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.
Answer:
v = 11 m/s is her final speed
Explanation:
work done by gravity = m g Δh = 40×9.8×10 = 3920 Joules
Work done by friction = - force×distance = - 20×100 = - 2000 Joules
[minus sign because friction force is opposite to the direction of motion]
Initial K.E. = (1/2) m u^2 = (1/2) × 40 × 5^2 = 500 Joules
Now, by work energy theorem
Work done = change in kinetic energy.
Final K.E. = initial K.E. + total work = 500 + 3920 - 2000 = 2420 Joules
Now, Final K.E. = (1/2) m v^2 [final speed being v= speed at the bottom]
⇒ 2420 = (1/2)×40×v^2
⇒ 121 = v^ 2
v = 11 m/s is her final speed
Answer:
Light provides brightness to see and also light carries energy
