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Amanda [17]
3 years ago
7

What frequency will you hear if a truck is driving toward you at 20 m/s sounding a horn of frequency 300 Hz. Assume the speed of

sound to be 340 m/s.
Physics
1 answer:
Damm [24]3 years ago
6 0

Answer:

=319Hz

Explanation:

This question is on sound waves

General formulae;

Frequency of observer= f of source*[ (v of sound+ v of observer)/(v of sound-v of source)]

where f is frequency and v is velocity

Given;

Velocity of sound = 340m/s

frequency of source = 300 Hz

Velocity of observer=0m/s

Velocity of source/truck= 20m/s

Substitute values in the formulae;

Frequency of observer= 300× {(340+0)/(340-20)}

                                       =300×{ 340/320}

                                       =300×340/320

                                       =318.75

                                       =319Hz

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Answer:

Explanation:

graph would be a straight line from (0, 0) to (400, 8)

Plot points are

PE = mgh

50(0) = 0 J

50(2) = 100 J

50(4) = 200 J

50(6) = 300 J

50(8) = 400 J

4 0
3 years ago
A spring stretches 0.150 m when a 0.30 kg mass is hung from it. The spring is then stretched an additional 0.100 m from this equ
DochEvi [55]

Answer:

a)  k=19.6N/m

b)  V_m=0.81m/s

c)  a_m=6.561m/s^2

d)  K.E=0.096J

e)  T=0.78sec &F=1.29sec

f)   mx'' + kx' =0

Explanation:

From the question we are told that:

Stretch Length L=0.150m

Mass m=0.30kg

Total stretch lengthL_t=0.150+0.100=>0.25

a)

Generally the equation for Force F on the spring is mathematically given by

F=-km\\\\k=F/m\\\\k=\frac{m*g}{x}\\\\k=\frac{0.30*9.8}{0.15}

k=19.6N/m

b)Generally the equation for Max Velocity of Mass on the spring is mathematically given by

V_m=A\omega

Where

A=Amplitude

A=0.100m

And

\omega=angulat Velocity\\\\\omega=\sqrt{\frac{k}{m}}\\\\\omega=\sqrt{\frac{19.6}{0.3}}\\\\\omega=8.1rad/s

Therefore

V_m=A\omega\\\\V_m=8.1*0.1

V_m=0.81m/s

c)

Generally the equation for Max Acceleration of Mass on the spring is mathematically given by

a_m=\omega^2A

a_m=8.1^2*0.1

a_m=6.561m/s^2

d)

Generally the equation for Total mechanical energy of Mass on the spring is mathematically given by

K.E=\frac{1}{2}mv^2

K.E=\frac{1}{2}*0.3*0.8^2

K.E=0.096J

e)

Generally the equation for  the period T is mathematically given by

\omega=\frac{2\pi}{T}

T=\frac{2*3.142}{8.1}

T=0.78sec

Generally the equation for  the Frequency is mathematically given by

F=\frac{1}{T}

F=1.29sec

f)

Generally the Equation of time-dependent vertical position of the mass is mathematically given by

mx'' + kx' =0

Where

'= signify order of differentiation

7 0
3 years ago
A hockey player uses her hockey stick to exert a force of 6.81 N on a stationary hockey puck. The hockey puck has a mass of 165
Anna007 [38]

Answer:

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Explanation:

force, F = 6.81 N

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Let a be the acceleration of the puck.

Use newtons' second law

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6.81 = 0.165 x a

a = 41.27 m/s^2

a = 41.3 m/s^2

Thus, the acceleration of the puck is 41.3 m/s^2.

5 0
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Sever21 [200]

Answer:

solar eclipse

Explanation:

because at that time, the moon completely covers the path and casts its shadow on earth because it is present between sun and earth's path. so, solar eclipse occurs.

hope it helps :)

8 0
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Answer:

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Explanation:

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