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otez555 [7]
2 years ago
10

Consider a 50-mH inductor. a. Express the voltage across the inductor and then evaluate it at t = 0.25 s if iL (t) = 5e −2t + 3t

e−2t −2 A. Be certain to simplify your expression.
Engineering
1 answer:
IrinaK [193]2 years ago
3 0

Answer:

V=L(di/dt) where i is current, V=0.208

Explanation:

using expression iL(t)=5e-2t+3te-2t-2 and L=0.05H(50/1000)

V=0.05*d(5e-2t+3te-2t-2)/dt

since there is no power of e, I'll assume the power to be 1

V=0.05*(-2+3e-2)

at t=0.25

V=0.15e-0.2

V=0.208

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What is the composition, in atom percent, of an alloy that consists of 4.5 wt% Pb and 95.5 wt% Sn?
jeka57 [31]

Answer: Option A is correct -- 2.6 at% Pb and 97.4 at% Sn.

Explanation:

Option A is the only correct option -- 2.6 at% Pb and 97.4 at% Sn. While option B, which is 7.6 at% Pb and 92.4 at% Sn. and option C, which is 97.4 at% Pb and 2.6 at% Sn. and option D, which is 92.4 at% Pb and 7.6 at% Sn. are wrong.

6 0
3 years ago
The inlet and exhaust flow processes are not included in the analysis of the Otto cycle. How do these processes affect the Otto
lara31 [8.8K]

Answer:

Suction and exhaust processes do not affect the performance of Otto cycle.

Explanation:

Step1

Inlet and exhaust flow processes are not including in the Otto cycle because the effect and nature of both the process are same in opposite direction.

Step2

Inlet process or the suction process is the process of suction of working fluid inside the cylinder. The suction process is the constant pressure process. The exhaust process is the process of exhaust out at constant pressure.

Step3

The suction and exhaust process have same work and heat in opposite direction. So, net effect of suction and exhaust processes cancels out. The suction and exhaust processes are shown below in P-V diagram of Otto cycle:

Process 0-1 is suction process and process 1-0 is exhaust process.

7 0
3 years ago
Sketches are a very efficient way to share ideas.<br> True<br> False
timurjin [86]

Answer:

yes

Explanation:

4 0
3 years ago
Read 2 more answers
Expalin the application of diesel cycle in detail.
mars1129 [50]

Explanation:

Diesel cycle:

        All diesel engine work on diesel cycle .In diesel cycle there are four process .These processes are as follows

1. Adiabatic reversible compression

2.Heat addition at constant pressure

3.Adiabatic reversible expansion

4.Constant volume heat rejection

In general compression ratio in diesel engine is high as compare to petrol engine.But the efficiency of diesel cycle is less as compare to petrol cycle for same compression ratio.

Applications of diesel cycle:

Generally diesel cycle used for heavy vehicle or equipment because heavy vehicle or equipment is required high initial torque.So this cycle have lots of applications such as in industrial machining,in trucks,power plant,in mining ,in defense or military,large motors ,compressor and pump etc.

   

5 0
3 years ago
The roof of a refrigerated truck compartment consists of a layer of foamed urethane insulation (t2 = 21 mm, ki = 0.026 W/m K) be
lakkis [162]

Answer:

Tso = 28.15°C

Explanation:

given data

t2 = 21 mm

ki = 0.026 W/m K

t1 = 9 mm

kp = 180 W/m K

length of the roof is L = 13 m

net solar radiation into the roof = 107 W/m²

temperature of the inner surface Ts,i = -4°C

air temperature is T[infinity] = 29°C

convective heat transfer coefficient h = 47 W/m² K

solution

As when energy on the outer surface at roof of a refrigerated truck that is balance as

Q = \frac{T \infty - T si }{\frac{1}{hA}+\frac{t1}{AKp}+\frac{t2}{AKi}+\frac{t1}{aKp}}       .....................1

Q = \frac{T \infty - Tso}{\frac{1}{hA}}                         .....................2

now we compare both equation 1 and 2 and put here value

\frac{29-(-4)}{\frac{1}{47}+\frac{2\times0.009}{180}+\frac{0.021}{0.026}} = \frac{29-Tso}{\frac{1}{47}}            

solve it and we get

Tso = 28.153113

so Tso = 28.15°C

3 0
2 years ago
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