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2 years ago

what would happen to the temperature on mercury during the day if a sunspot was directly in line with it's surface?

Physics

1 answer:

QveST [7]2 years ago

3 0

Answer:

mind-bogglingly hot

Explanation:

If a sunspot were pointed right at Mercury in this way, the planet would become even more mind-bogglingly hot. Sunspots look dark in our visual spectrum, so we might suspect that they are colder (or at least less hot) than the rest of the Sun; but in fact, they are hotter.

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Stars combine Hydrogen to make Helium during nuclear fusion. Living things are made of heavier elements like Carbon, Oxygen, Iro

alex41 [277]

They were formed in the nuclear<span> fusion reaction inside older </span><span>stars.

As a star burns, fusion reactions inside its core create heavier elements. Those materials are released when the star dies of old age in an explosion.</span>

8 0

3 years ago

Help plzzzz thank youu

dybincka [34]

The answer is B I hope this helps luv

5 0

3 years ago

light travels approximately 982,080,000 ft/s, and one year has approximately 32,000,000 seconds. A light year is the distance li

lapo4ka [179]

Answer:

The distance traveled in 1 year is: $3.143*10^{16}ft$

Explanation:

Given

$s = 982,080,000 ft/s$ --- speed

$t = 32,000,000 s$ --- time

Required

The distance traveled

This is calculated as:

$Speed = \frac{Distance}{Time}$

So, we have:

$Distance = Speed * Time$

This gives:

$Distance = 982,080,000 ft/s * 32,000,000 s$

$Distance = 982,080,000 * 32,000,000ft$

$Distance = 3.143*10^{16}ft$ -- approximated

5 0

2 years ago

An open 1-m-diameter tank contains water at a depth of 0.5 m when at rest. As the tank is rotated about its vertical axis the ce

Morgarella [4.7K]

Answer:

Angular velocity (w) = 8.86 rad/s

Explanation:

Angular velocity (w) = $\sqrt{} 4ghi/R^{2}$

g= 9.81 m/s

R= 0.5

hi (initial depth) = 0.5m

Hence= $\sqrt4* 9.81* 0.5/0.5^{2}$ = 8.86 rad/s

3 0

3 years ago

A 24 kg crate is moving at a constant speed because it is being pushed with a force of 53n. What would be the coefficient of kin

____ [38]

Answer:

The coefficient of kinetic friction between the crate and the floor can be calculated using the formula μ = Ff / N, where Ff is the frictional force, N is the normal force, and μ is the coefficient of kinetic friction.

In this case, the normal force is equal to the weight of the crate, which is 24 kg * 9.8 m/s2 = 235.2 N. The frictional force can be calculated using the formula Ff = μ * N, where μ is the coefficient of kinetic friction and N is the normal force.

If we substitute the values for N and Ff into the formula for the coefficient of kinetic friction, we get:μ = 53 N / 235.2 N = 0.225

Therefore, the coefficient of kinetic friction between the crate and the floor is 0.225.

6 0

1 year ago