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Bas_tet [7]
3 years ago
12

planets A & B are near each other but there is a large difference in their temperatures using the data in the table explain

how the atmosphere of these two planets can influence the average temperature​

Physics
1 answer:
Yakvenalex [24]3 years ago
7 0

Answer: Planet A is closer to the Sun and has much greater atmospheric pressure. This suggests that planet A has a thicker atmosphere. Planet A's atmosphere is also mostly made of carbon dioxide, which is a greenhouse gas. This gas retains heat, raising the surface temperature of planet A.

Explanation:

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It takes brooke 10minutes to run 1 mile what is her speed in miles per minutes
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If you divide miles by minutes, the answer will have units of
miles per minute, which is exactly what you want.

(1 mile) / (10 minutes) = 1/10  mile/minute = 0.1 mile per minute


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A positive charge is moved from point A to point B along an equipotential surface. How much work is performed or required in mov
Nitella [24]

Answer:

No work is performed or required in moving the positive charge from point A to point B.

Explanation:

Lets take

Q= Positive charge which move from  point A to point B along

Voltage difference,ΔV =V₁ - V₂  

The work done

W = Q . ΔV

Given that  charge is moved from point A to point B along an equipotential surface.It means that voltage  difference is zero.

ΔV = 0

So

W = Q . ΔV

W = Q x 0

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5 0
3 years ago
Suppose a 2.0kg bird is flying at a speed of of 1m/s. It’s kinetic energy would be what?
Ganezh [65]

As we know that the formula of kinetic energy will be

KE = \frac{1}{2} mv^2

now here we know that

m = 2 kg

v = 1 m/s

so from the above equation we have

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7 0
2 years ago
The latent heat of fusion for Aluminium is 3.97 x 105. How much energy would be required to melt 0.75 kg of it?
RoseWind [281]

Answer:

E = 2.9775\times10^5 J

Explanation:

Given:  The latent heat of fusion for Aluminum is L = 3.97\times10^5  J/Kg

mass to be malted m = 0.75 Kg

Energy require to melt E = mL

E = 3.97\times10^5\times0.75 = 2.9775\times10^5 J

Therefore, energy required to melt 0.75 Kg aluminum

E = 2.9775\times10^5 J

5 0
2 years ago
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