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Svetlanka [38]
2 years ago
15

You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you from behind and you perceive

the frequency as 1310 Hz. You are relieved that he is in pursuit of a different speeder when he continues past you, but now you perceive the frequency as 1240 Hz. What is the frequency of the siren in the police car? The speed of sound in air is 343 m/s.
Physics
1 answer:
padilas [110]2 years ago
7 0

Answer:

f_{police}=1268.7 Hz    

Explanation:

We can use Doppler equation to find the frequency of the siren.

First of all we have the police car moving behind the car. Hence, the frequency detected by the car will be:  

f_{car1}=f_{police}(\frac{v_{s}-v_{car}}{v_{s}-v_{police}})      (1)

Now, when the police car is moving in front of the car, the frequency detected by the car will be:

f_{car2}=f_{police}(\frac{v_{s}+v_{car}}{v_{s}+v_{police}})        (2)            

By solving equation (1) and equation (2) for v_{police} we have:

v_{police} = 44.67 m/s

Knowing that:

  • f(car1) = 1310 Hz
  • f(car2) = 1240 Hz
  • Vs = 343 m/s
  • V(car) = 35 m/s

Finally, we just need to put this value into the first equation to find frequency of the police car.

f_{police}=f_{car}(\frac{v_{s}-v_{police}}{v_{s}-v_{car}})    

f_{police}=1310(\frac{343-44.7}{343-35})  

f_{police}=1268.7 Hz    

I hope it helps you!

                                                                                                                                       

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mihalych1998 [28]

Answer:

a) 1.2*10^-7 m

b) 1.0*10^-7 m

c) 9.7*10^-8 m

d) ultraviolet region

Explanation:

To find the different wavelengths you use the following formula:

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\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(2)^2})=8227500m^{-1}\\\\\lambda=1.2*10^{-7}m

(b)

\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(3)^2})=9751111,1m^{-1}\\\\\lambda=1.0*10^{-7}m

(c)

\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(4)^2})=10284375m^{-1}\\\\\lambda=9.7*10^{-8}m

(d) The three lines belong to the ultraviolet region.

8 0
2 years ago
A 20~\mu F20 μF capacitor has previously charged up to contain a total charge of Q = 100~\mu CQ=100 μC on it. The capacitor is t
sertanlavr [38]

Explanation:

The given data is as follows.

       C = 20 \times 10^{-6} F

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        Q_{o} = 100 \times 10^{-6} C

          Q = 13.5 \times 10^{-6} C

Formula to calculate the time is as follows.

          Q_{t}  = Q_{o} [e^{\frac{-t}{\tau}]

       13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}]

               0.135 = e^{\frac{-t}{2}}

         e^{\frac{t}{2}} = \frac{1}{0.135}

                         = 7.407

           \frac{t}{2} = ln (7.407)

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Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.

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2 years ago
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B) equal to the gravitational force of the brick
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Emmy kicks a soccer ball up at an angle of 45° over a level field. She watches the ball's trajectory and notices that it lands,
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Let u be the initial velocity of the soccer ball at an angle of inclination of \theta_0 with the positive x-axis.

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\theta_0=45^{\circ}

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Acceleration due to gravity, g= 10 m/s^2 downward.

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So, g= -10 m/s^2

As the acceleration due to gravity is in the vertical direction, so the horizontal component of the initial velocity remains unchanged.

The x-component of the initial velocity, u_x=u\cos\theta_0.

The horizontal distance covered by the projectile = u_x\times t_f

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From the equation of motion,

v=u+at

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Joel uses a claw hammer to remove a nail from a wall. He applies a force of 40 newtons on the hammer. The hammer applies a force
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Hi!


Mechanical advantage is defined as the<em> ratio of force produced by an object to the force that is applied to it.</em>

In our case, this would be the ratio of the force applied by the claw hammer on the nail to the force Joel applies to the claw hammer, which is

160:40 or 4:1

So the mechanical advantage of the hammer is four.


Hope this helps!


3 0
2 years ago
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