Answer:
w = 5832.372 Joules
Explanation:
Mass of water, m = 20 kg
The water was pulled up to a height of 35 meters, i.e. h = 35 m
It takes 14 minutes to pull up the water through the height, 35 m
speed = distance/ time = 35/14 = 2.5 m/min
The bucket's height, y = speed * time = 2.5t meters
6 kg of water drips out of the bucket throughout the 14 minutes
The rate at which the water drips drips out = (6/14) = 0.4286 kg/min
Mass of water that drips out in time, t = 0.4286t kg
The mass of water remaining = (20 - 0.4286t) kg
Change in Workdone, Δw = mgΔy
Δy = 2.5 Δt
Δw = mg * 2.5 Δt
dw = (20 - 0.4286t)g2.5 dt
integrating both sides
dw = (50g - 1.07gt)dt
where b = 0, a = 14
w = 50gt - 1.07g(t²)/2 g = 9.8 m/s²
w = 490t - 5.243t²
w = (490*14 - 5.243*14²) - (490*0 - 5.243*0²)
w = 6860 - 1027.628
w = 5832.372 Joules
The answer is D-Testable
Hope this helps
Answer:
C
Explanation:
- Let acceleration due to gravity @ massive planet be a = 30 m/s^2
- Let acceleration due to gravity @ earth be g = 30 m/s^2
Solution:
- The average time taken for the ball to cover a distance h from chin to ground with acceleration a on massive planet is:
t = v / a
t = v / 30
- The average time taken for the ball to cover a distance h from chin to ground with acceleration g on earth is:
t = v / g
t = v / 9.81
- Hence, we can see the average time taken by the ball on massive planet is less than that on earth to reach back to its initial position. Hence, option C
The period of the pendulum is given by the following equation
T = 2π * sqrt (L/g)
Where g is the gravity (free fall acceleration)
L is the longitude of the pendulum
T is the period.
We find g.............> (T /2π)^2 = L/g
g = L/(T /2π)^2...........> g = 22.657 m/s^2