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ivolga24 [154]
3 years ago
11

While sitting in your car by the side of a country road, you see your friend, who happens to have an identical car with an ident

ical horn, approaching you. You blow your horn, which has a frequency of 260 Hz; your friend begins to blow his horn as well, and you hear a beat frequency of 5.0 Hz. Part A How fast is your friend approaching you
Physics
1 answer:
Feliz [49]3 years ago
3 0

Answer:

-6.49 m/s

Explanation:

This is doppler effect.

The equation is;

F_l = [(v + v_l)/(v + v_s)]F_s

Where;

F_l is frequency observed by the listener

v is speed of sound

v_l is speed of listener

v_s is speed of source of the sound

F_s is frequency of the source of the sound

In this question, the source of the sound is the moving vehicle.

Thus;

F_l = F_beat + F_s

We are given beat frequency (f_beat) as 5 Hz while source frequency (F_s) as 260 Hz.

So,

F_l = 5 + 260

F_l = 265 Hz

Since listener is sitting by car, thus; v_l = 0 m/s

Thus,from our doppler effect equation, let's make v_s the subject;

v_s = F_s[(v + v_l)/F_l] - v

Speed of sound has a value of v = 344 m/s

Thus;

v_s = 260[(344 + 0)/265] - 344

v_s = -6.49 m/s

This value is negative because the source is moving towards the listener

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A mass of 25.5 g of H2O(g) at 373 K is mixed with 325 g of H2O(l) at 285 K and 1 atm. Calculate the final temperature of the sys
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Answer:

331.28 K

Explanation:

To solve this problem, you need to know that the heat that the water at 373 K is equal to the heat that the water at 285 K gains.

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The heat that the steam (H20(g)) loses is equal to the heat lost because the change of phase plus the heat lost because of the decrease in temperature:

Q_g = c_l*m_{wg} + m_{wg}*c*(T_{og}-T{fg})

The specific Heat c of water at 298K is 4.18 kJ/K*kg.

The latent heat cl of water is equal to  2257 kJ/kg.

The heat that the cold water gains is equal to heat necessary to increase its temperature to its final value:

Q_l = m_{wl}*c*(T_{fl}-T_{ol})

Remember that in equilibrium, the final temperature of both bodies of water will be equal.

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Q_g = Q_l\\c_l*m_{wg} + m_{wg}*c*(T_{og}-T{fg}) = m_{wl}*c*(T_{fl}-T_{ol})\\2257 kJ/kg*0.0255 kg + 0.0255 kg*4.18 kJ/kg*K*(373K - T_f) = 0.325 kg*4.18kJ/kg*K*(T_f-285K)\\57.5535 kJ + 39.75807kJ - 0.10659T_f = 1.3585 T_f - 387.1725 kJ\\484.48407 kJ = 1.46244 T_f\\T_f = 484.48407 kJ /1.46244 = 331.28 K

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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

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2h/Re =1- g'/g

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