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Andrews [41]
2 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]2 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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The intensity at a distance of 6.0 m from a source that is radiating equally in all directions is 6.0 × 10-10 w/m2 . what is the
satela [25.4K]
The intensity is defined as the ratio between the power emitted by the source and the area through which the power is calculated:
I= \frac{P}{A} (1)
where
P is the power
A is the area

In our problem, the intensity is I=6.0 \cdot 10^{-10} W/m^2. At a distance of r=6.0 m from the source, the area intercepted by the radiation (which propagates in all directions) is equal to the area of a sphere of radius r, so:
A=4 \pi r^2 = 4 \pi (6.0 m)^2 = 452.2 m^2

And so if we re-arrange (1) we find the power emitted by the source:
P=IA = (6.0 \cdot 10^{-10}W/m^2)(452.2 m^2)=2.7 \cdot 10^{-7} W
3 0
2 years ago
In 1993, Ileana Salvador of Italy walked 3.0 km in under 12.0 min. Suppose that during 115 m of her walk Salvador is observed to
dexar [7]

Answer:

t = 25 seconds

Explanation:

Given that,

Distance, d = 115 m

Initial speed, u = 4.2 m/s

Final speed, v = 5 m/s

We need to find the time taken in increasing the speed.

We know that,

Acceleration, a=\dfrac{v-u}{t} ....(1)

The third equation of kinematics is as follows :

v^2-u^2=2ad\\\\\text{Put the value of a in above equation}\\\\v^2-u^2=2\times \dfrac{v-u}{t}\times d\\\\\because (a^2-b^2)=(a-b)(a+b)\\\\(v-u)(v+u)=\dfrac{2\times (v-u)d}{t}\\\\t=\dfrac{2d}{v+u}\\\\\text{Putting all the values}\\\\t=\dfrac{2\times 115}{4.2+5}\\\\t=25\ s

Hence, it will take 25 seconds to increase the speed.

6 0
1 year ago
An electron moving at right angles to a 0.1 T magnetic field experiences an acceleration of 6 × 1015 m.s-2. What is the speed of
GaryK [48]

Explanation:

It is given that,

Magnetic field, B = 0.1 T

Acceleration, a=6\times 10^{15}\ m/s^2

Charge on electron, q=1.6\times 10^{-19}\ C    

Mass of electron, m=9.1\times 10^{-31}\ kg    

(a) The force acting on the electron when it is accelerated is, F = ma

The force acting on the electron when it is in magnetic field, F=qvB\ sin\theta

Here, \theta=90

So, ma=qvB

Where

v is the velocity of the electron

B is the magnetic field

v=\dfrac{ma}{qB}

v=\dfrac{9.1\times 10^{-31}\ kg\times 6\times 10^{15}\ m/s^2}{1.6\times 10^{-19}\ C\times 0.1\ T}

v = 341250  m/s

or

v=3.41\times 10^5\ m/s

So, the speed of the electron is 3.41\times 10^5\ m/s

(b) In 1 ns, the speed of the electron remains the same as the force is perpendicular to the cross product of velocity and the magnetic field.

7 0
2 years ago
Two workhorses tow a barge along a straight canal. Each horse exerts a constant force of magnitude F, and the tow ropes make an
morpeh [17]

Answer:

<em>a) Fvt cosθ</em>

<em>b) Fv cosθ</em>

<em></em>

Explanation:

Each horse exerts a force = F

the rope is inclined at an angle = θ

speed of each horse = v

a) In time t, the distance traveled d = speed x time

i.e d = v x t = vt

also, the resultant force = F cosθ

Work done W = force x distance

W = F cosθ x vt = <em>Fvt cosθ</em>

<em></em>

b) Power provided by the horse P = force x speed

P = F cosθ x v

P = <em>Fv cosθ</em>

8 0
1 year ago
Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop safely. A truck enters a
Furkat [3]

Answer:

bfghhg

Explanation:

6 0
2 years ago
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