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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.

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A hot-air balloon with a mass of 400 kilograms moves across the sky with 3,200 joules of kinetic energy. The velocity of the bal

kotegsom [21]

Answer:

4 m/s

Explanation:

KE =

Velocity of balloon will be 4 m/s.

!! Hope It Helps !!

4 0

3 years ago

A Ping-Pong ball moving East at a speed of 4 m/s collides with a stationary bowling ball. The Ping-Pong ball bounces back to the

In-s [12.5K]

Answer:

They experience the same magnitude impulse

Explanation:

We have a ping-pong ball colliding with a stationary bowling ball. According to the law of conservation of momentum, we have that the total momentum before and after the collision must be conserved:

$p_i = p_f\\p_p + p_b = p'_p+p'_b$

where

$p_p$ is the initial momentum of the ping-poll ball

$p_b$ is the initial momentum of the bowling ball (which is zero, since the ball is stationary)

$p'_p$ is the final momentum of the ping-poll ball

$p'_f$ is the final momentum of the bowling ball

We can re-arrange the equation as follows

$p_p - p'_p = p_b'-p_b$

$-\Delta p_p = \Delta p_b$

which means

$|\Delta p_p | = |\Delta p_b|$ (1)

so the magnitude of the change in momentum of the ping-pong ball is equal to the magnitude of the change in momentum of the bowling ball.

However, we also know that the magnitude of the impulse on an object is equal to the change of momentum of the object:

$I=\Delta p$ (2)

Therefore, (1)+(2) tells us that the ping-pong ball and the bowling ball experiences the same magnitude impulse:

$|I_p| = |I_b|$

4 0

4 years ago

Which statement is true about the turbine in a hydroelectric dam? a. It is turned by wind and provides a direct current. b. It i

tresset_1 [31]

B) It is turned by water to produce a direct current....... hope it helps :)

8 0

3 years ago

A small steel wire of diameter 1.0mm is connected to an oscillator and is under a tension of 5.7N . The frequency of the oscilla

kotegsom [21]

Answer:

The power output of the oscillator is 0.350 watt.

Explanation:

Given that,

Diameter = 1.0 mm

Tension = 5.7 N

Frequency = 57.0 Hz

Amplitude = 0.54 cm

We need to calculate the power output of the oscillator

Using formula of the power

$P=\dfrac{1}{2}\times\mu\times\omega^2\times a^2\times v$

Put the value into the formula

$P=\dfrac{1}{2}\times A\times\rho\times\omega^2\times a^2\times\dfrac{\sqrt{T}}{\mu}$

$P=\dfrac{1}{2}\times3.14\times(0.0005)^2\times7850\times(2\times\pi\times57.0)^2\times(0.54\times10^{-2})^2\times\sqrt{\dfrac{5.7}{3.14\times(0.0005)^2\times7850}}$

$P=0.350\ Watt$

Hence, The power output of the oscillator is 0.350 watt.

3 0

3 years ago

Advancements in the area of agriculture has started to depend on technology sources for better yield and technology advancements

Ksenya-84 [330]

I don't know if my answer is right or not but on the test I picked ( D)

3 0

4 years ago