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.
Applying the Newton Second Law of motion that is F=ma
We have F=395.2N (North) and m=259.1Kg
Putting these values in the equation
395.2 = 259.1 x a than
for a = 395.2/259.1
a = 1.525 m/sec 2
Answer:
The electric field inside the wire will remain the same or constant, while the drift velocity will by a factor of four.
Explanation:
Electron mobility, μ = 
where
= Drift velocity
E = Electric field
Given that the electric field strength = 1.48 V/m,
Therefore since the electric potential depends on the length of the wire and the attached potential difference, then when the electron mobility is increased 4 times the Electric field E will be the same but the drift velocity will increase four times. That is
4·μ = 
