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>
Latitude, elevation, ocean currents, topography, and prevailing winds. There's probably a few others but these are the most important.
Increasing the number of bulbs in a series circuit decreases the brightness of the bulbs. In a series circuit, the voltage is equally distributed among all of the bulbs. Bulbs in parallel are brighter than bulbs in series. In a parallel circuit the voltage for each bulb is the same as the voltage in the circuit.
Answer:
The acceleration of man 1 and 2 is
and
.
Explanation:
Mass of man 1, m₁ = 80 kg
Mass of man 2, m₂ = 60 kg
One man pulls on the rope with a force of 250 N.
Let a₁ is acceleration of man 1,
F = m₁a₁
![a_1=\dfrac{F}{m_1}\\\\a_1=\dfrac{250}{80}\\\\a_1=3.125\ m/s^2](https://tex.z-dn.net/?f=a_1%3D%5Cdfrac%7BF%7D%7Bm_1%7D%5C%5C%5C%5Ca_1%3D%5Cdfrac%7B250%7D%7B80%7D%5C%5C%5C%5Ca_1%3D3.125%5C%20m%2Fs%5E2)
Let a₂ is acceleration of man 1,
F = m₂a₂
![a_2=\dfrac{F}{m_2}\\\\a_2=\dfrac{250}{60}\\\\a_2=4.167\ m/s^2](https://tex.z-dn.net/?f=a_2%3D%5Cdfrac%7BF%7D%7Bm_2%7D%5C%5C%5C%5Ca_2%3D%5Cdfrac%7B250%7D%7B60%7D%5C%5C%5C%5Ca_2%3D4.167%5C%20m%2Fs%5E2)
So, the acceleration of man 1 and 2 is
and
.