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Dahasolnce [82]

3 years ago

12

a fan is rotating clockwise and its acceleration has a positive sign. is the angular velocity of the fan speeding up, slowing do

wn, or constant?

1 answer:

balu736 [363]3 years ago

5 0

Answer:

The angular velocity is slowing down.

Explanation:

By convention, if a rigid body is rotating clockwise, the angular velocity is negative.
If the angular acceleration has a positive sign, since the angular acceleration and the angular velocity have opposite signs, this means that the angular velocity is slowing down.

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Explanation:

Please see the attached file.

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What is the pressure drop due to the bernoulli effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire

Paul [167]

Answer:

$\Delta P=1581357.92\ Pa$

Explanation:

Given:

diameter of hose pipe, $D=0.09\ m$

diameter of nozzle, $d=0.03\ m$

volume flow rate, $\dot{V}=40\ L.s^{-1}=0.04\ m^3.s^{-1}$

<u>Now, flow velocity in hose:</u>

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$v_h=\frac{0.04\times 4}{\pi\times 0.09^2}$

$v_h=6.2876\ m.s^{-1}$

<u>Now, flow velocity in nozzle:</u>

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$v_n=\frac{0.04\times 4}{\pi\times 0.03^2}$

$v_n=56.5884\ m.s^{-1}$

We know the Bernoulli's equation:

$\frac{P_1}{\rho.g}+\frac{v_1^2}{2g}+Z_1=\frac{P_2}{\rho.g}+\frac{v_2^2}{2g}+Z_2$

when the two points are at same height then the eq. becomes

$\frac{P_1}{\rho.g}+\frac{v_1^2}{2g}=\frac{P_2}{\rho.g}+\frac{v_2^2}{2g}$

$\Delta P=\frac{\rho(v_n^2-v_h^2)}{2}$

$\Delta P=\frac{1000(56.5884^2-6.2876^2)}{2}$

$\Delta P=1581357.92\ Pa$

8 0

3 years ago

Please need help this

snow_tiger [21]

Melting and pressure.

Weathering and erosion affect any surface rock and make sedimentary rocks, heat and pressure make metamorphic rocks, and melting and cooling make igneous rocks. Melting and pressure is not a step.

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3 years ago