. It takes 3 minutes an
lisabon 2012 [21]

Answer:

14.4kJ

Explanation:

Work = Force x distance

W × h = mgh

Given that,

mass m, = 59.5kg

acceleration due to gravity = 9.8m/s^2

height ,h = 16.2cm

convert to m is 0.162m

How much work = m x g x h

height is 0.162 x 152 steps

h = 24.624m

work = 59.5 x 9.8 x 24.624

= 14,358.25Joule

= 14.4kJ

7 0
3 years ago
1. Which of the following are examples of natural resources
wariber [46]

Answer:

1) Coal is a natural resource, so is wood, wind energy, soil, salt, fluorite, water, and copper

2) Solar energy, natural gas, and wind energy are all renewable

3) Salt is non - renewable as is synthetic diamonds and plastic  

4) I think it would be number

5) Based on their mineral, chemical, and textural composition

Explanation:

Hope this helps :)

4 0
2 years ago
A car changes velocity at a constant acceleration of 2.5m/s to reach 43.7m/s in 2.7 s how fast was the car moving when it began
Montano1993 [528]

The formula we can use in this case is:

v = v0 + a t

where v is final velocity, v0 is initial velocity, a is acceleration and t is time

So finding for v0:

v0 = v – a t

v0 = 43.7 – (2.5) 2.7

v0 =  36.95 m/s

8 0
3 years ago
An object of mass 300 g, moving with an initial velocity of 5.00i-3.20j m/s, collides with an sticks to an object of mass 400 g,
Alexus [3.1K]

Answer:

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

Explanation:

Mass of object 1 , m₁ = 300 g = 0.3 kg

Mass of object 2 , m₂ = 400 g = 0.4 kg

Initial velocity of object 1 , v₁ = 5.00i-3.20j m/s

Initial velocity of object 2 , v₂ = 3.00j m/s

Mass of composite = 0.7 kg

We need to find final velocity of composite.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = 0.3 x (5.00i-3.20j) + 0.4 x 3.00j = 1.5 i + 0.24 j kgm/s

Final momentum = 0.7 x v = 0.7v kgm/s

Comparing

1.5 i + 0.24 j = 0.7v

v = 2.14 i + 0.34 j

Magnitude of velocity      

       v=\sqrt{2.14^2+0.34^2}=2.17m/s

Direction,  

       \theta =tan^{-1}\left ( \frac{0.34}{2.14}\right )=9.03^0

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

7 0
3 years ago
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