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3 years ago

What causes standing wave?

1 answer:

6 0

Standing waves become produced/caused whenever two waves of identical frequency interfere with one another while traveling opposite directions along the same medium.

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From a vantage point very close to the track at a stock car race, you hear the sound emitted by a moving car. You detect a frequ

Alexandra [31]

Approximately 15 m/s is the speed of the car.

Explanation:

Given:

speed of sound - 343 m/s

You detect a frequency that is 0.959 times as small as the frequency emitted by the car when it is stationary. So, it can be written as,

$f^{\prime}=0.959 f$

$\frac{f^{\prime}}{f}=0.959$

If there is relative movement between an observer and source, the frequency heard by an observer differs from the actual frequency of the source. This changed frequency is called the apparent frequency. This variation in frequency of sound wave due to motion is called the Doppler shift (Doppler effect). In general,

$f^{\prime}=\left(\frac{v+v_{0}}{v-v_{s}}\right) \times f$

Where,

$f^'$ - Observed frequency

f – Actual frequency

v – Velocity of sound waves

$v_0$ – Velocity of observer

$v_s$ - velocity of source

When source moves away from an observer at rest ( $v_{0} = 0$ ), the equation would be

$f^{\prime}=\left(\frac{v}{v-\left(-v_{s}\right)}\right) \times f$

$f^{\prime}=\left(\frac{v}{v+v_{s}}\right) \times f$

$\frac{f^{\prime}}{f}=\left(\frac{v}{v+v_{s}}\right)$

By substituting the known values, we get

$0.959=\left(\frac{343}{343+v_{s}}\right)$

$0.959\left(343+v_{s}\right)=343$

$0.959(343)+0.959\left(v_{s}\right)=343$

$328.937+0.959 v_{s}=343$

$0.959 v_{s}=343-328.937=14.063$

$v_{s}=\frac{14.063}{0.959}=14.66 \mathrm{m} / \mathrm{s}$

Approximately 15 m/s is the speed of the car.

3 0

3 years ago

Which of the following is an oxidation-reduction reaction? ZnS(s) + 2O2(g) mc011-1.jpg ZnSO4(s) CaO(s) + H2O(l) mc011-2.jpg Ca(O

aliya0001 [1]

ZnS(s) + 2O2(g) mc011-1.jpg ZnSO4(s) CaO(s) + H2O(l)

4 0

3 years ago

Read 2 more answers

A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi

vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

$V=\frac {C_2V_2-C_1V_1}{C_1+C_2}$

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

$C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}$