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3 years ago

Describe in short the working system of a water pump

1 answer:

5 0

The pump is powered by an electric motor that drives an impeller, or centrifugal pump. The impeller moves water, called drive water, from the well through a narrow orifice, or jet, mounted in the housing in front of the impeller.

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A sinusoidal wave travels along a string. The time for a particular point to move from maximum displacement to zero is 0.13 s. W

vitfil [10]

Answer:

Part a)

T = 0.52 s

Part b)

$f = 1.92 Hz$

Part c)

$speed = 3.65 m/s$

Explanation:

As we know that the particle move from its maximum displacement to its mean position in t = 0.13 s

so total time period of the particle is given as

$T = 4\times 0.13 = 0.52 s$

now we have

Part a)

T = time to complete one oscillation

so here it will move to and fro for one complete oscillation

so T = 0.52 s

Part b)

As we know that frequency and time period related to each other as

$f = \frac{1}{T}$

$f = \frac{1}{0.52}$

$f = 1.92 Hz$

Part c)

As we know that

wavelength = 1.9 m

frequency = 1.92 Hz

so wave speed is given as

$speed = wavelength \times frequency$

$speed = 1.92 \times 1.9$

$speed = 3.65 m/s$

4 0

3 years ago

) A satellite of mass m has an orbital period T when it is in a circular orbit of radius R around the earth. If the satellite in

Mrrafil [7]

Answer:

A) T.

Explanation:

Kepler's third law states that the orbital period (T) of a satellite is related with the radius (R) and the mass of the object (M) it orbits:

$T=\frac{2\pi R^{\frac{3}{2}}}{\sqrt{GM}}$

So the orbital period is independent of the mass of the satellite, that means no matter the mass every satellite at a radius R around the earth have an orbital period A.

4 0

2 years ago

A man stands on a platform that is rotating (without friction) with an angular speed of 1.0 rev/s; his arms are outstretched and

Airida [17]

Answer:

Explanation:

Given

$N_1=1 rev/s$

angular velocity $\omega =2\pi N_1=6.284 rad/s$

Combined moment of inertia of stool,student and bricks $=6\ kg.m^2$

Now student pull off his hands so as to increase its speed to suppose $N_2$ rev/s

$\omega _2=2\pi N_2$

After Pulling off hands so final moment of inertia is

$I_2=2\ kg-m^2$

Conserving angular momentum as no external torque is applied

$I_1\omega _1=I_2\omega _2$

$6\times 6.284=2\times \omega _2$

$\omega _2=18.85\ rad/s$

$N_2=3 rev/s$

7 0

3 years ago

A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal

Gemiola [76]

Answer:

Time taken, $T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}$

Explanation:

It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.

From the figure,

The sum of forces in y direction is :

$T\ cos\theta-mg=0$

$T=\dfrac{mg}{cos\theta}$

Sum of forces in x direction,

$T\ sin\theta=\dfrac{mv^2}{r}$

$mg\ tan\theta=\dfrac{mv^2}{r}$ .............(1)

Also, $r=l\ sin\theta$

Equation (1) becomes :

$mg\ tan\theta=\dfrac{mv^2}{l\ sin\theta}$

$v=\sqrt{gl\ tan\theta.sin\theta}$ ...............(2)

Let t is the time taken for the ball to rotate once around the axis. It is given by :

$T=\dfrac{2\pi r}{v}$

Put the value of T from equation (2) to the above expression:

$T=\dfrac{2\pi r}{\sqrt{gl\ tan\theta.sin\theta}}$

$T=\dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta.sin\theta}}$

On solving above equation :

$T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}$

Hence, this is the required solution.

4 0

2 years ago

What were some major reasons for the american television from analog broadcast to digital broadcast?

Vika [28.1K]

Answer: It frees up valuable portions of the broadcast spectrum, it has better audio and picture quality, and there are more options on digital broadcasting

Explanation:

6 0

3 years ago

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Describe in short the working system of a water pump ​

Describe in short the working system of a water pump