A light spring obeys Hooke's law. The spring's unstretched length is 34.0 cm. One end of the spring is attached to the top of a
1 answer:
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Answer:
J of energy is available through fusion .
Explanation:
let us consider weight of the human body on an average be 70 kg .
since 0.8% of mass is used to convert into energy =
since according to einstein mass energy equation
The angular velocity depends on the length of the orbit and the orbital
speed of the telescope.
Response:
First question:
- The angular velocity of the telescope is approximately <u>0.199 rad/s</u>
Second question:
- The telescope should accelerates away by approximately F = <u>0.0005·m </u>
Third question:
- <u>The pulling force between the Earth and the satellite</u>
The distance of the James Webb telescope from the Sun = 1.5 million kilometers from Earth on the side facing away from the Sun
The orbital velocity of the telescope = The Earth's orbital velocity
First question:
The orbital velocity of the Earth = 29.8 km/s
The distance between the Earth and the Sun = 148.27 million km
The radius of the orbit of the telescope = 148.27 + 1.5 = 149.77
Radius of the orbit, r = 149.77 million kilometer from the Sun
The length of the orbit of the James Webb telescope = 2 × π × r
Which gives;
r = 2 × π × 149.77 million kilometers ≈ 941.03 million kilometers
Therefore;
- The angular velocity of the telescope, ω ≈ <u>0.199 rad/s</u>
Second question:
Centrifugal force force, = m·ω²·r
Which gives;
Universal gravitational constant, G = 6.67408 × 10⁻¹¹ m³·kg⁻¹·s⁻²
Mass of the Sun = 1.989 × 10³⁰ kg
Which gives;
Which gives;
<
, therefore, the James Webb telescope has to accelerate away from the Sun
F = -
The amount by which the telescope accelerates away is approximately 0.00592·m - 0.0054233·m ≈ <u>0.0005·m (away from the Sun)</u>
Third part:
Other forces include;
- <u>The force of attraction between the Earth and the telescope </u>which can contribute to the the telescope having a stable orbit at the given speed.
Learn more about orbital motion here: