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2 years ago

13

The same fluid flows through four different branching pipes. It enters each pipe from the left with the same speed, v0, and flow

s to the right. Assume that the fluid fills the pipes everywhere completely. You can also assume that all of these pipes are in the horizontal plane and that you are looking at them from above, so that there are no differences in height anywhere among the pipe branches. Finally, you can ignore viscosity. Which fluid speed is the largest?
All four speeds are the same.
a. v1
b. v4
c. v2
d. v3

1 answer:

Anvisha [2.4K]2 years ago

6 0

Answer:

The correct option is A.

Explanation:

Following the equation of continuum, AV remains constant.

Case a

(3A)(V0) = AV1 + AV1 + AV1

3AV0 = 3AV1

V1 = V0

Case b

(A)(V0) = (A/3)V2 + (A/3)V2 + (A/3)V2 + (A/3)V2

AV0 = 4V2/3

V2 = 3/4V0

Case c

(A/2)(V0) = AV3 + AV3 + AV3

AV0/2 = 3AV3

V3 = V0/6

Case d

(3A)(V0) = 2AV4 + 2AV4

3AV0 = 4AV4

V4 = 3V0/4

Comparing all the flow speeds, V1 is the largest.

Thus, the correct option is A.

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2 years ago

A pendulum is made of a small sphere of mass 0.250 kg attached to a lightweight string 1.20 m in length. As the pendulum swings

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Answer:v=2 m/s

Explanation:

Given

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A flywheel of mass M is rotating about a vertical axis with angular velocity ω0. A second flywheel of mass M/5 is not rotating a

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Answer:

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Explanation:

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let the final angular velocity of the system is ω'

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Moment of inertia of the second flywheel, I' = 0.5 x M/5 x R² = 0.1 MR²

use the conservation of angular momentum as no external torque is applied on the system.

I x ω = ( I + I') x ω'

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2 years ago

Suppose you first walk 12.0 m in a direction 20 owest of north and then 20.0 m in a direction 40.0osouth of west. How far are yo

alexgriva [62]

Answer:

R=19.5m

$\theta$ = 4.65° S of W

Explanation:

Refer the attached fig.

displacement of the x and y components

x-component displacement is ( $R_{x}$ ) = $A_{x}+B_{x}$

= A $\sin$ (20°) + B $\cos$ (40°)

= -12.0 $\sin$ (20°) + 20.0 $\cos$ (40°)

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x-component displacement is ( $R_{y}$ ) = $A_{y}+ B_{y}$

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resultant displacement

∴

R = $\sqrt{R_{x}^{2} +R_{y}^{2} } }$

= $\sqrt{(-19.425)^{2}+(-1.579)^{2} }$

=19.5m

$\theta$ = $\tan^{-1}\left | \frac{R_{x}}{R_{y}} \right |$

$\theta$ = $\tan^{-1}\left | \frac{1.579}{19.425} \right |$

$\theta$ = 4.65° S of W

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