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schepotkina [342]
3 years ago
10

Help with these please?

Physics
2 answers:
Anuta_ua [19.1K]3 years ago
5 0

E=kq/r^2

q=(E*r^2)/k

q=(.086N/C)(1.7m^2)/(8.99*10^9N*m^2/C^2)

q=2.76*10^-11 C

q=2.8*10^-11 C

Anettt [7]3 years ago
4 0

Another badly worded question. The Field is always directed away from the point charge. The question is the direction towards q or away from it and that's not much of an improvement.

Formula

E = k * q / r^2

Givens

k = 8.99 * 10^9 Nm^2/c^2

q = ??

E = 8.6*10^-2

r = 1.7 m

Solution

8.6*10^-2 * 1.7^2 / 8.99 * 10^9 = q

0.24 / 8.99 *10^9 = q

q = 2.8 * 10^-11

A but the wording is so bad that you have to guess what the question means. q is positive because the charge is going to the right away from the test charge.

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Answer:

The force becomes 16 times what it is now.

Explanation:

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F = G * m1 * m2 / r^2

When you do what you have described, you are setting a stage that not even the USS Enterprise (Star Trek) can get out of. The increase is huge.

If you double m1 and m2 and don't do anything to r, you've already increased the force by 4 times. (2m1 * 2m2 = 4 * m1 * m2)

But you are not finished. If you 1/2 the distance, you are again increasing the Force by 4 times. 1 / (2r) ^2 = 1/ 4* r^2

Because this is in the denominator, the 1/4 is going to flip to the numerator.

So the total increase is going to be 4 * (4 * m1 * m2) = 16 * m1 * m2.

Think about what that means. If you were out golfing, your drives would be roughly 1/16 times as far as they are now. Also you would be lugging around 16 times your weight around the golf course. My feeling is that you would never finish 5 holes at that rate.

3 0
2 years ago
Give a calculate answer to show that the two values (English system and metric system) for the Planck Constant are equivalent.
Nataly_w [17]

Answer:

Given values of Planck Constant are equivalent in English system and metric system.

Explanation:

Value of Planck's constant is given in English system as 4.14 x 10⁻¹⁵eV s.

Converting this in to metric system .

We have 1 eV = 1.6 x 10⁻¹⁹ J

Converting

     4.14 x 10⁻¹⁵eV s = 4.14 x 10⁻¹⁵x 1.6 x 10⁻¹⁹ = 6.63 x 10⁻³⁴ Joule s

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7 0
3 years ago
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Answer:

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5 0
3 years ago
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ki77a [65]
The answer is <em>Compressional Stress
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4 0
3 years ago
1 kg air in a piston-cylinder assembly is heated at constant pressure, resulting the expansion of the volume. the initial temper
rewona [7]

Answer:

57,42 KJ

Explanation:

By a isobaric proces, the expresion for the works in the jpg adjunt. Then:

W = Pa(Vb - Va) = Pa*Vb - Pa*Va ---(1)

By the ideal gases law: PV=RTn

Then, in (1): (remember Pa = Pb)

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Since we have 1 Kg air: How much is this in moles?

From bibliography: 28.96 g/mol

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4 0
3 years ago
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