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Andrews [41]
4 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]4 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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Complete question;

A plane circular loop of conducting wire of radius r-10 cm which possesses 15 turns is placed in a uniform magnetic field. The direction of the magnetic field makes an angle of 30° with respect to the normal direction to the loop. The magnetic field-strength is increased at a constant rate from IT to 5T in a time interval of 10 s.

a) What is the emf generated around the loop?

b) If the electrical resistance of the loop is 15Ω, what current flows around the loop as the magnetic field is increased?

Answer:

A) E.M.F generated around loop = 0.1632 Wb/s

B)Current in loop; I = 0.0109 A

Explanation:

A) We are given;

Initial magnetic field strength;B1 = 1T

Final magnetic field strength;B2 = 5T

Number of turns;N = 15 turns

Radius; r = 10cm = 0.1m

Angle;θ = 30°

Time interval; Δt = 10 s

Now, the formula for magnetic flux is: Φ = NABcosθ

Where;

N is number of turns

A is area = πr²

B is magnetic field strength

θ is angle

So, initial magnetic flux is;

Φ1 = NA(B1)cosθ

Plugging in the relevant values to obtain;

Φ1 = 15*(π*0.1²)(1)cos30

Φ1 = 0.4081 Wb

Similarly, final magnetic flux is;

Φ2 = NA(B2)cosθ

Plugging in the relevant values to obtain;

Φ2 = 15*(π*0.1²)(5)cos30

Φ2 = 2.0405 Wb

The time rate of change of the flux is;

dΦ_B/dt = (Φ2 - Φ1)/Δt

So, dΦ_B/dt = (2.0405 - 0.4081)/10

dΦ_B/dt = 0.1632 Wb/s

Thus, the emf generated around the loop is; E = dΦ_B/dt = 0.1632 Wb/s

B) from Ohm's law, the current which flows around the loop in response to the emf is given as;

I = E/R

We are given R =15Ω

Thus; I = 0.1632/15

I = 0.0109 A

θπΦ

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