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Crank
3 years ago
12

The average distance of the planet mercury from the sun is 0.39 times the average distance of the earth from the sun. How long i

s a year on mercury in units of earth years?
Physics
1 answer:
Stels [109]3 years ago
5 0

Answer:

T_1=0.24y

Explanation:

Using Kepler's third law, we can relate the orbital periods of the planets and their average distances from the Sun, as follows:

(\frac{T_1}{T_2})^2=(\frac{D_1}{D_2})^3

Where T_1 and T_2 are the orbital periods of Mercury and Earth respectively. We have D_1=0.39D_2 and T_2=1y. Replacing this and solving for

T_1^2=T_2^2(\frac{D_1}{D_2})^3\\T_1^2=(1y)^2(\frac{0.39D_2}{D_2})^3\\T_1^2=1y^2(0.39)^3\\T_1^2=0.059319y^2\\T_1=0.24y

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Describe a procedure that would increase the potential energy of two magnets if like poles are used. Explain why the energy of t
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Answer:

If you apply a force to separate 2 opposite poles, the potential energy of the system increases.

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2 years ago
A satellite circles the Earth in an orbit whose radius is twice the Earth’s radius. The Earth’s mass is 5.98 x 1024 kg, and its
gavmur [86]

Hello!

Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:

d = 2\pi r ← The circumference of the orbit

speed = orbital speed, we will solve for this later

time = period

Therefore:

T = \frac{2\pi r}{v}

Where 'r' is the orbital radius of the satellite.

First, let's solve for 'v' assuming a uniform orbit using the equation:
v = \sqrt{\frac{Gm}{r}}

G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)

m = mass of the earth (5.98 × 10²⁴ kg)

r = radius of orbit (1.276 × 10⁷ m)

Plug in the givens:
v = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{(1.276*10^7)}} = 5590.983 m/s

Now, we can solve for the period:

T = \frac{2\pi (1.276*10^7)}{5590.983} =\boxed{ 14339.776 s}

7 0
2 years ago
A box is being pulled to the right. What is the magnitude of the Kinect frictional force?
Anna35 [415]
The answer to this question is A - 25 N
3 0
3 years ago
Read 2 more answers
As a woman walks, her entire weight is momentarily placed on one heel of her high-heeled shoes. Calculate the pressure exerted o
STatiana [176]

Answer:

3335400 N/m² or 483.75889 lb/in²

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

A = Area = 1.5 cm²

m = Mass of woman = 51 kg

F = Force = mg

When we divide force by area we get pressure

P=\frac{F}{A}\\\Rightarrow P=\frac{mg}{A}\\\Rightarrow P=\frac{51\times 9.81}{1.5\times 10^{-4}}\\\Rightarrow P=3335400\ N/m^2

1\ N/m^2=\frac{1}{6894.757}\ lb/in^2

3335400\ N/m^2=3335400\times \frac{1}{6894.757}\ lb/in^2=483.75889\ lb/in^2

The pressure exerted on the floor is 3335400 N/m² or 483.75889 lb/in²

7 0
3 years ago
Identify evidence that supports the theories of continental drift and plate tectonics. put responses in the correct input to ans
GenaCL600 [577]

The evidence that supports continental drift and plate tectonics includes different fossils, the same rocks and the shapes of continents that fit together.

<h3>What is continental drift?</h3>

Continental drift is a theory that states continents once were part of one big landmass known as Pangea.

Nowadays, the theory of continental drift proposed by Alfred Wegener has been replaced by plate tectonics.

In conclusion, the evidence that supports continental drift and plate tectonics includes fossils, the same rocks and the shapes of continents that fit together.

Learn more on the continental drift here:

brainly.com/question/394265

#SPJ1

6 0
2 years ago
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