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ki77a [65]
3 years ago
9

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.A. What frequency fapproach is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction opposite to the first train and approaching it?B. What frequency frecede is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction opposite to the first train and receding from it?
Physics
1 answer:
Zigmanuir [339]3 years ago
8 0

Answer

given,

speed of sound = 344 m/s

speed of train = 30 m/s

frequency emitted by the train = 262 Hz

   Doppler's effect

    f_L = \dfrac{v + v_L}{v + v_s}\ f_S

f_L is the frequency of listener

f_S is the frequency of the source of the sound

v is the speed of the sound

v_L is the speed of listener.

v_S is the speed of the source

a) Speed of the passenger in another train , v = 18 m/s

   another train is moving in opposite direction and approaching

   v_L is positive as the listener is moving forward.

    v_S is negative at the source is moving toward the listener.

      f_L = \dfrac{344 + 18}{344 - 30}\times 262

     f_L = 302\ Hz

b) Speed of the passenger in another train , v = 18 m/s

   another train is moving in opposite direction and receding

    v_L is negative as the listener is moving away from source.

    v_S is positive at the source is moving away the listener.

      f_L = \dfrac{344 - 18}{344 + 30}\times 262

     f_L = 228.37\ Hz

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Answer:

1.04μT

Explanation:

Due to both wires have opposite currents, the magnitude of the total magnetic field is given by

B_T=\frac{\mu_o I}{2 \pi r_1}-\frac{\mu_o I}{2 \pi r_2}

I: electric current = 10A

mu_o: magnetic permeability of vacuum = 4pi*10^{-7} N/A^2

r1: distance from wire 1 to the point in which B is measured.

r2: distance from wire 2.

The distance between wires is 40cm = 0.4m. Hence, r1=0.2m r2=0.6m

By replacing in the formula you obtain:

B_T=\frac{(4\pi *10^{-7}N/A^2)(10A)}{2\pi}(\frac{1}{0.4m}-\frac{1}{0.6m})=1.04*10^{-6}T =1.04\mu T

hence, the magnitude of the magnetic field is 1.04μT

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The length of aluminum rods produced by a company are approximated by a Gaussian distribution with a mean of 10 cm and a standar
ExtremeBDS [4]

Given Information:

Mean length of aluminum rods = μ = 10 cm

Standard deviation of length of aluminum rods = σ = 0.02 cm

Required Information:

a) P(9.98 < X < 10.02) = ?

b) P(9.90 < X < 10.1) = ?

Answer:

a) P(9.98 < X < 10.02) = 68.27%

b) P(9.90 < X < 10.1) = 100%

Explanation:

What is Normal Distribution?

Normal Distribution or also known as Gaussian Distribution, is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability

a) We want to find out the probability that the length of aluminum rods is between 9.98 and 10.02 cm.

P(9.98 < X < 10.02) = P( \frac{x - \mu}{\sigma} < Z < \frac{x - \mu}{\sigma} )\\\\P(9.98 < X < 10.02) = P( \frac{9.98- 10}{0.02} < Z < \frac{10.02 - 10}{0.02} )\\\\P(9.98 < X < 10.02) = P( \frac{-0.02}{0.02} < Z < \frac{0.02}{0.02} )\\\\P(9.98 < X < 10.02) = P( -1 < Z < 1 )\\\\

The z-score corresponding to -1 is 0.15866 and 1 is 0.84134

P(9.98 < X < 10.02) = P( Z < 1 ) - P( Z < -1 ) \\\\P(9.98 < X < 10.02) = 0.84134 - 0.15866 \\\\P(9.98 < X < 10.02) = 0.6827\\\\P(9.98 < X < 10.02) = 68.27 \%

Therefore, the probability that the length of aluminum rods is between 9.98 and 10.02 cm is 68.27%

b) We want to find out the probability that the length of aluminum rods is between 9.90 and 10.1 cm.

P(9.90 < X < 10.1) = P( \frac{x - \mu}{\sigma} < Z < \frac{x - \mu}{\sigma} )\\\\P(9.90 < X < 10.1) = P( \frac{9.90- 10}{0.02} < Z < \frac{10.1 - 10}{0.02} )\\\\P(9.90 < X < 10.1) = P( \frac{-0.1}{0.02} < Z < \frac{0.1}{0.02} )\\\\P(9.90 < X < 10.1) = P( -5 < Z < 5 )\\\\

The z-score corresponding to -5 is 0 and 5 is 1

P(9.90 < X < 10.1) = P( Z < 5 ) - P( Z < -5 ) \\\\P(9.90 < X < 10.1) = 1 - 0 \\\\P(9.90 < X < 10.1) = 1\\\\P(9.90 < X < 10.1) = 100 \%

Therefore, the probability that the length of aluminum rods is between 9.90 and 10.1 cm is 100%

How to use z-table?

Step 1:

In the z-table, find the two-digit number on the left side corresponding to your z-score. (e.g 1.0, 2.2, 0.5 etc.)

Step 2:

Then look up at the top of z-table to find the remaining decimal point in the range of 0.00 to 0.09. (e.g. if you are looking for 1.00 then go for 0.00 column)

Step 3:

Finally, find the corresponding probability from the z-table at the intersection of step 1 and step 2.

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