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Anna11 [10]
3 years ago
15

What occurs when the pitch increases and the sound source is in motion? explain your answer.

Physics
1 answer:
Hoochie [10]3 years ago
3 0
If you're listening to a sound that has a steady pitch, and suddenly the
pitch goes up, then you know that two things could have happened:

EITHER ...

-- The person or other source making the sound could have
raised the pitch of the sound being produced.

OR ...

-- The person or other source making the sound could have
started moving toward you.

OR ...

-- both.

Even if the pitch of the sound leaving the source doesn't change,
you would still hear it increase if the source starts moving toward
you.  That's the so-called "Doppler effect".
You might be interested in
Which of the following is not a reason fluorescent lamps are advantages over incandescent lamps?
iren2701 [21]

It’s because flourecent lights operate at higher temperatures than incadecent lights.

3 0
3 years ago
A spring gun consists of a spring inside a plastic tube with spring constant, k. The spring can be compressed 20 cm from its equ
emmainna [20.7K]

Answer: The spring constant is K=392.4N/m

Explanation:

According to hook's law the applied force F will be directly proportional to the extension e produced provided the spring is not distorted

The force F=ke

Where k=spring constant

e= Extention produced

h=2m

Given that

e=20cm to meter 20/100= 0.2m

m=100g to kg m=100/1000= 0.1kg

But F=mg

Ignoring air resistance

assuming g=9.81m/s²

Since the compression causes the plastic ball to poses potential energy hence energy stored in the spring

E=1/2ke²=mgh

Substituting our values to find k

First we make k subject of formula

k=2mgh/e²

k=2*0.1*9.81*2/0.1²

K=3.921/0.01

K=392.4N/m

5 0
3 years ago
During a braking test, a car is brought to rest beginning from an initial speed of 60 mi/hr in a distance of 120 ft. With the sa
maw [93]

Answer:

Explanation:

Given

Initial speed u=60\ mi/hr\approx 88\ ft/s

distance traveled before coming to rest d_1=120\ ft

using equation of motion

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

0-(88)^2=2\times a\times 120---1

for u_2=80\ mi/hr\approx 117.33\ ft/s

using same relation we get

0-(117.33)^2=2\times a\times (d_2)----2

divide 1 and 2 we get

(\frac{88}{117.33})^2=\frac{120}{d_2}

d_2=213.32\ ft

So a distance if 213.32 ft is required to stop the vehicle with 80 mph speed

8 0
3 years ago
Calculate the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m
otez555 [7]

Answer: The period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

Explanation:

Given: Mass = 5 kg

Spring constant = 6 N/m

Formula used to calculate period is as follows.

T = 2 \pi \sqrt\frac{m}{k}

where,

T = period

m = mass

k = spring constant

Substitute the values into above formula as follows.

T = 2 \pi \sqrt\frac{m}{k}\\= 2 \times 3.14 \times \sqrt\frac{5}{6}\\= 5.73 s

Thus, we can conclude that the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

5 0
3 years ago
DUE TODAY PLEASE HELP.....WILL GIVE BRANIST!!!!!!!
Karo-lina-s [1.5K]

girl how someone supposed to help you with that?!

7 0
3 years ago
Read 2 more answers
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