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ollegr [7]
3 years ago
14

The process of heat radiation is the ONLY method of heat transfer that can occur

Physics
2 answers:
AysviL [449]3 years ago
6 0

Answer:

Through space

Explanation:

The radiation from the sun heats the outer atmosphere

lina2011 [118]3 years ago
5 0
Radiation emitted by a body is a consequence of thermal agitation of its composing molecules. so...<span> electromagnetic waves ?</span>
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1. Water boils at 100°C at sea level. If the water in this experiment did not boil at 100°C, what could be the reason?
umka2103 [35]

Answer:

It may not be at the sea level

Explanation:

The reason here is water only boils at sea level. This means that if you move water to a different height, say top of a mountain, the boiling temperature of water would change. This is due to the pressure drop at high place. The drop of pressure would make it harder to transform water liquid to gas, thus requiring more temperature.

8 0
3 years ago
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A 720 kg roller-coaster starts off from Location A. Assuming friction does not impede the car's motion, what will be the change
vaieri [72.5K]
We know, Potential Energy = m * g * h
Here, mass & gravity would be same, but their height will change so it will be:

ΔU = U₂ - U₁
ΔU = mgh₂ - mgh₁
ΔU = mg (h₂ - h₁)

Hope this helps!
7 0
3 years ago
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Which part of the water cycle would follow step D in the diagram shown?
Dmitry [639]

Condensation.


iGreen, more like iNOTGreen amirite?

5 0
3 years ago
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1. The process of producing energy by utilizing heat trapped inside the earth's
babymother [125]

Answer:

B

Explanation:

The process is called Geo-Thermal energy because it's an inexhaustible source of energy

6 0
3 years ago
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A stone is thrown towards a wall with an initial velocity of v0=19m/s and an angle = 71 with the horizontal, as illustrated in t
HACTEHA [7]

Answer:

(a) 2.85 m

(b) 16.5 m

(c) 21.7 m

(d) 22.7 m

Explanation:

Given:

v₀ₓ = 19 cos 71° m/s

v₀ᵧ = 19 sin 71° m/s

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

(a) Find Δy when t = 3.5 s.

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²

Δy = 2.85 m

(b) Find Δy when vᵧ = 0 m/s.

vᵧ² = v₀ᵧ² + 2 aᵧ Δy

(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy

Δy = 16.5 m

(c) Find Δx when t = 3.5 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²

Δx = 21.7 m

(d) Find Δx when Δy = 0 m.

First, find t when Δy = 0 m.

Δy = v₀ᵧ t + ½ aᵧ t²

(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²

0 = t (18.0 − 4.9 t)

t = 3.67

Next, find Δx when t = 3.67 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²

Δx = 22.7 m

7 0
3 years ago
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