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GuDViN [60]
3 years ago
10

What is the observed frequency of a siren with a frequency of

Physics
1 answer:
Yakvenalex [24]3 years ago
7 0

Answer: The observed frequency is 518.6 Hz

Explanation:

By the Doppler Effect, we know that:

f' = ((v + v0)/(v - vs))*f

where f' is the perceived frequency.

v is the velocity of the wave, in this case 330m/s

v0 is the velocity of the receptor, in this case 0m/s

vs is the velocity of the source, in this case -50m/s (the minus sign is because it is coming thowards us)

f is the original frequency, f = 440hz.

then we have:

f' = 440hz*(330m/s)/(330m/s - 50m/s) = 518.6 Hz

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Ira Lisetskai [31]

Answer:

s = 37.81 m

Explanation:

Given that,

Mass of an apple, m = 1.5 kg

The apple falls for 2.75s before striking the ground.

We need to find how far the apple dropped before hitting the ground. It means we need to find the distance covered. Using the second equation of motion to find it :

s=ut+\dfrac{1}{2}gt^2

u is initial velocity, u = 0 (at rest)

s=\dfrac{1}{2}gt^2\\\\s=\dfrac{1}{2}\times 10\times (2.75)^2\\\\=37.81\ m

So, Apple dropped 37.81 m before hitting the ground.

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At what height from the surface of the earth does the value of acceleration due to gravity be 2.45 m/s square where the radius o
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Answer:

Explanation: RADIUS OF EARTH = 6400X1000m =

ACC DUE GRAVITY ABOVE SURFACE OF EARTH = g' =2.45 m/s^2

ACC DUE GRAVITY ON SURFACE OF EARTH =g= 9.8 m/s^2

A/C TO FORmULA

g'/g=1-2h/Re

g'/g +2h/Re = 1

2h/Re =1- g'/g

2h= (1- g'/g)Re

2h=(1-2.45 /9.8)

6400X1000

2h = (0.75)6400X1000

2h = 4800000

h= 2400000

m

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