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faltersainse [42]

3 years ago

8

The diagram illustrates the movement of sound waves between an observer and a race car. As the race car drives away from the obs

erver, notes that the sound from the car gets quieter and the pitch lowers. Which statement is the MOST plausible question the observer might ask to explain the
phenomenon she experiences?
A. Why does the sound of the engine change as the car drives away?
B. What is the effect of the change in speed of the car and the pitch of the sound?
C. What is the relationship between the volume of the sound and the pitch of the sound
of the car?
D. How do changes in frequency and amplitude affect the pitch of the sound generated
by the car?

1 answer:

Andrei [34K]3 years ago

6 0

I believe the answer is d

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How do rocks within Earth change as P waves pass? How do rocks within Earth change as P waves pass? Rocks within Earth both expa

ohaa [14]

Rocks within Earth both expand and contract as P waves pass

Explanation:

Rocks within the earth both expands and contracts as P-waves passes through them. P-waves are elastic waves.

Elastic waves behaves in such a way that they do not cause permanent deformation of rocks.
They can be said to cause elastic deformation when they travel through rocks.
They simply temporarily expand and contract the rock within a short period by causing the vibration of particles of the medium.
After a short while, the rock returns back to its original position as if nothing has happened to it.
These elastic waves are better called seismic waves.
P-waves are primary waves that can travel through any medium.

Learn more:

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6 0

4 years ago

If only 10 pounds is required to lift a 500-lb block, how much chain must be played out to lift the engine 3.0 inches?

adelina 88 [10]

Answer:

150 inches (12.5 ft)

Explanation:

The work done to lift the 500 pound block 3 inches should be the same as that to lift the 10 pond object a given distance.

The following is the equation one needs to solve:

$10 \,lb\,* \,x\,=\,500\,lb\,*\,3\,in\\10 \,lb\,* \,x\,=\,1500\,lb\,in\\$

therefore solving for the distance "x" gives as the answer (in inches):

$10 \,lb\,* \,x\,=\,1500\,lb\,in\\x\,=\,\frac{1500\,lb\,in}{10\,lb} \\\\x\,=150\,in$

which can also be given in feet as: 12.5 ft

3 0

3 years ago

What is carbon dioxide?

Bezzdna [24]

Answer:

acidic colorless gas

Explanation:

Carbon dioxide is an acidic colorless gas with a density of about 53% higher than that of dry air. Carbon dioxide molecules consist of a carbon atom covalently double bonded to two oxygen atoms. It occurs naturally in Earth's atmosphere as a trace gas.

4 0

3 years ago

Read 2 more answers

If you decide you want to meet someone you met online, what should you do first?

seraphim [82]

c tell your parents it could be someone very dangerous and meeting people online can be very risky I would also tell my bsf just for fun tho

6 0

3 years ago

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p

igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

$B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

$B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

$B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

$F = qvB$

$F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

$F = 21.3 \times 10^{-16}$ N

From the newton's second law

$F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

$a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

7 0

3 years ago