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romanna [79]
3 years ago
11

Two trains on separate tracks move toward each other. Train 1 has a speed of 125 km/h; train 2, a speed of 94.0 km/h. Train 2 bl

ows its horn, emitting a frequency of 500 Hz. What is the frequency heard by the engineer on train 1?
Physics
1 answer:
dexar [7]3 years ago
4 0

Answer:

f_{o}=595.98Hz

Explanation:

Given data

Train 1 speed v₀=125 km/h =34.7222 m/s

Train 2 speed vs=94.0 km/h=26.1111 m/s

Speed of sound v=343 m/s

Train 2 frequency fs=500 Hz

To find

Frequency f₀ heard by engineer on train 1

Solution

We can use below formula to find the required frequency

f_{o}=f_{s}(\frac{v+v_{o}}{v-v_{s}} )\\f_{o}=(500Hz)(\frac{343m/s+34.722m/s}{343m/s-26.111m/s} ) \\f_{o}=595.98Hz

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This motion is exchanging potential energy (stored) for kinetic energy (in motion).

Remember that for motion, potential energy is proportional to the height of an object. E_P = gmz where z is the height, m is the mass, and g is the acceleration of gravity.

So when the pendulum mass is at its lowest point, it has converted all of the potential energy into kinetic energy.

The why is because it has fallen as far as the string will let it along its path of motion - thereby converting all the potential energy to kinetic energy.

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3 years ago
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Explanation:

We know that the standard lapse rate equals -2^{o}C per 1000 feet

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Neporo4naja [7]

Answer:

The center line temperature of the beam is 164^{circ}C

Solution:

As per the question:

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Initial temperature, T_{o} = 20^{\circ}C

Convection coefficient of heat flow, h = 24 W/m^{2}

Time, t = 46 min

k = 1.5\ W/mK

Now,

Biot no. is given by:

B_{i} = \frac{hr'}{2k}

B_{i} = \frac{24\times 0.25}{2\times 1.5} = 2

Now, Fourier no. is given by:

\frac{\alpha t}{r^{2}} = \frac{k}{C}\times t

\frac{\alpha t}{r^{2}} = \frac{k\times t}{rC_{p}r'^{2}} = \frac{1.5\times 46\times 60}{1495\times 880} = 0.05

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In a closed system with no friction, a red sphere of 2.5 kg stands stationary. A blue sphere with a mass of 5.8 kg approaches th
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Answer:

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v2 = initial velocity of the red sphere before the collision = 0 m/s

v'1 = final velocity of the blue sphere after the collision = 1.3 m/s

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23.78 = 7.54 + (2.5) (v'2)

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Explanation:

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