Answer:
ΔP.E = 6.48 x 10⁸ J
Explanation:
First we need to calculate the acceleration due to gravity on the surface of moon:
g = GM/R²
where,
g = acceleration due to gravity on the surface of moon = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of moon = 7.36 x 10²² kg
R = Radius of Moon = 1740 km = 1.74 x 10⁶ m
Therefore,
g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²
g = 2.82 m/s²
now the change in gravitational potential energy of rocket is calculated by:
ΔP.E = mgΔh
where,
ΔP.E = Change in Gravitational Potential Energy = ?
m = mass of rocket = 1090 kg
Δh = altitude = 211 km = 2.11 x 10⁵ m
Therefore,
ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)
<u>ΔP.E = 6.48 x 10⁸ J</u>
Answer:
no sabo por q no mi entender
Given speed and the distance that must be covered, the time it will take the ultraviolet light to reach the earth is 3.7 × 10⁴ hours.
<h3>
What is Speed?</h3>
Speed is simply referred to as distance traveled per unit time.
Mathematically, Speed = Distance ÷ time.
Given the data in the question;
- Speed of the Ultraviolet light c = 3.0 × 10⁸m/s = 1.08 × 10⁹km/h
- Distance it must cover d = 4.0 × 10¹³km
We substitute our given values into the expression above.
Speed = Distance ÷ time
1.08 × 10⁹km/h = 4.0 × 10¹³km ÷ t
t = 4.0 × 10¹³km ÷ 1.08 × 10⁹km/h
t = 3.7 × 10⁴ hrs
Therefore, given speed and the distance that must be covered, the time it will take the ultraviolet light to reach the earth is 3.7 × 10⁴ hours.
Learn more about speed here: brainly.com/question/7359669
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The gravitation acceleration on the moon is different than on Earth. It is 1.6 m/s^2. If you weigh 120 lbs, then you would multiply 120 pounds by the gravitational acceleration on the moon and then divide by the acceleration on Earth.
(120 lbs * 1.6) / 9.8 = 20 pounds.
The mass will always be the same no matter what planet you’re on, so it’s still 54 kg.
Answer:
5.024 years
Explanation:
T1 = 1 year
r1 = 150 million km
r2 = 440 million km
let the period of asteroid orbit is T2.
Use Kepler's third law
T² ∝ r³
So,


T2 = 5.024 years
Thus, the period of the asteroid's orbit is 5.024 years.