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inessss [21]
3 years ago
7

Several different loads are going to be used with the voltage divider from Part A. If the load resistances are 300 kΩkΩ , 200 kΩ

kΩ , and 100 kΩkΩ , what is the output voltage that is the most different from the design output voltage vovov_o = 6 VV ? Express your answer in V to three significant figures.

Engineering
1 answer:
harina [27]3 years ago
7 0

Answer:

attached below

Explanation:

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If you’re designing a new pair of shoes, what are some things you’ll have to think about ??
shusha [124]

Answer:

leather, textiles, synthetics, shoelaces, glue, reinforcements, hardware, logos and midsoles

4 0
3 years ago
Read 2 more answers
A weighted, frictionless piston-cylinder device initially contains 5.25 kg of R134a as saturated vapor at 500 kPa. The container
kykrilka [37]

Answer:

-6.326 KJ/K

Explanation:

A) the entropy change is defined as:

delta S_{12}=\int\limits^2_1  \, \frac{dQ}{T}

In an isobaric process heat (Q) is defined as:

Q= m*Cp*dT

Replacing in the equation for entropy  

delta S_{12}=\int\limits^2_1 \frac{m*Cp*dT}{T}

m is the mass and Cp is the specific heat of R134a. We can considerer these values as constants so the expression for entropy would be:  

delta S_{12}= m*Cp*\int\limits^2_1 \frac{ dT }{T}  

Solving the integral we get the expression to estimate the entropy change in the system  

delta S_{12}= m*Cp *ln(\frac{T_{2}}{T_{1}})

The mass is 5.25 Kg and Cp for R134a vapor can be consulted in tables, this value is 0.85\frac{kJ}{Kg*K}

We can get the temperature at the beginning knowing that is saturated vapor at 500 KPa. Consulting the thermodynamic tables, we get that temperature of saturation at this pressure is: 288.86 K

The temperature in the final state we can get it from the heat expression, since we know how much heat was lost in the process (-976.71 kJ). By convention when heat is released by the system a negative sign is used to express it.

Q= m*Cp*dT

With dt=T_{2}-T_{1} clearing for T2 we get:

T_{2}=\frac{Q}{m*Cp}+T1= \frac{-976.71kJ}{5.25Kg*0.85\frac{kJ}{Kg*K}}+288.86 K =69.98 K

Now we can estimate the entropy change in the system

delta S_{12}= m*Cp*ln(\frac{T_{2}}{T_{1}})= 5.25Kg*0.85\frac{kJ}{Kg*K}*ln(\frac{69.98}{288.861})= -6.326\frac{kJ}{K}

The entropy change in the system is negative because we are going from a state with a lot of disorder (high temperature) to one more organize (less temperature. This was done increasing the entropy of the surroundings.  

b) see picture.

3 0
3 years ago
Consider a simple ideal Rankine cycle and an ideal regenerative Rankine cycle with one open feedwater heater. The two cycles are
borishaifa [10]

Answer:

They both have the same efficiency.

Explanation:

The simple ideal Rankine cycle and an ideal regenerative Rankine cycle with one open feedwater heater would both have the same efficiency because the extraction steam would just create a mini cycle that recirculates. The energy given to the feedwater heater is proportional to the added heat in the boiler to the feedwater in the simple cycle to raise its temperature to the same boiler inlet condition.

Therefore in comparison, the efficiency is the same for both.

4 0
3 years ago
Consider the circuit below where R1 = R4 = 5 Ohms, R2 = R3 = 10 Ohms, Vs1 = 9V, and Vs2 = 6V. Use superposition to solve for the
VladimirAG [237]

Answer:

The value of v2 in each case is:

A) V2=3v for only Vs1

B) V2=2v for only Vs2

C) V2=5v for both Vs1 and Vs2

Explanation:

In the attached graphic we draw the currents in the circuit. If we consider only one of the batteries, we can consider the other shorted.

Also, what the problem asks is the value V2 in each case, where:

V_2=I_2R_2=V_{ab}

If we use superposition, we passivate a battery and consider the circuit affected only by the other battery.

In the first case we can use an equivalent resistance between R2 and R3:

V_{ab}'=I_1'R_{2||3}=I_1'\cdot(\frac{1}{R_2}+\frac{1}{R_3})^{-1}

And

V_{S1}-I_1'R_1-I_1'R_4-I_1'R_{2||3}=0 \rightarrow I_1'=0.6A

V_{ab}'=I_1'R_{2||3}=3V=V_{2}'

In the second case we can use an equivalent resistance between R2 and (R1+R4):

V_{ab}''=I_3'R_{2||1-4}=I_3'\cdot(\frac{1}{R_2}+\frac{1}{R_1+R_4})^{-1}

And

V_{S2}-I_3'R_3-I_3'R_{2||1-4}=0 \rightarrow I_3'=0.4A

V_{ab}''=I_3'R_{2||1-4}=2V

If we consider both batteries:

V_2=I_2R_2=V_{ab}=V_{ab}'+V_{ab}''=5V

7 0
4 years ago
What substance do humans give to livestock to help them stay healthy?
Scorpion4ik [409]

Answer: nutrients

Explanation: to help the cells have energy

3 0
3 years ago
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