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Dimas [21]
3 years ago
10

As you stand near a railroad track, a train passes by at a speed of 31.7 m/s while sounding its horn at a frequency of 218 Hz. W

hat frequency do you hear as the train approaches you and what frequency while it recedes?
Physics
1 answer:
Darya [45]3 years ago
6 0

Explanation:

Given that,

Frequency of train horn, f = 218 Hz

Speed of train, v_t = 31.7 m/s

The speed of sound, V = 344 m/s (say)

The speed of the observed person, V_o=0\ m/s

(a) When the train approaches you, the Doppler's effect gives the frequency as follows :

f'=f(\dfrac{V}{V-v_t})\\\\f'=218\times (\dfrac{344}{344-31.7})\\\\f'=240.12\ Hz

(b) When the train moves away from you, the Doppler's effect gives the frequency as follows :

f'=f(\dfrac{V}{V+v_t})\\\\f'=218\times (\dfrac{344}{344+31.7})\\\\f'=199.6\ Hz

Hence, this is the required solution.

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The universe was 5 percent its current size when light left objects observed now at redshift of ______________
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The redshift of distant galaxy are larger than those of closer galaxies, which indicates that the galaxy is receding at a faster rate.

  • The Universe was 5 percent its current size when light left objects now at redshift of <u>19</u>.

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The size of the universe represented as a scale factor with relation to the redshift can be presented as follows;

\displaystyle \frac{a}{a_0}  =\mathbf{ \frac{1}{1 + z}}

Where;

a₀ = The current size of the Universe

a = The size of the early Universe = 5% of a

Therefore;

\displaystyle \frac{a}{a_0}  =5\% = 0.05=  \frac{1}{1 + z}

\displaystyle 0.05  = \frac{1}{1 + z}

0.05 + 0.05·z = 1

\displaystyle z = \mathbf{ \frac{1 - 0.05}{0.05} } = 19

  • The redshift is of the observed light is, z = <u>19</u>

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5 0
3 years ago
At t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 26.0 rad/s2 until a cir
Agata [3.3K]

Answer:

a) The total angle of the grinding wheel is 569.88 radians, b) The grinding wheel stop at t = 12.354 seconds, c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

Explanation:

Since the grinding wheel accelerates and decelerates at constant rate, motion can be represented by the following kinematic equations:

\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

\omega = \omega_{o} + \alpha \cdot t

\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})

Where:

\theta_{o}, \theta - Initial and final angular position, measured in radians.

\omega_{o}, \omega - Initial and final angular speed, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

t - Time, measured in seconds.

Likewise, the grinding wheel experiments two different regimes:

1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

If \omega_{o} = 25\,\frac{rad}{s}, t = 2.40\,s and \alpha = 26\,\frac{rad}{s^{2}}, then:

\theta-\theta_{o} = \left(25\,\frac{rad}{s} \right)\cdot (2.40\,s) + \frac{1}{2}\cdot \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)^{2}

\theta-\theta_{o} = 134.88\,rad

The final angular angular speed can be found by the equation:

\omega = \omega_{o} + \alpha \cdot t

If  \omega_{o} = 25\,\frac{rad}{s}, t = 2.40\,s and \alpha = 26\,\frac{rad}{s^{2}}, then:

\omega = 25\,\frac{rad}{s} + \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)

\omega = 87.4\,\frac{rad}{s}

The total angle that grinding wheel did from t = 0 s and the time it stopped is:

\Delta \theta = 134.88\,rad + 435\,rad

\Delta \theta = 569.88\,rad

The total angle of the grinding wheel is 569.88 radians.

b) Before finding the instant when the grinding wheel stops, it is needed to find the value of angular deceleration, which can be determined from the following kinematic expression:

\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})

The angular acceleration is now cleared:

\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}

Given that \omega_{o} = 87.4\,\frac{rad}{s}, \omega = 0\,\frac{rad}{s} and \theta-\theta_{o} = 435\,rad, the angular deceleration is:

\alpha = \frac{ \left(0\,\frac{rad}{s}\right)^{2}-\left(87.4\,\frac{rad}{s} \right)^{2}}{2\cdot \left(435\,rad\right)}

\alpha = -8.780\,\frac{rad}{s^{2}}

Now, the time interval of the Deceleration Phase is obtained from this formula:

\omega = \omega_{o} + \alpha \cdot t

t = \frac{\omega - \omega_{o}}{\alpha}

If \omega_{o} = 87.4\,\frac{rad}{s}, \omega = 0\,\frac{rad}{s}  and \alpha = -8.780\,\frac{rad}{s^{2}}, the time interval is:

t = \frac{0\,\frac{rad}{s} - 87.4\,\frac{rad}{s} }{-8.780\,\frac{rad}{s^{2}} }

t = 9.954\,s

The total time needed for the grinding wheel before stopping is:

t_{T} = 2.40\,s + 9.954\,s

t_{T} = 12.354\,s

The grinding wheel stop at t = 12.354 seconds.

c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

4 0
3 years ago
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