Answer:
Beat frequency is determined by the number of beats per second one hears from the sound. It can also be determined by taking the difference in frequency of the two notes that interfere to produce the beat.
Explanation:
Beat frequency is determined by the number of beats per second one hears from the sound.
Assuming the sound pulses is coming from two tuning forks;
- Frequency of the first tuning fork (F₁) is known as 240Hz
- Frequency of the second tuning fork (F₂) is not known
- there are 8 beats in 2 seconds
Beat Frequency = number of beats/time
Beat Frequency = 8/2 = 4Hz
Beat Frequency = F₂ - F₁
±4 = F₂ - 240
4 = F₂ - 240
4+240 = F₂
244 = F₂
F₂ = 244 Hz
-4 = F₂ - 240
-4 + 240 = F₂
F₂ = 236 Hz
The percentage uncertainty in the beat frequency is
4 /240 = 0.0167 = 1.67%
To reduce the percentage uncertainty in the beat frequency
A tuning fork of almost the same frequency as the first tuning fork are used to produce sound pulses, let's say the frequency of the second tuning fork is 241 Hz.
The beat frequency = 241Hz -240Hz = I Hz
Then, the percentage uncertainty = 1/240 = 0.00416 =0.416%
Haploid and diploid cells
Answer:
La palabra silueta se deriva del nombre de Étienne de Silhouette, una ministra de finanzas francesa que, en 1759, se vio obligada por la crisis crediticia de Francia durante la Guerra de los Siete Años a imponer severas demandas económicas al pueblo francés, particularmente a los ricos.
Explanation:
The area of Piston A is 13.88 in^2
What is a hydraulic system?
The hydraulic system functions by using pressurized fluid. The power of liquid fuel in hydraulics is significant and as a result, they are commonly used in heavy equipment.
Hydraulic system simply means a mechanical function that operates through the force of liquid pressure.
Here
A hydraulic system contains 3 pistons.
Pressure, P = 1440 psi
for piston A:
Area, A= 0.625 in^2
Force, F = 20,000 lb
We know that,
Force = Pressure x Area
F = P x A
Area ( piston A) = F / P
A = 20,000 / 1440
A = 13.88 in^2
Hence,
The area of Piston A is 13.88 in^2
Learn more about hydraulic system here:
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Answer:
(a) burnout speed at apogee in the external tank disposal orbit=7.82 km/sec
(b) ΔV required for the OMS-1 =0.066 km/sec
(c) ΔV required for the OMS-2 =0.045 km/sec
(d) This part of the question is explained in detailed way in the attached file.
Explanation:
Detailed explanation of all the parts of the question are given in attached files.