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Yuki888 [10]
3 years ago
7

However, had it been a real sound, the sound's pitch would have been increased by the Doppler Effect, since the Falcon was movin

g _____ the source of the sound.
perpendicular to

away from

towards

at the same speed as
Physics
2 answers:
grandymaker [24]3 years ago
5 0

Answer:

Towards

Explanation:

According to Doppler Effect, there is change in the apparent frequency of the sound heard by the listener, when the source and the listener have relative motion.

The apparent frequency (pitch) increases when the source and the listener approach while it decreases when the source and listener recede away from each other.

Sound's pitch is related to frequency. With increase in frequency, pitch increases. If the Falcon was moving <u>towards</u> the source, the sound's pitch would have increased.

kkurt [141]3 years ago
3 0
I believe the answer is C.
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damaskus [11]

Upward and downward forces cancel out. Net force is 8 newtons to the right

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3 years ago
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A proton is moving in a circular orbit of radius 12 cm in a uniform 0.31-T magnetic field perpendicular to the velocity of the p
PIT_PIT [208]

Answer:

Explanation:

A proton of charge

q=+1.609×10^-19C

Orbit a radius of 12cm

r=0.12m

Magnetic Field of 0.31T

Angle between velocity and field is 90°

a. Because the magnetic force F supplies the centripetal force Fc.

The magnitude of the magnetic force F on a charge q moving at a speed v in a magnetic field of strength B is given by

F = qvB sin θ

And the centripetal force is given as

Fc=mv²/r

Where m is mass of proton

m=1.673×10^-27kg

Then, F=Fc

qvB sin θ=mv²/r

qBSin90=mv/r

rqB=mv

Then, v=rqB/m

v=0.12×1.609×10^-19×0.31/1.673×10^-23

v=3577692.78m/s

v=3.58×10^6m/s

b. Since,

F=qVBSin90

F=1.609×10^-19×3.58×10^6×0.31

F=1.785×10^-13 N.

6 0
4 years ago
What is a normal force?
Flauer [41]
Is the component perpendicular to the surface on contact  of the  contact force  <span />
3 0
4 years ago
A truck traveling at a velocity of 33m/s comes to a halt by decelerating at 11m/s^2. How far does the truck travel in the proces
snow_lady [41]

Answer:

<u><em>The truck was moving 16.5 m/s during the time it took to stop, which was 3 seconds. </em></u>

  • <u><em>Initial velocity = 33 m/s</em></u>
  • <u><em>Final velocity = 0 m/s</em></u>
  • <u><em>Average velocity = (33 + 0) / 2  m/s = 16.5 m/s</em></u>

Explanation:

  1. <u><em>First, how long does it take the truck to come to a complete stop?</em></u>
  1. <u><em>( 33 m/s ) / ( 11 m / s^2 ) = 3 seconds</em></u>
  1. <u><em>Then we can look at the average velocity between when the truck started decelerating and when it came to a complete stop. Because the deceleration is constant (always 11m/s^2) we can use this trick.</em></u>
4 0
3 years ago
Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton
pishuonlain [190]

Hello!

Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.

Data:

Hooke represented mathematically his theory with the equation:

F = K * Δx  

On what:

F (elastic force) = 2 N

K (elastic constant) = 4 N/cm

Δx (deformation or elongation of the elastic medium or distance from a spring) = ?

Solving:


F = K * \Delta{x}

2\:N = 4\:N/cm*\Delta{x}

4\:N/cm*\Delta{x} = 2\:N

\Delta{x} = \dfrac{2\:\diagup\!\!\!\!\!N}{4\:\diagup\!\!\!\!\!N/cm}

simplify by 2

\Delta{x} = \dfrac{2}{4}\frac{\div2}{\div2}

\boxed{\boxed{\Delta{x} = \dfrac{1}{2}\:cm}}\Longleftarrow(distance)\end{array}}\qquad\checkmark

Answer:

B.) 1/2 cm

_______________________

I Hope this helps, greetings ... Dexteright02! =)

7 0
3 years ago
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