One exciting fairground ride acts like a gaint catapuit. The capsule which the 'rider' is strapped in is fired high into the sky
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Stephen`s Law:
P = (Sigma) · A · e · T^4
P in = P out
e = 1 for blacktop;
1150 W = (Sigma) · T^4
(Sigma) = 5.669 · 10 ^(-8) W/m²K^4
T^4 = 1150 : ( 5.669 · 10^(-8) )
T^4 = 202.875 · 10^8
![T = \sqrt[4]{202.857 * 10 ^{8} }](https://tex.z-dn.net/?f=T%20%3D%20%20%5Csqrt%5B4%5D%7B202.857%20%2A%2010%20%5E%7B8%7D%20%7D%20)
T = 3.774 · 10² = 377.4 K
Answer: Equilibrium temperature is 377.4 K.
P = (Sigma) · A · e · T^4
P in = P out
e = 1 for blacktop;
1150 W = (Sigma) · T^4
(Sigma) = 5.669 · 10 ^(-8) W/m²K^4
T^4 = 1150 : ( 5.669 · 10^(-8) )
T^4 = 202.875 · 10^8
T = 3.774 · 10² = 377.4 K
Answer: Equilibrium temperature is 377.4 K.
(a) 2 Hz
The frequency of the nth-harmonic is given by
where
is the fundamental frequency
Therefore, the frequency of the third harmonic of the A () is
while the frequency of the second harmonic of the E () is
So the frequency difference is
(b) 2 Hz
The beat frequency between two harmonics of different frequencies f, f' is given by
In this case, when the strings are properly tuned, we have
- Frequency of the 3rd harmonic of A-note: 1320 Hz
- Frequency of the 2nd harmonic of E-note: 1318 Hz
So, the beat frequency should be (if the strings are properly tuned)
(c) 1324 Hz
The fundamental frequency on a string is proportional to the square root of the tension in the string:
this means that by tightening the string (increasing the tension), will increase the fundamental frequency also*, and therefore will increase also the frequency of the 2nd harmonic.
In this situation, the beat frequency is 4 Hz (four beats per second):
And since the beat frequency is equal to the absolute value of the difference between the 3rd harmonic of the A-note and the 2nd harmonic of the E-note,
and , we have two possible solutions for
:
However, we said that increasing the tension will increase also the frequency of the harmonics (*), therefore the correct frequency in this case will be
1324 Hz