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mixer [17]
3 years ago
5

How are electric potential difference and electric potential energy related?

Physics
1 answer:
Anna007 [38]3 years ago
3 0

Answer: NOT B) Electric potential energy is the charge divided by the electric potential difference.

Explanation:

just missed it on edg

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he membrane that surrounds a certain type of living cell has a surface area of 6.0 x 10-9 m2 and a thickness of 1.6 x 10-8 m. As
k0ka [10]

Answer:

1.54481175\times 10^{-12}\ C

Explanation:

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

A = Area = 6\times 10^{-9}\ m^2

d = Thickness = 1.6\times 10^{-8}\ m

k = Dielectric constant = 5.4

V = Voltage = 86.2 mV

Charge is given by

Q=CV\\\Rightarrow Q=k\epsilon\dfrac{A}{d}V\\\Rightarrow Q=5.4\times 8.85\times 10^{-12}\times \dfrac{6\times 10^{-9}}{1.6\times 10^{-8}}\times 86.2\times 10^{-3}\\\Rightarrow Q=1.54481175\times 10^{-12}\ C

The charge on the outer surface is 1.54481175\times 10^{-12}\ C

4 0
3 years ago
A uniform thin circular rubber band of mass M and spring constant k has an original radius R?
Gemiola [76]

Answer: Yes

Explanation:

A ruber band can be described by his mas, M, the spring constant K, and his rest radius, wich is the radius of the circular rubber band when there is no force applied to it, so yes, a uniform thin circular rubber band of mass M and spring constant k has an original radius R

6 0
3 years ago
What factor affects the speed of a water wave?
8090 [49]
The speed of a wave is dependant on four factors: wavelength, frequency, medium, and temperature.
6 0
2 years ago
A compound microscope has an objective lens of focal length 1.40 cm and an eyepiece with a focal length of 2.20 cm. The objectiv
olchik [2.2K]

Answer:

magnification is - 159

objective distance is 3.85 cm

Explanation:

Given data

focal length f1 = 1.40 cm

focal length f2 = 2.20 cm

separated d = 19.6 cm

to find out

angular magnification and How far from the objective

solution

we know magnification formula that is

magnification = ( - L / f1 ) (D/f2)

here D = 25 cm put all value

magnification = ( - 19.6 / 1.40 ) (25/2.20)

magnification = - 159

and

now we apply lens formula

i/f = 1/q + 1/p

p = f2 = 2.20

so

q = f2 p / p -f2

q = 1.4(2.20) / ( 2.2 - 1.4 )

q = 3.85 cm

so objective distance is 3.85 cm

3 0
3 years ago
How much heat energy is tequired to raise the temperature of 3 kg of copper from 0
rusak2 [61]

2 would be the answer

6 0
4 years ago
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