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Tema [17]
3 years ago
14

A bird is flying directly toward a stationary bird-watcher and emits a frequency of 1420 Hz. The bird-watcher, however, hears a

frequency of 1456 Hz. What is the speed of the bird, expressed as a percentage of the speed of sound
Physics
1 answer:
kogti [31]3 years ago
7 0

Answer:

2.47 %

Explanation:

We are given;

Frequency emitted by source(bird); f_s = 1420 Hz

Frequency heard by observer; f_o = 1456 Hz

From doppler shift frequency, we know that;

f_o = f_s [c/(c - c_s)]

Where c_s is speed of source which is the bird and c is speed of sound.

Thus;

Rearranging the equation, we have;

f_s/f_o = (c - c_s)/c

f_s/f_o = 1 - (c_s/c)

Plugging in the relevant values to get ;

1420/1456 = 1 - (c_s/c)

0.9753 = 1 - (c_s/c)

1 - 0.9753 = (c_s/c)

(c_s/c) = 0.0247

Since we want it expressed im percentage,

Thus, (c_s/c) % = 0.0247 x 100 = 2.47 %

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3 years ago
A proton is moving in a circular orbit of radius 12 cm in a uniform 0.31-T magnetic field perpendicular to the velocity of the p
PIT_PIT [208]

Answer:

Explanation:

A proton of charge

q=+1.609×10^-19C

Orbit a radius of 12cm

r=0.12m

Magnetic Field of 0.31T

Angle between velocity and field is 90°

a. Because the magnetic force F supplies the centripetal force Fc.

The magnitude of the magnetic force F on a charge q moving at a speed v in a magnetic field of strength B is given by

F = qvB sin θ

And the centripetal force is given as

Fc=mv²/r

Where m is mass of proton

m=1.673×10^-27kg

Then, F=Fc

qvB sin θ=mv²/r

qBSin90=mv/r

rqB=mv

Then, v=rqB/m

v=0.12×1.609×10^-19×0.31/1.673×10^-23

v=3577692.78m/s

v=3.58×10^6m/s

b. Since,

F=qVBSin90

F=1.609×10^-19×3.58×10^6×0.31

F=1.785×10^-13 N.

6 0
3 years ago
What is the minimum amount of data points an experiment should gather?
nydimaria [60]

The answer is "Three".

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7 0
2 years ago
To calibrate your calorimeter cup, you first put 45 mL of cold water in the cup, and measure its temperature to be 24.7 °C. You
drek231 [11]

Answer : The heat change of the cold water in Joules is, 1.6\times 10^3J

Explanation :

First we have to calculate the mass of cold water.

As we know that the density of water is 1 g/mL. The volume of cold water is 45 mL.

Density=\frac{Mass}{Volume}

Mass=Density\times Volume=1g/mL\times 45mL=45g

Now we have to calculate the heat change of cold water.

Formula used :

Q=m\times c\times (T_2-T_1)

where,

Q = heat change of cold water = ?

m = mass of cold water = 45 g

c = specific heat of water = 4.184J/g^oC

T_1 = initial temperature of cold water = 24.7^oC

T_2 = final temperature  = 33.4^oC

Now put all the given value in the above formula, we get:

Q=45g\times 4.184J/g^oC\times (33.4-24.7)^oC

Q=1638.036J=1.6\times 10^3J

Therefore, the heat change of cold water is 1.6\times 10^3J

4 0
3 years ago
A watermelon is dropped from the top of a 80m tall building. We want to find the velocity of the watermelon after it falls for 1
White raven [17]

Answer:

v_f = v_i + at

v_f = 13.23 m/s

Explanation:

Height Of the watermelon when it is dropped is given as

h = 80 m

time of fall under gravity

t = 1.35 s

now if water melon start from rest then we have

v_i = 0

acceleration due to gravity for watermelon

a = 9.81 m/s^2

now we need to find the final speed of watermelon

v_f = v_i + at

so we will have

v_f = 0 + (9.81)(1.35)

v_f = 13.23 m/s

7 0
3 years ago
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