Answer:
a) The module's acceleration in a vertical takeoff from the Moon will be 
b) Then we can say that a thrust of 
won't be able to lift off the module from the Earth because it's smaller than the module's weight (
).
Explanation:
a) During a vertical takeoff, the sum of the forces in the vertical axis will be equal to mass times the module's acceleration. In this this case, the thrust of the module's engines and the total module's weight are the only vertical forces. (In the Moon, the module's weight will be equal to its mass times the Moon's gravity acceleration)
Where:
thrust 
module's mass 
moon's gravity acceleration 
module's acceleration during takeoff
Then, we can find the acceleration like this:
The module's acceleration in a vertical takeoff from the Moon will be
b) To takeoff, the module's engines must generate a thrust bigger than the module's weight, which will be its mass times the Earth's gravity acceleration.
Then we can say that a thrust of won't be able to lift off the module from the Earth because it's smaller than the module's weight ().
Answer:
(a) 8 m/s
(b) 5 s
Explanation:
(a)
Using,
V² = U²+2gh ......................... Equation 1
Where V = final velocity, U = Initial velocity, g = acceleration due to gravity on the surface of the moon, h = height reached.
Given: V = 0 m/s ( At it's maximum height), g = -1.6 m/s² ( as its moves against gravity), h = 20 m.
Substitute into equation 1
0 = U²+[2×20×(-1.6)]
-U² = - 64
U² = 64
U = √64
U = 8 m/s.
(b)
V = U +gt.................... Equation 2
Where t = time to reach the maximum height.
Given: V = 0 m/s ( At the maximum height), g = -1.6 m/s² ( Moving against gravity), U = 8 m/s.
Substitute into equation 2
0 = 8+(-1.6t)
-8 = -1.6t
-1.6t = -8
t = -8/-1.6
t = 5 s.
Is that the full question?
Between the two asymptotic gigantic branches, the Sun changes the greatest in size, brightness, and temperature.
Beats only occur when the frequency of two objects is interfere with one another and when those objects are made to identical amplitudes together.
