Organ pipe A, with both ends open, has a fundamental frequency of 360 Hz. The third harmonic of organ pipe B, with one end open,
1 answer:
Answer:
The length of the pipe A is = 0.4763 m & the length of the pipe B is
0.357 m
Explanation:
Fundamental frequency = 360 Hz
Velocity = 343
(a). Length of the pipe is given by
Put all the values in above equation we get
= 0.4763 m
(b). Given that
The third harmonic of organ pipe B = the second harmonic of pipe A
Thus
Put all the values in above formula we get
0.357 m
Therefore the length of the pipe A is = 0.4763 m & the length of the pipe B is
0.357 m
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Answer:
Q = 160.36[kJ], is the heat lost.
Explanation:
This is a thermodynamic problem where we can find the latent heat of vaporization, at a constant temperature of 100 [°C].
We know that for steam at 100[C], the enthalpy is
For liquid water at 100[C], the enthalpy is
Therefore
The amout of heat is given by:
Answer:
4.8 m/s
Explanation:
When she catches the train,
- They will have travelled the same distance.and
- Their speeds will be equal
The formula for the distance covered by the train is
d = ½at² = ½ × 0.40t² = 0.20t²
The passenger starts running at a constant speed 6 s later, so her formula is
d = v(t - 6.0)
The passenger and the train will have covered the same distance when she has caught it, so
(1) 0.20t² = v(t - 6.0)
The speed of the train is
v = at = 0.40t
The speed of the passenger is v.
(2) 0.40t = v
Substitute (2) into (1)
0.20t² = 0.40t(t - 6.0) = 0.40t² - 2.4 t
Subtract 0.20t² from each side
0.20t² - 2.4t = 0
Factor the quadratic
t(0.20t - 2.4) = 0
Apply the zero-product rule
t =0 0.20t - 2.4 = 0
0.20t = 2.4
(3) t = 12
We reject t = 0 s.
Substitute (3) into (2)
0.40 × 12 = v
v = 4.8 m/s
The slowest constant speed at which she can run and catch the train is 4.8 m/s.
A plot of distance vs time shows that she will catch the train 6 s after starting. Both she and the train will have travelled 28.8 m. Her average speed is 28.8 m/6 s = 4.8 m/s.
