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3 years ago

Why do some "people" still believe that the earth is "flat".

1 answer:

7 0

Answer: Members of the Flat Earth Society claim to believe the Earth is flat. Walking around on the planet's surface, it looks and feels flat, so they deem all evidence to the contrary, such as satellite photos of Earth as a sphere, to be fabrications of a "round Earth conspiracy" orchestrated by NASA and other government agencies.

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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;

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

where $f_{2}$ is the frequency of the reflected sound;

$f_{1}$ is the frequency which the wave strikes the hill = 242.61 Hz;

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

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

$f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]$

$f_{2}$ = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound and the reflected sound. This can be obtained as follows;

Δf = $f_{2}$ - $f_{1}$

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

3 0

3 years ago

John recently suffered a blow to his head. Since then, he finds it difficult to comprehend what others say to him. He also finds

katovenus [111]

Answer: Wernicke's aphasia

Explanation:

John recently suffered a blow to his head. Since then, he finds it difficult to comprehend what others say to him. He also finds it difficult to express his thoughts and cannot seem to find the right words to say while speaking. However, he can speak freely with proper syntax. In this scenario, John is most likely suffering from Wernicke's aphasia.

Wernicke's aphasia occurs when the leftward side of the middle of the brain is damaged or has been altered. An individual who suffers from Wernicke's aphasia will have difficulty in speaking in meaningful and coherent sentences or may have difficulty in understanding the speech of others.

6 0

3 years ago

The greater the of an object the more force is needed to cause acceleration

d1i1m1o1n [39]

the greater the mass of an object the more force is needed to cause acceleration

8 0

3 years ago

For a ∆x of 0.1mm, what is ∆px, the uncertainty in the transverse momentum of a photon passing through a slit (where uncertainty

Colt1911 [192]

Answer:

$0.53\times 10^{-30}kgms^{-1}$