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3 years ago

(BRAINLIEST W/ WORK SHOWN!!!) Help with Physics question?

Physics

1 answer:

VashaNatasha [74]3 years ago

6 0

Answer:

$\boxed {3.43 x 10^{3}}$

Explanation:

We know that speed is defined as distance moved per unit time hence expressed as $v=\frac {d}{t}$ where v is speed in m/s, d is distance in m and t is time in seconds. Making d the subject of the above formula then

$d=vt$

Substituting 343 m/s for d and 10 s for t then

$d= 343\times10= 3430= 3.43 x 10^{3}$

Therefore, the distance between speaker and deter is $\boxed {3.43 x 10^{3}}$

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3 years ago

Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp

slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

$Q=Av$

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

$Q=1.20 m^3/s$

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

$A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2$

So we can re-arrange the equation to find the speed of the water:

$v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s$

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

$Q_1 = Q_2\\A_1 v_1 = A_2 v_2$

where we have

$A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s$

and where $A_2$ is the cross-sectional area of the pipe at the second point.

Solving for A2,

$A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2$

And finally we can find the radius of the pipe at that point:

$A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m$

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3 years ago