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2 years ago

8

If 200 amperes flow from the positive terminal of a battery and operate the starter motor, how many amperes will flow back to th

e negative terminal?

1 answer:

garri49 [273]2 years ago

5 0

Answer:

$200$ Amperes WB =<em> </em> $200J$

Explanation:

$200$ Amperes will flow back to the negative terminal because both the positive terminal and negative terminal contain the equivalent amount of current that will flow and it works in the starter motor.

To learn more about it, refer

to brainly.com/question/6561461

#SPJ4

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A converging-diverging nozzle has an area ratio of 5.9. (1) Determine the (P0/Pt) values corresponding to the 1st, 2nd, and 3rd

nata0808 [166]

Answer:

Check the explanation

Explanation:

The Total pressure is the overall of fixed or static pressure p, the dynamic pressure q, as well as gravitational head. Total pressure can also be referred to as the measure of the overall energy of the airstream, and is the same to static pressure plus velocity pressure.

kindly check the step by step solution in the attached image below to Determine the (P0/Pt) values corresponding to the 1st, 2nd, and 3rd critical points.

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3 years ago

A counter-flow double pipe heat exchanger is heat heat water from 20 degrees Celsius to 80 degrees Celsius at the rate of 1.2 kg

lakkis [162]

Answer:

$L=107.6m$

Explanation:

Cold water in: $m_{c}=1.2kg/s, C_{c}=4.18kJ/kg\°C, T_{c,in}=20\°C, T_{c,out}=80\°C$

Hot water in: $m_{h}=2kg/s, C_{h}=4.18kJ/kg\°C, T_{h,in}=160\°C, T_{h,out}=?\°C$

$D=1.5cm=0.015m, U=649W/m^{2}K, LMTD=?\°C, A_{s}=?m^{2},L=?m$

Step 1: Determine the rate of heat transfer in the heat exchanger

$Q=m_{c}C_{c}(T_{c,out}-T_{c,in})$

$Q=1.2*4.18*(80-20)$

$Q=1.2*4.18*(80-20)$

$Q=300.96kW$

Step 2: Determine outlet temperature of hot water

$Q=m_{h}C_{h}(T_{h,in}-T_{h,out})$

$300.96=2*4.18*(160-T_{h,out})$

$T_{h,out}=124\°C$

Step 3: Determine the Logarithmic Mean Temperature Difference (LMTD)

$dT_{1}=T_{h,in}-T_{c,out}$

$dT_{1}=160-80$

$dT_{1}=80\°C$

$dT_{2}=T_{h,out}-T_{c,in}$

$dT_{2}=124-20$

$dT_{2}=104\°C$

$LMTD = \frac{dT_{2}-dT_{1}}{ln(\frac{dT_{2}}{dT_{1}})}$

$LMTD = \frac{104-80}{ln(\frac{104}{80})}$

$LMTD = \frac{24}{ln(1.3)}$

$LMTD = 91.48\°C$

Step 4: Determine required surface area of heat exchanger

$Q=UA_{s}LMTD$

$300.96*10^{3}=649*A_{s}*91.48$

$A_{s}=5.07m^{2}$

Step 5: Determine length of heat exchanger

$A_{s}=piDL$

$5.07=pi*0.015*L$

$L=107.57m$

7 0

2 years ago

lisabon 2012 [21]

Answer:

That's a really nice question sadly I don't know the answer I'm replying to you cuz I'm tryna get points so... Sorry

3 0

3 years ago

What is the De Broglie wavelength of an electron under 150 V acceleration?

yanalaym [24]

Answer:

0.1 nm

Explanation

Potential deference of the electron is given as V =150 V

Mass of electron $m=9.1\times 10^{-31}$

Let the velocity of electron = v

Charge on the electron $=1.6\times 10^{-19}C$

plank's constant h = $6.67\times 10^{-34}$

According to energy conservation $eV =\frac{1}{2}mv^2$

$v=\sqrt{\frac{2eV}{M}}=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 150}{9.1\times 10^{-31}}}=7.2627\times 10^{-6}m/sec$

Now we know that De Broglie wavelength $\lambda =\frac{h}{mv}=\frac{6.67\times 10^{-34}}{9.1\times 10^{-31}\times 7.2627\times10^6 }=0.100\times 10^{-9}m=0.1nm$

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3 years ago

Read 2 more answers

since your car had a cylinder volume of 275cm what is its engine displacement if the engine had 4 cylinders

RUDIKE [14]

Answer:

$$\begin{align*}

P(Y−X=m|Y>X)=∑kP(Y−X=m,X=k|Y>X)=∑kP(Y−X=m|X=k,Y>X)P(X=k|Y>X)=∑kP(Y−k=m|Y>k)P(X=k|Y>X).

Explanation:

P(Y−X=m|Y>X)=∑kP(Y−X=m,X=k|Y>X)=∑kP(Y−X=m|X=k,Y>X)P(X=k|Y>X)=∑kP(Y−k=m|Y>k)P(X=k|Y>X).

5 0

2 years ago