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AysviL [449]
3 years ago
10

According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about

_________ year(s).
Physics
1 answer:
NARA [144]3 years ago
6 0

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, r=4\ AU=1.496\times 10^{11}\ m

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

T^2=\dfrac{4\pi^2}{GM}r^3

M is the mass of the sun

T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3    

T^2=\sqrt{9.96\times 10^{14}}\ s

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

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Answer:

Explanation:

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8 0
3 years ago
National Instant Criminal Background Check System is used for a person trying to access a government website.
Paladinen [302]

The given statement "National Instant Criminal Background Check System is used for a person trying to access a government website" is false.

Answer: Option B

<u>Explanation:</u>

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4 0
4 years ago
The intensity of sunlight reaching the earth is 1360 w/m2. Assuming all the sunligh is absorbed, what is the radiation pressure
krok68 [10]

(a) 1.15\cdot 10^9 N

For an electromagnetic wave incident on a surface, the radiation pressure is given by (assuming all the radiation is absorbed)

p=\frac{I}{c}

where

I is the intensity

c is the speed of light

In this problem, I=1360 W/m^2; substituting this value, we find the radiation pressure:

p=\frac{1360 W/m^2}{3\cdot 10^8 m/s}=4.5\cdot 10^{-6} Pa

the force exerted on the Earth depends on the surface considered. Assuming that the sunlight hits half of the Earth's surface (the half illuminated by the Sun), we have to consider the area of a hemisphere, which is

A=2pi R^2

where

R=6.37\cdot 10^6 m

is the Earth's radius. Substituting,

A=2\pi (6.37\cdot 10^6 m)^2=2.55\cdot 10^{14}m^2

And so the force exerted by the sunlight is

F=pA=(4.5\cdot 10^{-6} Pa)(2.55\cdot 10^{14} m^2)=1.15\cdot 10^9 N

(b) 3.2\cdot 10^{-14}

The gravitational force exerted by the Sun on the Earth is

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M=1.99\cdot 10^{30}kg is the Sun's mass

m=5.98\cdot 10^{24}kg is the Earth's mass

r=1.49\cdot 10^{11} m is the distance between the Sun and the Earth

Substituting,

F=(6.67\cdot 10^{-11} )\frac{(1.99\cdot 10^{30}kg)(5.98\cdot 10^{24} kg)}{(1.49\cdot 10^{11} m)^2}=3.58\cdot 10^{22}N

And so, the radiation pressure force on Earth as a fraction of the sun's gravitational force on Earth is

\frac{1.15\cdot 10^9 N}{3.58\cdot 10^{22}N}=3.2\cdot 10^{-14}

6 0
4 years ago
What is the cheetah's average speed for the first 3.11 s of its sprint? for the second 3.11 s of its sprint?
photoshop1234 [79]
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5 0
3 years ago
Two point charges are separated by a certain distance. How does the strength of the electric field produced by the first charge,
olga55 [171]

Answer:

The field will remain the same

Explanation:

This is because electric field given as

E1= kq1/r²

And that of second charge

E² = kq2/r²

Is not affected by the size of the second charge q2

8 0
3 years ago
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