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Rina8888 [55]
3 years ago
14

How many earth years does it take mercury to go around the sun?

Physics
2 answers:
Bad White [126]3 years ago
5 0
<h2>Answer:</h2><h3>0.24 Earth years</h3><h3 /><h2>Explanation:</h2>

However, it's proximity to the Sun means that its average orbital velocity is a speedy 47.362 kilometers a second or 29.429 miles per second – approximately 170,500 km/h; 105,945 mph. At this rate, it takes Mercury 87.969 days, or the equivalent of 0.24 Earth years, to complete a single orbit of the Sun.

sertanlavr [38]3 years ago
5 0
88 days to orbit length of day 58d 15h 30 m
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When a boy throws a ball and accidentally breaks a window, the momentum of the ball and all the pieces of glass taken together after the collision is THE SAME as the momentum of the ball before the collision

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Which of the following represents a possible magnitude for the force of static friction when Xavier applied 72.1 Newtons of forc
lana66690 [7]

The possible magnitude for the force of static friction on the stationary cart is 72.1 N.

The given parameters:

  • <em>Applied force on the cart, F = 72.1 N</em>

<em />

Based on Newton's second law of motion, the force applied to object is directly proportional to the product of mass and acceleration of the object.

F = ma

Static frictional force is the force resisting the motion of an object at rest.

\Sigma F = 0\\\\F -F_f = 0

where;

F_f is the frictional force

F= F_f \\\\72.1 = F_f\\\\F_f = 72.1\  N

Thus, the possible magnitude for the force of static friction on the stationary cart is 72.1 N.

Learn more about Newton's second law of motion: brainly.com/question/25307325

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2 years ago
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Which of the following is a subsurface event takes place during the rock cycle
alukav5142 [94]

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On a hot summer day, 3.50 ✕ 106 J of heat transfer into a parked car takes place, increasing its temperature from 36.5°C to 44.4
anygoal [31]

Answer:

a) \Delta s=443037.9747\ J.K^{-1}

b) \Delta s=31868131.8681\ J.K^{-1}

Explanation:

a)

Given:

amount of heat transfer occurred,dQ=3.5\times 10^6\ J

initial temperature of car, T_i=36.5+273=309.5\ K

final temperature of car, T_f=44.4+273=317.4\ K

We know that the change in entropy is given by:

\Delta s=\frac{dQ}{T_f-T_i}

\Delta s=\frac{3.5\times 10^6}{(44.4-36.5)} (heat is transferred into the system of car)

\Delta s=443037.9747\ J.K^{-1}

b)

amount of heat transfer form the system of house, dQ=5.8\times 10^8\ J

initial temperature of house, T_i=23.5+273=296.5\ K

final temperature of house, T_f=5.3+273=278.3\ K

\Delta s=\frac{dQ}{T_f-T_i}

\Delta s=\frac{5.8\times 10^8}{278.3-296.5}

\Delta s=31868131.8681\ J.K^{-1}

6 0
3 years ago
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