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3 years ago

Under the Doppler effect, an observer behind a moving wave source observes

Physics

2 answers:

kondaur [170]3 years ago

6 0

Answer: The correct option is C.

Explanation:

Doppler effect is the phenomenon in which there is an apparent change in the frequency or increase or decrease when there is a relative motion between the listener and the observer.

When the listener and the observer are approaching each other then the frequency of the wave will get increased.

When the listener and the observer are moving away from each other then the frequency of the wave will get decreased.

Therefore, under the Doppler effect, an observer behind a moving wave source observes decreased wave frequency.

Likurg_2 [28]3 years ago

5 0

IT should be c a decreased wave frequency.

I don't know man I could be wrong. papa bless

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An elevator filled with passengers has a mass of 1.70×103kg. (a) The elevator accelerates upward from rest at a rate of 1.20m/s2

Anton [14]

Answer with Explanation:

We are given that mass of an elevator filled with passengers = $1.70\times 10^3 Kg$

a.a= $1.2 m/s^2$

t=1.50 s

Initial velocity=0

We have to calculate the tension in the cable supporting the elevator.

According to newton's second law,the net force is given by

$F_net=\sum F=ma$

$F-mg=ma$

Substitute the values then we get

$F=ma+mg=m(a+g)=1.7\times 10^3(1.2+9.8)=1.87\times 10^4 N$

Hence, the tension in the cable supporting the elevator= $1.87\times 10^4 N$

b.t=8.5 s

We have to find the tension in the table during this time.

Fnet=0

F=W= $1.7\times 10^3\times 9.8=1.67\times 10^4 N$

Hence, the tension in the cable during this time = $1.67\times 10^4 N$

c. Deceleration= $0.6 m/s^2$

t=3 s

We have to find the tension in the cable during deceleration.

$F-mg=-ma$

$F=mg-ma=m(g-a)$

$F=1.7\times 10^3(9.8-0.6)=1.56\times 10^4 N$

Hence, the tension in the cable during deceleration= $1.56\times 10^4 N$ .

d.We have to find that the height of elevator moved above its original staring point .

We have to find the final velocity.

$y_1=ut_1+\frac{1}{2}a_1t_1^2$

$y_1=0(1.50+\frac{1}{2}(1.2)(1.5)=1.35 m$

Final velocity

$v_1=a_1t_1$

$v_1=(1.2)(1.5)=1.8 m/s$

$y_2=v_1t_1=(1.8)(8.5)=15.3 m$

The third distance $y_3$ is up in the direction because the elevator decelerate upward which mean the distance is positive and is given by

$y_3=(1.8)(3)-\frac{1}{2}(0.6)(3)^2=2.7 m$

The total distance moved by elevator from its original position

$d=1.35+1.8+2.7=19.35 m$

Final velocity=0

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3 years ago

A small sphere with mass 5.00×10−7 kg and charge +8.00 μC is released from rest a distance of 0.500 m above a large horizontal i

Damm [24]

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andreev551 [17]

Answer:

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Explanation:

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GalinKa [24]

Answer:

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Explanation:

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zepelin [54]

Answer: D Medium B is denser than Medium A because light bends more in Medium B.

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3 years ago