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Hatshy [7]
3 years ago
9

PART ONE

Physics
1 answer:
Andrej [43]3 years ago
8 0

Answer:\Delta L=0.0101\ m

Explanation:

Given

Length of track L_o=28\ m when

T_o=2^{\circ}C

Coefficient of linear expansion \alpha =11\times 10^{-6}\ ^{\circ}C^{-1}

When Temperature rises to T=35^{\circ}C

\Delta T=35-2=33^{\circ}C

and we know length expand on increasing temperature

L=L_o[1+\alpha \Delta T]

L-L_o=L_o\alpha \Delta T

\Delta L=28\times 11\times 10^{-6}\times (33)

\Delta L=0.0101=10.164\ mm

(b)When rails are clamped thermal stress induced

we know E=\frac{stress}{strain}

Stress=E\times strain

Stress=20\times 10^{10}\times \frac{\Delta L}{L_o}

Stress=20\times 10^{10}\times \frac{0.0101}{28}

Stress=72.14\ MPa

Stress=72.14\times 10^{6}\ N/m^2

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Answer:

5. 9GmM/(10R)

Explanation:

m is the mass of the satellite

M is the mass of the earth

W is the energy required to launch the satellite

Energy at earth surface = Potential energy (PE) + W

W = Energy at earth surface - Potential energy (PE)

But PE = -\frac{GMm}{R}

Therefore: W = Energy at earth surface - \frac{GMm}{R}

Energy at earth surface (E) at an altitude of 5R = -\frac{GMm}{5r} +\frac{1}{2}mV^2

But V=\sqrt{\frac{GM}{5R} }

Therefore: E=-\frac{GMm}{5R}+\frac{1}{2}m(\sqrt{\frac{GM}{5R} } )^2=  -\frac{GMm}{5R}+\frac{GMm}{10R}  = -\frac{GMm}{10R}

W = E - PE

W=-\frac{GMm}{10R}-(-\frac{GMm}{R})=-\frac{GMm}{10R}+\frac{GMm}{R}=\frac{9GMm}{10R} \\W=\frac{9GMm}{10R}

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3 years ago
Which of the following is right?
barxatty [35]
Scientific theory because it's a theory it's already an answer but it might change depending on the condition.
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A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7
aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

f'=(\frac{v+v_r}{v+v_s})f

where

f is the original frequency

f is the apparent frequency

v is the velocity of the wave

v_r is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

v_s is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

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v_r = +62.2 m/s (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz

(b) 0.232 m

The wavelength of a wave is given by

\lambda=\frac{v}{f}

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m

(c) 1481.2 Hz

Again, we can use the same formula

f'=(\frac{v+v_r}{v+v_s})f

In the source frame (= on surface A), we have

v_s = v_B = -62.2 m/s (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

v_r = +28.7 m/s (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz

(d) 0.225 m

The wavelength of the wave is given by

\lambda=\frac{v}{f}

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m

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