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3 years ago

Why are temperature variations greater over land than they are over water?

Physics

1 answer:

Stells [14]3 years ago

5 0

Because over land they wont be cooled by the water, and the deeper water is, the colder it is.

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3. What is the force of sliding friction

ale4655 [162]

Answer:

the force of the friction is A-0.52

3 0

3 years ago

Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 5.0

e-lub [12.9K]

Answer:

$f = 421.8 Hz$

Explanation:

When she moved a distance of 1 m from mid point she observe first destructive interference due to two speakers

so we can say that path difference of sound due to two speakers will be equal to half of the wavelength

so path difference is given as

$\Delta L = {3.5^2 + 12^2}^{0.5} - {1.5^2 + 12^2}^{0.5}$

so it will be

$\Delta L = 12.5 - 12.093$

$\Delta L = 0.4066$

now we know that

$\frac{\lambda}{2} = 0.4066$

$\lambda = 0.813$

now frequency of sound is given as

$f = \frac{v}{\lambda}$

$f = \frac{343}{0.813}$

$f = 421.8 Hz$

4 0

3 years ago

You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov

Marina CMI [18]

Answer:

a. The frequency with which the waves strike the hill is 242.61 Hz

b. The frequency of the reflected sound wave is 254.23 Hz

c. The beat frequency produced by the direct and reflected sound is

11.62 Hz

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

$f_{1} = f[\frac{v_{s} }{v_{s} -v} ]$ ..............................1

where $f_{1}$ is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = $36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}$

v = 16.27 m/s

Substituting into equation 1 we have

$f_{1}$ = $231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})$

$f_{1}$ = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

$f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]$