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2 years ago

Describe the procedure of using pani ghatt

Physics

1 answer:

Ahat [919]2 years ago

5 0

Answer:

The water from head of 20m or above is brought in open or closed conduit. Traditionally wooden blades are used as turbine on which water jets are strike upon, jets rotates the bigger wheel mounted on smaller wheel. There is continuos feed of grains in between grinds into finer particles.

You might be interested in

The key difference between the binomial and hypergeometric distribution is that

Ivenika [448]

Explanation:

Both distributions describe the number of times an event occurs in a givn number of trials. In the binomial distribution, the probability is the same for each trial. While in the hypergeometric distribution, each trial changes the probability of each subsequent trial, since there is no replacement.

5 0

3 years ago

The primary difference between a barometer and a manometer is

Nitella [24]

Answer:

a barometer is used to measure atmospheric pressure, and a manometer is used to measure gauge pressure.

Explanation:

A barometer measures air pressure at any locality with sea level as the reference.

However, a manometer is used to measure all pressures especially gauge pressures. Thus, if the aim is to measure the pressure at any point below a fluid surface, a barometer is used to determine the air pressure. The manometer may now be used to determine the gauge pressure

The algebraic sum of these two values gives the absolute pressure.

4 0

3 years ago

The tape in a videotape cassette has a total

xxMikexx [17]

Answer:

$w=19.76 \ rad/s$

Explanation:

<u>Circular Motion </u>

Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:

$w=w_o+\alpha \ t$

Where $\alpha$ is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:

$\displaystyle v_t=w.r$

We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call $w_1$ the angular speed of the loaded reel and $w_2$ that from the empty reel. We have

$w_1=w_{o1}+\alpha_1 \ t$

$w_2=w_{o2}+\alpha_2 \ t$

If $r_f=35\ mm=0.035\ m$ is the radius of the reel when it's full of tape, the angular speed for the loaded reel is computed by

$\displaystyle w_{01}=\frac{v_t}{r_f}$

The tangent speed is computed by knowing the length of the tape and the time needed to fully play it.

$t=1.8\ h=1.8*3600=6480\ sec$

$\displaystyle v_t=\frac{x}{t}=\frac{249\ m}{6480\ sec}=0.0384 \ m/s$

$\displaystyle w_{01}=\frac{0.0384}{0.035}=1,098 \ rad/s$

If $r_e=10\ mm=0.001\ m$ is the radius of the reel when it's empty, the angular speed for the empty reel is computed by

$\displaystyle w_{02}=\frac{0.0384}{0.001}=38.426 \ rad/s$

The full reel goes from $w_{01}$ to $w_{02}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_1=\frac{w_{02}-w_{01}}{6480}$

$\displaystyle \alpha_1=\frac{38.426-1.098}{6480}=0.00576 \ rad/sec^2$

The empty reel goes from $w_{02}$ to $w_{01}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_2=\frac{1.098-38.426}{6480}=-0.00576 \ rad/sec^2$

So the equations for both reels are

$w_1=1.098+0.00576 \ t$

$w_2=38.426-0.00576 \ t$

They will be the same when

$1.098+0.00576 \ t=38.426-0.00576 \ t$

Solving for t

$\displaystyle t=\frac{38.426-1.098}{0.0115}$

$t=3240 \ sec$

The common angular speed is

$w_1=1.098+0.00576 \ 3240=19.76 \ rad/s$

$w_2=38.426-0.00576 \ 3240=19.76 \ rad/s$

They both result in the same, as expected

$\boxed{w=19.76 \ rad/s}$

6 0

3 years ago

How much force is needed to accelerate a 15kg bowling ball at 2 m/s^2

olga_2 [115]

Golf no doing what I iuroeurir to do my homework and my homework

7 0

3 years ago

Read 2 more answers

two punds of water vapor at 30 psia fill the 4ft3 left chmaber of a partitioned system. The right chmaber has twice the volume o

tamaranim1 [39]

Answer:

pressure of water will be 49.7 atm

Explanation:

given data

pressure = 30 psi = 2.04 atm

water = 2 pound = 907.18

mole of water vapor = 907.19 /2 = 50.4 mole

volume = 4 ft³ = 113.2 L

temperature = 40 F = 277.59 K

to find out

pressure of water

solution

we will apply here ideal gas condition

that is

PV = nRT .......................1

put here all value and here R = 0.0821 , T temperature and V volume and P pressure and n is no of mole

and we get here temperature

PV = nRT

2.04 × 113.2 = 50.4×0.0821×T

solve it and we get

T = 55.8 K

so we have given right chamber has twice the volume of the left chamber i.e

volume = twice of volume + volume

volume = 2(113.2) + 113.2

volume = 339.6 L

so from equation 1 pressure will be

PV = nRT

P(339.6) = 50.4 × ( 0.0821) × (277.59)

P = 3.3822 atm = 49.7 atm

so pressure of water will be 49.7 atm

7 0

3 years ago