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makvit [3.9K]
3 years ago
15

A train is traveling at 30.0 m/sm/s relative to the ground in still air. The frequency of the note emitted by the train whistle

is 262 HzHz. The speed of sound in air should be taken as 344 m/sm/s. Part A What frequency fapproachfapproachf_approach is heard by a passenger on a train moving at a speed of 18.0 m/sm/s relative to the ground in a direction opposite to the first train and approaching it
Physics
1 answer:
REY [17]3 years ago
3 0

Answer:

Explanation:

The problem is based on Doppler's effect , for which expression is as follows

n = \frac{n_o(V+v_o)}{(V-v_s)}

n is apparent frequency , n₀ is original frequency , V is velocity of light , v₀ is velocity of observer and v_s is velocity of source , both coming towards each other.

Putting the values

n = \frac{262(344+18)}{(344-30)}

n = 302 .

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Waldo gets stopped by the police, and the police officer tells Waldo that she caught him on radar doing 50 miles per hour in a 4
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Answer:

He calculated his average speed.

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3 years ago
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28. Magnesium bromide is a binary ionic compound. From its formula, MgBr2, how do you know that Mg is the metal? (Pages 567-568)
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Answer:

See Below

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8 0
3 years ago
In an insulated container, 0.50 kg of water at 80°C is mixed with 0.050 kg of ice at −5.0°C. After a while, all the ice melts, l
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Explanation:

When ice tends to absorb heat energy then it will change into liquid state. Now,  change in the temperature of cold water dT = T_{f} - 0^{o}C = T_{f}

Change in the temperature of original water in the container dT' = 80 -T_{f
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Latent heat of melting of ice L = 80 cal/g

Specific heat of water C = 1 cal/g C

Heat absorbed by ice to melt = mass(m_{i}) x latent heat(L)

Heat absorbed by water = mass(m_{w}) x specific heat(C) x change in temperature (dT')

m_{i} = 0.050 kg = 50 g and m_{w} = 0.50 kg = 500 g

Now, heat lost by water is as follows.

heat lost by hot water( at 80^{o}C)  = heat absorbed by the ice + heat absorbed by cold water

         m_{i} \times L + m_{i} \times C dT = m_{w} \times C dT'

            50 \times 80 + 50 \times 1 \times (T_{f}) = 500 \times 1 \times (80 - T_{f})

So, we will cancel out 50 from each term given above. Hence,

              80 + T_{f} = 800 - 10T_{f}  

               11 T_{f} = 720

               T_{f} = 65.45^{o}C

Thus, we can conclude that final temperature of the water is 65.45^{o}C.

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