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Agata [3.3K]
3 years ago
10

For a Van de Graaff generator to acquire a charge on its surface what must happen?

Physics
1 answer:
horsena [70]3 years ago
3 0
The answer is D. A Van de Graaff generator is an electrostatic generator which utilizes a moving belt to aggregate an electric charge on an empty metal globe on the highest point of a protected section, making high electric possibilities. It delivers high voltage coordinate current (DC) power at low current levels.
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Calculate the force of an object that has a mass of 10kg and an acceleration of 4m/s²
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The answer is A-40N.
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What is the speed of sound at the atmospheric temperature of 30°C?
12345 [234]

The parcel will undergo projectile motion, which means that it will have motion in both the horizontal and vertical direction.


First, we determine how long the parcel will fall using:


s = ut + 1/2 at²


where s will be the height, u is the initial vertical velocity of the parcel (0), t is the time of fall and a is the acceleration due to gravity.


5.5 = (0)(t) + 1/2 (9.81)(t)²

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A

4 0
3 years ago
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A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate
Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is \bf{2.81 \times 10^{4}~n~C^{-1}}.

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and '\epsilon_{0}' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm

Part(c):

If we apply Gauss' law of electrostatics, then

&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}

3 0
3 years ago
What phytophotodermatitis caused by?
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Phytophotodermatitis happens when certain plant chemicals cause the skin to become inflamed following exposure to sunlight.

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3 years ago
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