Answer:
Length of the pipe = 53.125 cm
Explanation:
given data
harmonic frequency f1 = 800 Hz
harmonic frequency f2 = 1120 Hz
harmonic frequency f3 = 1440 Hz
solution
first we get here fundamental frequency that is express as
2F = f2 - f1 ...............1
put here value
2F = 1120 - 800
F = 160 Hz
and
Wavelength is express as
Wavelength = Speed ÷ Fundamental frequency ................2
here speed of waves in air = 340 m/s
so put here value
Wavelength =340 ÷ 160
Wavelength = 2.125 m
so
Length of the pipe will be
Length of the pipe = 0.25 × wavelength ......................3
put here value
Length of the pipe = 0.25 × 2.125
Length of the pipe = 0.53125 m
Length of the pipe = 53.125 cm
Answer:
C2, C1, C4, C5 and C6 are in parallel. Therefore, we use the formula Cp = C1 + C2 + ....
Cp = C2 + C1 + C4 + C5 + C6 = ( 7 * 10 ^-3) + (18 * 10^-6) + (0.8F) + (200 * 10^-3 F) + (750 * 10^-6) = 1.008F
Now, Cp will become one capacitor and it will be aligned with C3, therefore it will now become a circuit in series.
We use the formula: 1/Cs = 1/C1 + 1/C2 + .... + ....1/Cn
Thus,
1/Cs = 1/C3 + 1/Cp
1/Cs = 1/(14 * 10^-3 F) + 1/(1.008F)
Cs = 1.4 * 10 ^-2 or if we do not round too much it will give exactly 0.0138 F
So the answer should be a)
Answer: Next time you create a question, add an image or PDF. Because I do not know the question. So, may you please create a new question?
Their relative speed is the sum of 60 and 40 or 100km/hr. They will travel the 150km in 1.5 hrs. When two object approach each other, the closing speed is just the sum of the speeds, therefore, the closing speed is your case is 100kph. So they will meet in 1.5 hours.
The answer is D. time really does pass more slowly in a rest frame of reference relative to a frame of reference that is moving
