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Varvara68 [4.7K]
2 years ago
13

FD=CD*((P*(V^2)*A)/2)Please solve for V

Engineering
1 answer:
REY [17]2 years ago
7 0

Answer:

The answer is V = √2FD ÷ CD × PA

Explanation:

FD = CD × PV²A ÷ 2

V² = 2FD ÷ CD × PA

V = √2FD ÷ CD × PA

Thus, The value of V is V = √2FD ÷ CD × PA

 

<u>-TheUnknownScientist 72</u>

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What is the primary function of NCEES?
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3 years ago
A sensor produces a signal with amplitude 15 mV. A voltage amplifier must amplify the signal such that the amplitude of the outp
Nady [450]

Answer:

42.50 dB

Explanation:

Determine the minimum voltage gain

amplitude of input signal ( Vi ) = 15 mV

amplitude of output signal ( Vo) = 2 V

Vo = 2 v

therefore ; minimum gain = Vo / Vi =  2 / ( 15 * 10^-3 )

                                                         = 133.33

Minimum gain in DB = 20 log ( 133.33 )

                                  = 42.498 ≈ 42.50 dB

8 0
3 years ago
What are the optical properties of steel
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Answer:

A selective surface with large absorption for solar radiation and high reflectance for thermal infrared radiation was produced by use of surface oxidation of stainless steel. The surfaces were studied for use with concentrated light in a solar power plant at temperatures of 400°C and higher.

In order to investigate the relation between surface treatment and optical properties, stainless steels (AISI 304 and 430) which were submitted to different chemical and mechanical surface treatments, were used. To increase the spectral selectivity, these surfaces were treated in air and in vacuum at different temperatures and times. The optical properties of these films were investigated. Visual and infrared spectral absorptances were measured at room temperature. The thermal hemispherical emittance and absorptance were obtained by a calorimetric method at 200°C. It was noticed that these chemically and mechanically treated stainless steel surfaces have good spectral properties without further oxidations. This is very important for high temperature uses. The best values are found for samples 7 and 8 under vacuum and air. These two samples with mechanically ground surfaces retained their selectivity and specularity after several hours oxidation. One can conclude that the surface ground treatment confers good selectivity on the steel surfaces for use in concentrating solar collectors with a working temperature of 500°C.

Sample surfaces were subjected to long temperature ageing tests in order to gain some idea of the thermal stability of the surfaces. The results promise better-performing surface and the production of durable selective finishes at, possibly, lower cost than competing processes.

Explanation:

3 0
3 years ago
P10.12. A certain amplifier has an open-circuit voltage gain of unity, an input resistance of and an output resistance of The si
klio [65]

complete question

A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?

Answer:

3.03 V  0.184 W

2.499 mV  125*10^-9 W

Explanation:

First, apply voltage-divider principle to the input circuit: 1

V_{i}= (R_i/R_i+R_s) *V_s = 10^6/10^6+(0.1*10^6)\\*5

    = 4.545 V

The voltage produced by the voltage-controlled source is:

A_voc*V_i = 4.545 V

We can find voltage across the load, again by using voltage-divider principle:  

V_o = A_voc*V_i*(R_o/R_l+R_o)

      = 4.545*(100/100+50)

      = 3.03 V  

Now we can determine delivered power:  

P_L = V_o^2/R_L

      = 0.184 W

Apply voltage-divider principle to the circuit:  

V_o = (R_o/R_o+R_s)*V_s

       = 50/50+100*10^3*5

       = 2.499 mV

Now we can determine delivered power:  

P_l = V_o^2/R_l

     = 125*10^-9 W

Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.  

4 0
3 years ago
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