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2 years ago

About light and energy

1 answer:

7 0

E = MC^2. Albert Einstein's proven formula. When mass travels at the square of speed of light, the mass gets converted into energy

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Tính lực tương tác giữa 2 điện tích q1=4.10^-6C và q2=-3.10^-6C cách nhau một 3cm đặt trong dầu hoả = 2

olga2289 [7]

Explanation:

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2 years ago

How do i find acceleration due to gravitational force?

Alexxandr [17]

Answer:

a = 9.8 m/s²

Explanation:

Acceleration due to gravity on Earth is constant, which is 9.8 m/s²

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2 years ago

2.What is the weight of an object that has a mass of 5 kg?<br>

Artemon [7]

Answer:

5 kg

or 11 lbs

Explanation:

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2 years ago

What feelings do learned helplessness and depression have in common?

8_murik_8 [283]

Answer:

Learned helplessness can have a profound impact on mental health and well-being. People who experience learned helplessness are also likely to experience symptoms of depression, elevated stress levels, and less motivation to take care of their physical health. Not everyone responds to experiences the same way.

Explanation:

Brainlist will be very accepted ;D *wink*

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2 years ago

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

2 years ago