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3 years ago

9

A stationary source S generates circular outgoing waves on a lake. The wave speed is 5.0 m/s and the crest-to-crest distance is

2.0 m. A person in a motorboat heads directly toward S at 3.0 m/s. To this person, the frequency of these waves is:

1 answer:

Aleonysh [2.5K]3 years ago

4 0

Answer:4 Hz

Explanation:

Speed of wave $v=5 m/s$

crest to crest distance $\lambda =2 m$

velocity of observer $v_0=3 m/s$

actual frequency $f=\frac{velocity}{\lambda }$

$f=\frac{5}{2}=2.5 Hz$

Apparent frequency $f'=f(\frac{v+v_0}{v})$

$f'=2.5\times \frac{5+3}{5}$

$f'=4 Hz$

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umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

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since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$

$\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}$

$3(r^2+1+2r)=5r^2$

$2r^2-6r-3=0$

$r=3.4365 \&\ r=-0.4365$

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

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