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Lelu [443]
3 years ago
8

You are standing on a sidewalk. There is a car in the distance. The horn on the car sounds. You hear it at a pitch that correspo

nds to a sound frequency lower than the frequency emitted by the horn. What can you conclude?A) Either A or C must be true.B) The car is moving on a line directly away from your present location.C) The wind has a component of its velocity in a direction away from you.D) The car has a component of its velocity in a direction away from you.E) The wind has a component of its velocity in a direction toward you.

Physics
1 answer:
AnnyKZ [126]3 years ago
4 0

Answer: The answer: The car is moving away from you.

Both A and C are true as Car can be moving in line away from you or has component of velocity in opposite direction.

Explanation:The decrease in the frequency of the sound is the result of Doppler's effect. A/c to Doppler's effect the frequency of received sound of source is changed if it is moving relative to the receiver, i.e. the distance between them is changing due to motion.

The general formula of Doppler's Effect is attached as the picture.

In this formula v_D is the velocity of Detector i.e the receiver relative to wind. While v_s is the velocity of source relative to wind and v is the velocity of sound.

The Doppler's effect is not effected by the velocity of wind as the wind itself could not change the distance between the two objects i.e. you and the car. Wind velocity can change the speed of sound and its wavelength but the change does not effect the frequency.

Hence if we assume the car to be moving with velocity v_c and you are stationary

f'=f_s*\frac{v}{v-v_c}

hence the frequency is reduced.

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Rudik [331]

Answer:

1)   P₁ = -2 D,   2) P₂ = 6 D

Explanation:

for this exercise in geometric optics let's use the equation of the constructor

          \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distance to the object and the image, respectively

1) to see a distant object it must be at infinity (p = ∞)

          \frac{1}{f_1} = \frac{1}{q}

           q = f₁

2) for an object located at p = 25 cm

            \frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}

We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm

we substitute in the equations

1) f₁ = -50 cm

2)  

        \frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}

        \frac{1}{f_2} = 0.06

         f₂ = 16.67 cm

the expression for the power of the lenses is

          P = \frac{1}{f}

where the focal length is in meters

           

1)       P₁ = 1/0.50

        P₁ = -2 D

2)     P₂ = 1 /0.16667

        P₂ = 6 D

4 0
2 years ago
Calculate the kinetic energy of a 0.032 kg ball as it leaves a hand to be thrown upwards at 6.2 m/s
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Answer:

The ball will have a kinetic energy of 0.615 Joules.

Explanation:

Use the kinetic energy formula

E_k = \frac{1}{2}mv^2 = \frac{1}{2}0.032kg\cdot 6.2^2 \frac{m^2}{s^2}= 0.615J

The kinetic energy at the moment of leaving the hand will be 0.615 Joules. (From there on, as it ball is traveling upwards, this energy will be gradually traded off with potential energy until the ball's velocity becomes zero at the apex of the flight)

3 0
3 years ago
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pav-90 [236]
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7 0
3 years ago
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Brrunno [24]

Explanation:

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4 0
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kherson [118]

Answer:

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Explanation:

easy question to be honest

7 0
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