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forsale [732]
2 years ago
5

If changing employment what do you need to do? Email your new employer information to the Deptartment of International Graduate

Services Nothing, you do not need to submit anything Submit a new CPT application and attach offer letter Call the Department of International Graduate Services
Engineering
1 answer:
likoan [24]2 years ago
7 0

Answer:Submit a new CPT application and attach offer letter.

Explanation:CPT(curricular practical training) is a practical training program for international students in the United States of America it is offered by employers who sponsor students through cooperative agreements with the schools. Their are different sub o organs making up the CPT they include Internship,corporative education etc.

For one to be eligible you have the following

(1) Legal Academic enrollment on a full-time basis for one academic year(two consecutive semesters or terms). Some academic programs requires immediate enrollment.

(2) Be in lawful F-1(visa issued to international students studying in American Colleges or universities) status.

(3) Have U-M approved health insurance.

(4) Have an employment offer.

Certain states may have some additional requirements.

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A square loop of wire surrounds a solenoid. The side of the square is 0.1 m, while the radius of the solenoid is 0.025 m. The sq
Semmy [17]

Answer:

I=9.6×e^{-8}  A

Explanation:

The magnetic field inside the solenoid.

B=I*500*muy0/0.3=2.1×e ^-3×I.

so the total flux go through the square loop.

B×π×r^2=I×2.1×e^-3π×0.025^2

=4.11×e^-6×I

we have that

(flux)'= -U

so differentiating flux we get

so the inducted emf in the loop.

U=4.11×e^{-6}×dI/dt=4.11×e^-6×0.7=2.9×e^-6 (V)

so, I=2.9×e^{-6}÷30

I=9.6×e^{-8}  A

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3 years ago
A teenager was pulling a prank and placed a large stuffed penguin in the middle of a roadway. A driver is traveling on this leve
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3 years ago
A 150 MVA, 24 kV, 123% three-phase synchronous generator supplies a large network. The network voltage is 27 kV. The phase angle
Aleks04 [339]

Answer:

the generator induced voltage is 60.59 kV

Explanation:

Given:

S = 150 MVA

Vline = 24 kV = 24000 V

X_{s} =1.23(\frac{V_{line}^{2}  }{s} )=1.23\frac{24000^{2} }{1500} =4723.2 ohms

the network voltage phase is

V_{phase} =\frac{V_{nline} }{\sqrt{3} } =\frac{27}{\sqrt{3} } =15.58kV

the power transmitted is equal to:

|E|=\frac{P*X_{s} }{3*|V_{phase}|sinO } ;if-O=60\\|E|=\frac{300*4.723}{3*15.58*sin60} =34.98kV

the line induced voltage is

|E_{line} |=\sqrt{3} *|E|=\sqrt{3} *34.98=60.59kV

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C

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