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Contact [7]
3 years ago
11

An asteroid, whose mass is 3.6 × 10-4 times the mass of Earth, revolves in a circular orbit around the Sun at a distance that is

1.6 times the Earth's distance from the Sun. (a) Calculate the period of revolution of the asteroid in years. (b) What is the ratio of the kinetic energy of the asteroid to the kinetic energy of Earth?
Physics
1 answer:
frosja888 [35]3 years ago
7 0

Answer:

a) 2.02 years

b) 8.1 x 10⁻⁸.

Explanation:

Time period of a rotating body T² is proportional to radius of orbit R³ So

T₁² / T₂² = R₁³ /R₂³ ( T₁ and R₁ is time period and radius of orbit of the earth .)

1² / T₂² =( 1/1.6)³

T₂ = 2.02 years.

Kinetic energy of an orbiting body = 1/2 m v₀² ( v₀ is orbital speed)

= 1/2 m x 2 g R = m x G m/R² X R= m² x G /R

Kinetic energy of asteroid K₁ / kinetic energy of earth K₂ =

(mass of asteroid/mass of earth)² x( radius of earth / radius of asteroid)

=( 3.6 x 10⁻⁴)² x 1/1.6 = 8.1 x 10⁻⁸ .

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What is the study of kinematics based upon ?<br><br>thankyou ~​
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2 years ago
A wall has a negative charge distribution producing a uniformhorizontal electric field. A small plastic ball of mass .01kg carry
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Answer:

a)  E = -4 10² N / C , b) x = 0.093 m, c)     a = 10.31 m / s², θ=-71.9⁰

Explanation:

For that exercise we use Newton's second Law, in the attached we can see a free body diagram of the ball

X axis

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Axis y

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            Fe = T_{x}  

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Let us search with trigonometry the components of the tendency

            cos θ = T_{y} / T

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           T_{y} = cos θ

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                F_{e} = q E

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Let's calculate

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              θ = tan⁻¹ 0.3265

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The Angle meet him with trigonometry

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               θ = tan⁻¹ (-9.8) / 3.2

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professor190 [17]

Complete Question

The complete question is shown on the first uploaded image

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a

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Nostrana [21]

Answer:

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