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
Answer:
Solving for time :
(There are 4 formulas from linear motion. These formulas are very helpful as it allows us to prevent complicated calculations. Choose among the four that has : 1. The most constants known
2. The unknown constant that we want to solve)
s = (1/2)(u+v)t <--- one of the formulas
from linear motion
s (distance) = 0.05m
u (initial velocity) = 100m/s
v (final velocity) = 0 m/s (it stops)
t (time taken for change in velocity) = to be found
0.05 = (1/2)(100+0)t
t = 0.001 seconds
Solving for the resistant force :
Since the bullet hits the bag with an impulsive force and stops, the force that stops the bullet is the resistant force.
When the bullet stops :
F net = 0
F r = F imp
F r = (mu -mv)/t
F r = (0.01x100-0.01x0)/0.001
F r = 1/0.001
F r = 1000N
Answer:
No, just because the electric field is zero at a particular point, it does not necessarily mean that the electric potential is zero at that point. ... At the midpoint between the charges, the electric field due to the charges is zero, but the electric potential due to the charges at that same point is non-zero.
Explanation:
