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qaws [65]
3 years ago
15

According to Bernoulli's fluid formula a An increase in the speed will lower the internal pressure b An increase in the speed wi

ll increase the internal pressure c The internal pressure doesn't change with the speed
Physics
1 answer:
lorasvet [3.4K]3 years ago
5 0

Answer:

a An increase in the speed will lower the internal pressure

Explanation:

Bernoulli's fluid formula

P_1+\frac{1}{2}\rho v_1^2+\rho gh_1=P_1+\frac{1}{2}\rho v_2^2+\rho gh_2

where

P = Pressure

ρ = Density of fluid

g = Acceleration due to gravity

h = Height

v = Velocity of fluid

If there is no change in height then we get

P_1+\frac{1}{2}\rho v_1^2=P_1+\frac{1}{2}\rho v_2^2\\\Rightarrow P+\frac{1}{2}\rho v^2=constant

According to the Bernoulli's principle when the speed of the fluid is larger in a region of streamline flow the pressure is smaller in that region. From the above equation it can be seen that increase in speed should simultaneously reduce pressure in order for their sum to be constant.

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You are standing at a subway platform when a subway passes, sounding its whistle. If you observe the whistle to be at 12750Hz wh
adell [148]

As per doppler's Effect of sound we can say when subway is approaching the platform we will have

f_1 = f_o* \frac{v}{v- v_s}

12750 = f_o* \frac{340}{340 - v_s}

Similarly we can find the frequency when subway is passing away with same speed from us

f_2 = f_o* \frac{v}{v + v_s}

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5 0
3 years ago
The angular momentum of a flywheel having a rotational inertia of 0.142 kg·m2 about its central axis decreases from 7.20 to 0.14
luda_lava [24]

Answer:

3.3619 Nm

54.27472 rad

182.46618 J

86.88 W

Explanation:

L_i = Initial angular momentum = 7.2 kgm²/s

L_f = Final angular momentum = 0.14 kgm²/s

I = Moment of inertia = 0.142 kgm²

t = Time taken

Average torque is given by

\tau_{av}=\frac{L_f-L_i}{\Delta t}\\\Rightarrow \tau_{av}=\frac{0.14-7.2}{2.1}\\\Rightarrow \tau_{av}=-3.3619\ Nm

Magnitude of the average torque acting on the flywheel is 3.3619 Nm

Angular speed is given by

\omega_i=\frac{L_i}{I}

Angular acceleration is given by

\alpha=\frac{\tau}{I}

From the equation of rotational motion

\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=\frac{L_i}{I}\times t+\frac{1}{2}\times \frac{\tau}{I}\times t^2\\\Rightarrow \theta=\frac{7.2}{0.142}\times 2.1+\frac{1}{2}\times \frac{-3.3619}{0.142}\times 2.1^2\\\Rightarrow \theta=54.27472\ rad

The angle the flywheel turns is 54.27472 rad

Work done is given by

W=\tau\theta\\\Rightarrow W=-3.3619\times 54.27472\\\Rightarrow W=-182.46618\ J

Work done on the wheel is 182.46618 J

Power is given by

P=\frac{W}{t}\\\Rightarrow P=\frac{-182.46618}{2.1}\\\Rightarrow P=-86.88\ W

The magnitude of the average power done on the flywheel is 86.88 W

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in the attachment

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