When placed at a certain point, a 4.20 C charge feels an electric force of 2.55 N. What is the magnitude of the electric field a
t that point?
(Remember, magnitudes are always positive.) (Unit = N/C)
(Remember, magnitudes are always positive.) (Unit = N/C)
2 answers:
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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:
a) 3.43 m/s
Explanation:
Due to the law of conservation of momentum, the total momentum of the bullet - rifle system must be conserved.
The total momentum before the bullet is shot is zero, because they are both at rest, so:
Instead the total momentum of the system after the shot is:
where:
m = 0.006 kg is the mass of the bullet
M = 1.4 kg is the mass of the rifle
v = 800 m/s is the velocity of the bullet
V is the recoil velocity of the rifle
The total momentum is conserved, therefore we can write:
Which means:
Solving for V, we can find the recoil velocity of the rifle:
where the negative sign indicates that the velocity is opposite to direction of the bullet: so the recoil speed is
a) 3.43 m/s