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umka21 [38]
3 years ago
5

n a distant solar system, a planet of mass 5.0 x 1024 kg orbits a sun of mass 3.0 x 1030 kg at a constant distance of 2.0 x1011

m. How many earth days does it take for the planet ot execute one complete orbit about the sun
Physics
1 answer:
Anna35 [415]3 years ago
4 0

Answer:

F = M2 ω^2 R       centripetal force of sun on planet

ω = (F / (M2 R))^1/2 = 2 pi f = 2 pi / P        where P is the period

P = 2 pi (M2 * R / F)^1/2

F = G M1 M2 / R^2        gravitational force on planet

P = 2 pi {R^3 / (G M1)]^1/2

P = 6.28 [(2.0E11)^3 / (6.67E-11 * 3.0E30)]^1/2

P = 6.28 (8 / 20)^1/2 E7 = 3.9E7 sec

1 yr = 3600 * 24 * 365 = 3.15E7 sec

P = 3.9 / 3.2 = 1.2 years

You might be interested in
Kinetic Energy RE-
tatiyna

Answer:

1) The potential energy, P.E  = 4527.6 J

2) The mass, m = 1.28 kg

3) The height of the bell, h = 4.31 m

4)  The kinetic energy, K.E = 385 J

5) The velocity of the car, V = 22.53 m/s

6) The mass of the car,  m = 17.59 kg

Explanation:

1) The height of the hill, h = 21 m

The carriage with a baby has mass of, m = 22 kg

The potential energy,

                             P.E = mgh

                                   = 22 x 9.8 x 21

                                  = 4527.6 J

2) The height of the platform, h = 20 m

The P.E of the cinder block, P.E = 250 J

The mass,

                                m = P.E / gh

                                     = 250 / (9.8 x 20)

                                    = 1.28 kg

3) The mass of the bell, m = 18 kg

The P.E of the bell, P.E = 760 J

The height of the bell,

                            h = P.E/mg

                               = 760 / 18 x 9.8

                               = 4.31 m

4) The mass of the runner, m = 55 kg

The velocity of the runner, v = 14 m/s

The kinetic energy,

                              K.E = ½ mv²

                                     = ½ x 55 x 14

                                     = 385 J

5)The kinetic energy of the car, K.E = 69,759 J

The mass of the car, m = 275 kg

The velocity of the car,

                                 V = √(2K.E/m)

                                  V = √( 2 x 69759 / 275)

                                      = 22.53 m/s

6) The velocity of the car, v = 38 m/s

The kinetic energy of the car, K.E = 12700 J

The mass of the car,

                                m = 2 K.E/ v²

                                     = 2 x 12700 / 38²      

                                 m = 17.59 kg

8 0
3 years ago
To make them gas, you have to give liquids yes or no?
jeka94

Answer:

Yes

Explanation:

6 0
3 years ago
Read 2 more answers
To a stationary observer, a man jogs east at 2.5 m/s and a woman jogs west at 1.5 m/s. from the woman's frame of reference, what
neonofarm [45]
T o a stationary observer, a man jogs east at 2.5 m/s and a woman jogs west at 1.5 m/s. from the woman's frame of reference, what is the man's velocity? it is 4m/s east
5 0
3 years ago
Read 2 more answers
A grandfather clock works by swinging a pendulum back and forth with a
allochka39001 [22]

Answer:

For example, a wave with a time period of 2 seconds has a frequency of

1 ÷ 2 = 0.5 Hz.

Explanation:

8 0
3 years ago
"The drawing shows three layers of different materials, with air above and below the layers. The interfaces between the layers a
My name is Ann [436]

Answer:

Angle of incidence that entered material b= 63.1°

Angle of incidence between a and b = 55.9°

Explanation: Using the formular:

n1sintheta1= n2sintheta2

The light ray which enters material B will be

1.4Sin72.8° = 1.5Sin theta

1.3373= 1.5Sintheta

sintheta = 1.3373/1.5

Sin^-1 0.8916 = Theta

63.1 = theta

When the ray hits interface with material a

1.5Sin63.1 = 1.3 Sin theta

1.3374 = 1.3Sin theta

Sintheta= 1.3374/1.3

Sin theta = 1.0877

There will be total reflection off the boundary b c because sin theta exceeded 1 in value.

The equation should be

1.4sin63.1 = 1.4 sin theta

Sin theta=72.8°

When the ray hits air-c boundary:

1.4sin72.8=1.00sin theta

Sin theta=1.3374/1 =1.3374

There is total reflection.

In material a,the ray will:

1.3sin72.8° = 1.00sin theta

There will be total reflection when the ray hits a-b boundary.

1.3sin72.8= 1.5sintheta

Sin theta= 1.2419/ 1.5

Sin theta =0.8279

Theta= Sin^-10.8279= 55.88°

When ray hits c-air boundary

1.4sin63.1= 1.00sintheta

1.2485= sin theta = Toal reflection.

Therefore when the ray of light pass through the layers of material a, b and c the boundary with air on top and bottom will be total reflection.

7 0
3 years ago
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