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madam [21]
3 years ago
9

If sound travels at 5600 m/s through a steel rod,what is the wavelength,given a wave frequency of 2480 Hz

Physics
1 answer:
Feliz [49]3 years ago
6 0

Answer:

wavelength = 5600/2480= 2.25m

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a sleepy atudent drops a calculator out of a window yhata 20.7 m off the geound. we can ignore air resistance. what js the veloc
frez [133]

The acceleration of gravity is

9.8 m/s^2 down.

When an object falls out of a hand, its speed after 1.8s is

(9.8)x(1.8) = 17.6 m/s down.

It doesn't matter what it is, how much it weighs, or how high it was dropped from.

If it's more than 17.6 m/s, then this happened on a different, bigger planet.

If it's less than 17.6 m/s, then it must have hit something on the way down, like some air or something.

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What 2 parts of your foot do you use to dribble a soccer ball
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im not 100% sure but i think its the base of your big toe or the arch of your foot

Explanation:

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2 years ago
Gases have two specific heat values .....?
kipiarov [429]
A At one constant temp and another at a constant pressure
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2 years ago
We humans are on track to increase the amount of CO2 in the atmosphere so that the concentration in the future more than double
Lapatulllka [165]

Answer:

Rapid increase in the temperature of the earth.

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3 years ago
A motorcycle and a police car are moving toward one another. The police car emits sound with a frequency of 523 Hz and has a spe
Firdavs [7]

To solve this problem it is necessary to apply the concepts related to Dopler's Law. Dopler describes the change in frequency of a wave in relation to that of an observer who is in motion relative to the Source of the Wave.

It can be described as

f = \frac{c\pm v_r}{c\pm v_s}f_0

c = Propagation speed of waves in the medium

v_r= Speed of the receiver relative to the medium

v_s= Speed of the source relative to the medium

f_0 =Frequency emited by the source

The sign depends on whether the receiver or the source approach or move away from each other.

Our values are given by,

v_s = 32.2m/s \rightarrow Velocity of car

v_r = 14.8 m/s \rightarrow velocity of motor

c = 343m/s \rightarrow Velocity of sound

f_0 = 523Hz \rightarrowFrequency emited by the source

Replacing we have that

f = \frac{c + v_r}{c - v_s}f_0

f = \frac{343 + 14.8}{343 - 32}(523)

f = 601.7Hz

Therefore the frequency that hear the motorcyclist is 601.7Hz

8 0
3 years ago
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