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Andrews [41]
3 years ago
5

A bucket of mass M (when empty) initially at rest and containing a mass of water is being pulled up a well by a rope exerting a

steady force P. The water is leaking out of the bucket at a steady rate such that the bucket is empty after a time T. Find the velocity of the bucket at the instant it becomes empty. Express your answer in terms of P, M, m, T, and g, the acceleration due to avily. Constant Rate Leak"
Physics
1 answer:
Naily [24]3 years ago
6 0

Answer:

V=\dfrac{PT}{m}\ ln\dfrac{M+m}{M}-gT

Explanation:

Given that

Constant rate of leak =R

Mass at time T ,m=RT

At any time t

The mass = Rt

So the total mass in downward direction=(M+Rt)

Now force equation

(M+Rt) a =P- (M+Rt) g

a=\dfrac{P}{M+Rt}-g

We know that

a=\dfrac{dV}{dt}

\dfrac{dV}{dt}=\dfrac{P}{M+Rt}-g

\int_{0}^{V}V=\int_0^T \left(\dfrac{P}{M+Rt}-g\right)dt

V=\dfrac{P}{R}\ ln\dfrac{M+RT}{M}-gT

V=\dfrac{PT}{m}\ ln\dfrac{M+m}{M}-gT

This is the velocity of bucket at the instance when it become empty.

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6 0
3 years ago
A girl lifts a 160-N load to a height of 1 min 0.5 s. How much power is used to lift the load?
Aleonysh [2.5K]

Answer: 320 W

Explanation:

P = W/t

W = F * d

F = 160 N

d = 1 m

t = 0.5 s

W = 160 * 1

W = 160 J

P = 160 J / 0.5 s

P = 320 W

8 0
3 years ago
How does area affect the pressure?​
dybincka [34]

The smaller the area the greater the pressure, while the bigger the smaller the pressure. So they are inversely proportional

5 0
3 years ago
Martina has a sample of an unknown substance. She measures the substance. It’s mass is 13.5 grams, and it’s volume is 5 cm3 .Whi
Dominik [7]
Cm^3 is same as mL

13.5 g / 5 mL = 2.7 g/mL
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6 0
3 years ago
On a hot summer day in the state of Washington while kayaking, I saw several swimmers jump from a railroad bridge into the Snoho
Svetlanka [38]

Answer: part a: 19.62m

part b: 19.62 m/s

part a: 2.83 secs

Explanation:If the air resistance is ignored then the swimmer experience free fall under gravity hence

u=0

a=9.81 m/s2

t=2 secs

s=ut+0.5at^2

s=h

h=0*2+0.5*9.81*2^2\\h=19.62 meters

Part b

v=u+at\\v=0+9.81*2\\v=19.62m/s

Part c

now we have h=2*19.62=39.24

39.24=0+0.5*9.81*t^2\\t^2=8\\t=2.83 secs

4 0
3 years ago
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