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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 is winter in Puerto Rico. Compare the air temperatures beachgoers feel near the water
Leviafan [203]
Large amounts of water do have a big impact on the weather: indeed, it takes less energy to warm/cool land than water.
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7 0
3 years ago
In a bi-prism experiment the eye-piece was placed at a distance 1.5m from the source. The distance between the virtual sources w
Vladimir [108]

Answer:

λ = 1.4 × 10^(-7) m

Explanation:

We are given;

distance of eye piece from the source;D = 1.5 m

distance between the virtual sources;d = 7.5 × 10^(-4) m

To find the wavelength, we will use the formula for fringe width;

X = λD/d

Where X is fringe width, λ is wavelength, while d and D remain as before.

Now, fringe width = eye-piece distance moved transversely/number of fringes

Eye piece distance moved transversely = 1.88 cm = 1.88 × 10^(-2) m

Thus,

Fringe width = (1.88 × 10^(-2))/10 = 1.88 × 10^(-3) m

Thus;

1.88 × 10^(-3) = λ(1.5)/(7.5 × 10^(-4))

λ = [1.88 × 10^(-3) × (7.5 × 10^(-4))]/1.5

λ = 1.4 × 10^(-7) m

6 0
3 years ago
The initial volume reading in a graduated cylinder is 30 mL. The mass of an irregular shape of an unknown metal piece is 55.3 g.
Vadim26 [7]

Answer:

7.9\frac{gr}{cm^3}

Explanation:

When we put the metal piece in the liquid (which is in the graduated cylinder), how much it goes up is equal to the volume of the piece we inserted.

So now we know that the volume of that piece of unknown metal is 7mL (which is the same as 7cm^3).

Density is \frac{mass}{volume}.

So the density of that piece of metal is \frac{55.3g}{7cm^3}

Which leaves us with a final density of 7.9\frac{gr}{cm^3}

6 0
3 years ago
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