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noname [10]
2 years ago
11

Two planets with the same mass and atmospheric conditions orbit a single star. Planet A is closer to the star than Planet B. Whi

ch description explains the expected average surface temperatures of Planets A and B?
Planet B is expected to be cooler than Planet A.

Planet B is expected to be hotter than Planet A.

Both planets are the same temperature.

Distance has no influence on a planet's average surface temperature.
Physics
1 answer:
8_murik_8 [283]2 years ago
3 0

Answer:

is not this one Planet B is expected to be hotter than Planet A.

Explanation:

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The specific heat of aluminum is greater.

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It lost the most heat.

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You toss a ball straight up with an initial speed of 30m/s. How high does it go, and how long is it in the air (neglecting air r
Brut [27]

Explanation:

Given that,

A ball is tossed straight up with an initial speed of 30 m/s

We need to find the height it will go and the time it takes in the air.

At its maximum height, its final speed, v = 0 and it will move under the action of gravity. Using equation of motion :

v = u +at

Here, a = -g

v = u -gt

i.e. u = gt

t=\dfrac{u}{g}\\\\t=\dfrac{30\ m/s}{9.8\ m/s^2}\\\\t=3.06\ s

So, the time for upward motion is 3.06 seconds. It means that it will in air for 3.06×2 = 6.12 seconds

Let d is the maximum distance covered by it.

d=ut-\dfrac{1}{2}gt^2

Putting all values

d=30(3.06)-\dfrac{1}{2}\times 9.8\times (3.06)^2\\\\d=45.91\ m

Hence, it will go to a height of 45.91 m and it will in the air for 6.12 seconds.

8 0
3 years ago
What happens to balloon filled with air when it goes very high attitude from surface of earth why​
vlada-n [284]

Answer:

The balloon will continue to expand and eventually burst.

Explanation:

Simply, the reason for this is because the density of the atmosphere decreases gradually as you increase in altitude closer to space. This means that the air on the outside of the balloon can't provide enough pressure over the surface of the balloon in order to counteract the gas on the inside of the balloon from expanding.

3 0
3 years ago
a projectile is launched at an angle of 30 degrees and lands later at the same level. if it's initial speed is 50 m/s, solve for
Mrrafil [7]
using \: the \: formula \\ t = \frac{2u \sin( \alpha ) }{g} where \: u = initial \: speed \: \\ \alpha = angle \: of \: projection \\ g = acceleration \: due \: to \: gravity \\ \frac{2 \times 50 \times \sin(30) }{10} \\ \frac{100 \times 0.5}{10} = \frac{50}{10} = 5seconds

Maximum height
= (Usinα)^2/2g
(50*0.5)^2/20
25^2/20
625/20
=31.25metres
horizontal distance = Range= [U^2 * sin2α]/g
[50^2 * sin60]/10
2500 * 0.8660/10
2165/10=216.5metres
3 0
3 years ago
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