Space Station Suppose a space-station is designed in s shape of a torus such as the one depicted in Stanley Kubrick's "2001: A s
1 answer:
Answer:
w = 3.2 rev / min
Explanation:
For this exercise we will use the centrine acceleration equal to the acceleration of gravity
a = v² / r
Angular and linear variables are related.
v = w r
Let's replace
a = w² r = g
w = √ g / r
r = d / 2
r = 175/2 = 87.5 m
w = √( 9.8 / 87.5)
w = 0.3347 rad / s
Let's reduce to rotations per min
w = 0.3347 rad / s (1 rov / 2pi rad) (60 s / 1 min)
w = 3.2 rev / min
Suppose the space station rotates counterclockwise, we have two possibilities for the car
The first car turns counterclockwise (same direction of the station
=
r
[texwv_{c}[/tex] = / r
[texwv_{c}[/tex] = 25.0 / 87.5
[texwv_{c}[/tex] = 0.286 rad / s
When the two rotate in the same direction their angular speeds are subtracted
w total = w -[texwv_{c}[/tex]
w total = 0.3347 - 0.286
w total= 0.487 rad / s
The car goes in the opposite direction of the station the speeds add up
w = 0.3347 + 0.286
w = 0.62 rad / s
From this values we can see that the person feels a variation of the acceleration of gravity, feels that he has less weight when he goes in the same direction of the season and that his weight increases when he goes in the opposite direction to the season.
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Answer:
<em>a. The frequency with which the waves strike the hill is 242.61 Hz</em>
<em>b. The frequency of the reflected sound wave is 254.23 Hz</em>
<em>c. The beat frequency produced by the direct and reflected sound is </em>
<em> 11.62 Hz</em>
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;
..............................1
where 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;
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 =
v = 16.27 m/s
Substituting into equation 1 we have
=
= 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;
where is the frequency of the reflected sound;
is the frequency which the wave strikes the hill = 242.61 Hz;
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;
= 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 = -
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
Answer:
a). 100.2 MHz (typical frequency for FM radio broadcasting)
The wavelength of a frequency of 100.2 Mhz is 2.99m.
b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)
The wavelength of a frequency of 1070 khz is 280.3 m.
c. 835.6 MHz (common frequency used for cell phone communication)
The wavelength of a frequency of 835.6 Mhz is 0.35m.
Explanation:
The wavelength can be determined by the following equation:
(1)
Where c is the speed of light, is the wavelength and
is the frequency.
Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave.
<em>a). 100.2 MHz (typical frequency for FM radio broadcasting)</em>
Then, can be isolated from equation 1:
(2)
since the value of c is . It is necessary to express the frequency in units of hertz.
⇒
But
Finally, equation 2 can be used:
Hence, the wavelength of a frequency of 100.2 Mhz is 2.99m.
<em>b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)</em>
<em> </em>
⇒
But
Finally, equation 2 can be used:
Hence, the wavelength of a frequency of 1070 khz is 280.3 m.
<em>c. 835.6 MHz (common frequency used for cell phone communication) </em>
⇒
But
Finally, equation 2 can be used:
Hence, the wavelength of a frequency of 835.6 Mhz is 0.35m.