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DENIUS [597]
2 years ago
12

Fill in the blank with the correct response.

Physics
1 answer:
viktelen [127]2 years ago
7 0

Answer:

the answer is portfolio

Explanation:

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How Can You Earn Your GED?

  • You must be at least 16 years old to take the GED Test, and you must not be enrolled in high school. ...
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>Rules

You must be at least 16 years old to take the GED Test, and you must not be enrolled in high school. You should also meet state eligibility requirements for the amount of time you've been out of high school. In some states, you have to be out of high school for at least 60 consecutive days before you're allowed to take the test. Contact the program administrator in your state for more details.

You must pass all four subject tests in the GED battery of tests, which means scoring at least a 145 on each test (New Jersey requires a score of at least 150 to pass). Scores range from 100 to 200 on the 2014 version). A score of 165 on each test is the benchmark for college and career readiness, and you will receive an Honors distinction if you reach this score. The four tests take seven and a half hours total and are administered by computer. You don't have to take them all on the same day, but some states enforce specific time frames for completing the tests. Most questions are multiple choice, although there is also a mix of fill-in-the-blank, drag and drop, and select-an-area questions, as well as an essay question on the Reasoning Through Language Arts section.

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Explanation:

I hope I helped and I hope u get what your looking for!!

-Good Luck and Plz mark me brainliest

8 0
2 years ago
A long cylindrical insulating shell has an inner radius of a = 1.41 m and an outer radius of b = 1.67 m. The shell has a constan
Natasha2012 [34]

Answer:

a. E = 122.4 N/C

b. E = 58.2 N/C

c. E = 0

Explanation:

The electric field at an arbitrary point away from the axis of the cylinder can found by applying Gauss’ Law, which states that an electric flux through a closed surface is equal to the total charge enclosed by this surface divided by electric permittivity.

In order to apply this law, we have to draw an imaginary cylindrical surface of arbitrary height ‘h’ and radius ‘r’, which is equal to the point where the E-field is asked.

A. For the outside of the cylinder, we will draw our imaginary surface with r = 1.97.

E2\pi rh = \frac{\lambda V}{\epsilon_0} = \frac{\lambda \pi (b^2 - a^2)h}{\epsilon_0}\\E2\pi (1.97)h = \frac{(5.3\times 10^{-9})\pi(1.67^2 - 1.41^2)h}{\epsilon_0}\\E = 122.4~N/C

B. This time our imaginary surface should be inside the cylinder, therefore the enclosed charge will be less than that of part A.

E2\pi rh = \frac{\lambda V_{enc}}{\epsilon_0} = \frac{\lambda \pi (r^2 - a^2}h{\epsilon_0}\\E2\pi (1.51)h = \frac{5.3\times 10^{-9})\pi(1.51^2 - 1.41^2)h}{\epsilon_0}\\E = 58.2~N/C

C. In this case our imaginary surface will be inside the cylinder, where there is no charge at all. Therefore, the enclosed charge will be zero and the electric field will be zero.

6 0
3 years ago
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