At Hoover Dam, the distance the water effectively falls before encountering the electric generators depends on the water levels
1 answer:
Answer:
Explanation:
recall that power is energy carried out or work done per time
P=W/t
P=2*10^6*35
t=6*60=420S
W=Energy
E=2*10^6*35*360S
E=25200000000
Energy stored by water from rest is called potential energy. Since the water is falling from a height , we calculate potential energy as thus
E=M*g*h
Assume that the water intakes are effectively 175 m above the electric generators. How much water must pass through the generators to power 2 million 35-W Las Vegas light bulbs for 6.0 minutes?
M=mass of water
g=acceleration due to gravity 9.81m/s^2
h=height ,175m
25200000000=M*9.81*175
M=
M=1716.75kg
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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) y = 2.4 x 10⁻³ m = 0.24 cm
b) y = 3.2 x 10⁻³ m = 0.32 cm
Explanation:
The formula of Young's Double Slit experiment will be used here:
where,
y = distance between dark spots = ?
λ = wavelength
L = distance of screen = 2 m
d = slit width = 4 x 10⁻⁴ m
a) FOR λ = 480 nm = 4.8 x 10⁻⁷ m:
<u>y = 2.4 x 10⁻³ m = 0.24 cm</u>
<u></u>
a) FOR λ = 640 nm = 6.4 x 10⁻⁷ m:
<u>y = 3.2 x 10⁻³ m = 0.32 cm</u>