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uranmaximum [27]
4 years ago
11

The switch has been in position a for a long time. At t = 0, the switch moves from position a to position b. The switch is a mak

e-before-break type so there is no interruption of the inductor current. a. Find the expression for i(t) for t ≥ 0. b. What is the initial voltage across the inductor after the switch has been moved to position b? c. Does this initial voltage make sense in terms of circuit behavior? d. How many milliseconds after the switch has been put in position b does the inductor voltage equal 24V? e. Plot both i(t) and v(t) versus t

Engineering
1 answer:
SashulF [63]4 years ago
6 0

Answer:

Hello there, please check explanations for step by step procedures to get answers.

Explanation:

Given that;

mkasblog

College Engineering 10+5 pts

The switch has been in position a for a long time. At t = 0, the switch moves from position a to position b. The switch is a make-before-break type so there is no interruption of the inductor current. a. Find the expression for i(t) for t ≥ 0. b. What is the initial voltage across the inductor after the switch has been moved to position b? c. Does this initial voltage make sense in terms of circuit behavior? d. How many milliseconds after the switch has been put in position b does the inductor voltage equal 24V? e. Plot both i(t) and v(t) versus t

See attachment for more clearity answer.

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Explanation:

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3 years ago
How can you use the IPDE Process to help protect motorcyclists while driving?
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3 years ago
3. (5%) you would like to physically separate different materials in a scrap recycling plant. describe at least one method that
Likurg_2 [28]

One of the methods that are used to separate polymers, aluminium alloys, and steels from one another is the Gravitation Separation method.

One straightforward technique is to run the mixture through a magnet, which will keep the steel particles on the magnet and separate them from the polymer.

What is the Gravitation Separation method?

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To learn more about the Polymer from the given link

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8 0
2 years ago
A rocket is launched vertically from rest with a constant thrust until the rocket reaches an altitude of 25 m and the thrust end
hjlf

Answer:

a) v=19.6 m/s

b) H=19.58 m

c) v_{f}=29.57 m/s  

Explanation:

a) Let's calculate the work done by the rocket until the thrust ends.

W=F_{tot}h=(F_{thrust}-mg)h=(35-(2*9.81))*25=384.5 J

But we know the work is equal to change of kinetic energy, so:

W=\Delta K=\frac{1}{2}mv^{2}

v=\sqrt{\frac{2W}{m}}=19.6 m/s

b) Here we have a free fall motion, because there is not external forces acting, that is way we can use the free-fall equations.

v_{f}^{2}=v_{i}^{2}-2gh

At the maximum height the velocity is 0, so v(f) = 0.

0=v_{i}^{2}-2gH

H=\frac{19.6^{2}}{2*9.81}=19.58 m  

c) Here we can evaluate the motion equation between the rocket at 25 m from the ground and the instant before the rocket touch the ground.

Using the same equation of part b)

v_{f}^{2}=v_{i}^{2}-2gh

v_{f}=\sqrt{19.6^{2}-(2*9.81*(-25))}=29.57 m/s

The minus sign of 25 means the zero of the reference system is at the pint when the thrust ends.

I hope it helps you!

6 0
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Mera bhai bhot pad marta h loll usko bvasir h kya koi upaye h pls tell pls​
ycow [4]

Answer:

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