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joja [24]
3 years ago
6

An empty parallel plate capacitor is connected between the terminals of a 9.0-V battery and charged up. The capacitor is then di

sconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor
Physics
1 answer:
DedPeter [7]3 years ago
3 0

Answer:

The new voltage between the plates of the capacitor is 18 V

Explanation:

The charge on parallel plate capacitor is calculated as;

q = CV

Where;

V is the battery voltage

C is the capacitance of the capacitor, calculated as;

C = \frac{\epsilon _0A}{d} \\\\q =CV = (\frac{\epsilon _0A}{d})V = \frac{\epsilon _0A V}{d}

q = \frac{\epsilon _0A V}{d}

where;

ε₀ is permittivity of free space

A is the area of the capacitor

d is the space between the parallel plate capacitors

If only the space between the capacitors is doubled and every other parameter is kept constant, the new voltage will be calculated as;

q = \frac{\epsilon _0A V}{d} \\\\\frac{\epsilon _0A V}{d}  = \frac{\epsilon _0A V}{d} \\\\\frac{V_1}{d_1}  = \frac{V_2}{d_2} \\\\V_2 = \frac{V_1d_2}{d_1} \\\\(d_2 = 2d_1)\\\\V_2 = \frac{V_1*2d_1}{d_1} \\\\(V_1 = 9V)\\\\V_2 = \frac{9*2d_1}{d_1} \\\\V_2 = 9*2\\\\V_2 = 18 \ V

Therefore, the new voltage between the plates of the capacitor is 18 V

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Answer:

a = 1.20m\s^{2}

Explanation:

225 x 9.8

= 441N

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710 - 441

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\frac{269}{225} = 1.19

a = 1.20

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For a wave, the _____ the amplitude, the _____ energy the wave carries.
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Answer:

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Explanation:

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A radioactive material has a count rate of 400 per minute. It has a half life of 40 years. How long will it take to decay to a r
cestrela7 [59]

Answer:

160 years.

Explanation:

From the question given above, the following data were obtained:

Initial count rate (Cᵢ) = 400 count/min

Half-life (t½) = 40 years

Final count rate (Cբ) = 25 count/min

Time (t) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Initial count rate (Cᵢ) = 400 count/min

Final count rate (Cբ) = 25 count/min

Number of half-lives (n) =?

Cբ = 1/2ⁿ × Cᵢ

25 = 1/2ⁿ × 400

Cross multiply

25 × 2ⁿ = 400

Divide both side by 25

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2ⁿ = 16

Express 16 in index form with 2 as the base

2ⁿ = 2⁴

n = 4

Thus, 4 half-lives has elapsed.

Finally, we shall determine the time taken for the radioactive material to decay to the rate of 25 counts per minute. This can be obtained as follow:

Half-life (t½) = 40 years

Number of half-lives (n) = 4

Time (t) =?

n = t / t½

4 = t / 40

Cross multiply

t = 4 × 40

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Thus, it will take 160 years for the radioactive material to decay to the rate of 25 counts per minute.

7 0
2 years ago
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sladkih [1.3K]

Answer:

q₃ = -4.81 nC

Explanation:

We can use the Gauss Law here:

∅ = q/∈₀

where,

∅ = Net Flux = - 216 N.m²/C

q = total charge enclosed inside sphere = ?

∈₀ = permittivity of free space = 8.85 x 10⁻¹² C/N.m²

Therefore,

- 216 N.m²/C = q / 8.85 x 10⁻¹² C²/N.m²

q = (-216 N.m²/C)(8.85 x 10⁻¹² C²/N.m²)

q = - 1.91 nC

So, the total charge will be sum of all three charges:

q = q₁ + q₂ + q₃

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q₃ = - 1.91 nC - 1.74 nC - 1.16 nC

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2 years ago
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nevsk [136]
We can use the kinematic equation
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(Vf)^2 = 2000
Vf = about 44.721
or 44.7 m/s   [if you are rounding this by significant figures]
8 0
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