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sp2606 [1]
3 years ago
10

A pressure sensor was used to measure the unsteady pressure in a cylinder. The sensor output was acquired for 15 seconds at a ra

te of 202 Hz and spectral analysis was performed using FFT. What is the maximum frequency that will be displayed on the power spectrum plot
Physics
1 answer:
4vir4ik [10]3 years ago
8 0

Answer:

Maximum frequency on power spectrum plot = 101 Hz

Explanation:

Given:

Time taken for output = 15 seconds

Frequency rate = 202 Hz

Find:

Maximum frequency on power spectrum plot

Computation:

Maximum frequency = Given frequency rate / 2

Maximum frequency on power spectrum plot = Frequency rate / 2

Maximum frequency on power spectrum plot = 202 / 2

Maximum frequency on power spectrum plot = 101 Hz

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A pure jet engine propels an aircraft at 240 m/s through air at 45 kPa and −13°C. The inlet diameter of this engine is 1.6 m, th
erica [24]
QUESTION: A pure jet engine propels and aircraft at 340 m/s through air at 45 kPa and -13C. The inlet diameter of this engine is 1.6 m, the compressor pressure ratio is 13, and the temperature at the turbine inlet is 557C. Determine the velocity at the exit of this engines nozzle and the thrust produced.

ANSWER: Due to the propulsion from the inlet diameter of this engine bring 1.6 m allows the compressor rations to radiate allowing thrust propultion above all velocitic rebisomes.
6 0
2 years ago
Hey can anyone please help me with this it’s due in few hours and I’m stuck with ittt
Ray Of Light [21]

Answer:

Check body of the explanation

Explanation:

Ooook, quick theory rushdown. if you're at a depth of h in a tank of a fluid, the pressure is the sum of the atmosferic pressure (if the tank is open on top) plus a term which is the product of acceleration of gravity - about 10 ms^-^2, the density of whatever you're sinking in, and the depth at which you are. In formula, p(h) = p_0 + \rho g h, and the pressure is the same for every point of the tank at the same depth.

At this point, we can start answering!

1a. The pressure at A is - not counting atmosferic pressure - 1000 * 10 * 1 = 10^4 Pa, while in B is 1000*10*2 = 2*10^4 Pa, so it's half of it.

1b. The two points are at the same depth, so the pressure is the same - they would be even if the two cilinders weren't linked!

1c. Ditto. Same depth? same pressure!

1d. Usual equation, this time density is 800. Pressure is 800*10*2 = 1,6*10^4 Pa: Since the density is 4/5 of water, the pressure is also 4/5 of the one exerted by water

2a. The volume is simply the product, so 4m*3m*2m = 24m^3

2b. Density is defined as mass over volume, so you simply multiply the volume you found earlier by the density of paraffine: 800* 24 = 1,92 *10^4kg

2c. Weight is defined as the mass of something times the acceleration due to gravity, in our case it's 1.92 *10^4 kg * 10 ms^{-2} = 1.92 * 10^5 N

2d. \rho gh again, what a surprise! 800 {kg \over m^3} * 10 {N \over kg}} * 2 m = 1,6* 10^4 {N\over m^2} =1.6*10^4 Pa

3. Yet again, \rho gh. 1000 {kg \over m^3} * 10 {N \over kg}} * 2 m = 2* 10^4 {N\over m^2} =2*10^4 Pa

4 0
2 years ago
What is the magnitude of your displacement when you follow directions that tell you to walk 225 m in one direction, make a 90° t
Gekata [30.6K]

Start by facing East. Your first displacement is the vector

<em>d</em>₁ = (225 m) <em>i</em>

Turning 90º to the left makes you face North, and walking 350 m in this direction gives the second displacement,

<em>d</em>₂ = (350 m) <em>j</em>

Turning 30º to the right would have you making an angle of 60º North of East, so that walking 125 m gives the third displacement,

<em>d</em>₃ = (125 m) (cos(60º) <em>i</em> + sin(60º) <em>j</em> )

<em>d</em>₃ ≈ (62.5 m) <em>i</em> + (108.25 m) <em>j</em>

The net displacement is

<em>d</em> = <em>d</em>₁ + <em>d</em>₂ + <em>d</em>₃

<em>d</em> ≈ (287.5 m) <em>i</em> + (458.25 m) <em>j</em>

and its magnitude is

|| <em>d</em> || = √[ (287.5 m)² + (458.25 m)² ] ≈ 540.973 m ≈ 541 m

7 0
3 years ago
A string is tied between two posts separated by 2.4 m. When the string is driven by an oscillator at frequency 567 Hz, 5 points
Alex787 [66]

Explanation:

The given data is as follows.

       Length (l) = 2.4 m

       Frequency (f) = 567 Hz

Formula to calculate the speed of a transverse wave is as follows.

                  f = \frac{5}{2l} \times v

Putting the gicven values into the above formula as follows.

                  f = \frac{5}{2l} \times v

                 567 Hz = \frac{5}{2 \times 2.4 m} \times v

                      v = 544.32 m/s

Thus, we can conclude that the speed (in m/s) of a transverse wave on this string is 544.32 m/s.

5 0
3 years ago
BRAINLIEST FOR YOU IF YOU ANSWER
luda_lava [24]

Answer:

d

Explanation:

i think

5 0
2 years ago
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