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lora16 [44]
1 year ago
11

What is the potential difference between the plates of a 3.7-f capacitor that stores sufficient energy to operate a 75.0-w light

bulb for one minute?
Physics
1 answer:
Elena L [17]1 year ago
4 0

The potential difference between the plates will be 1552 Volts.

<h3>What is a potential difference?</h3>

Voltage, or the difference in electric potential between two points, is defined as the amount of labor per unit of charge needed to move a test charge between the two points.

Given that a 3.7-f capacitor that stores sufficient energy to operate a 75.0-w

The potential difference will be calculated by the formula below:-

Q = I t            

Where I = charge / time

Q = V * C    

V C = I t

V = I t / C

V = 75 C/s x 60 sec / 2.9 faraday

V =  1552 Volts

To know more about potential differences follow

brainly.com/question/25923373

#SPJ4

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Two parallel plates 19 cm on a side are given equal and opposite charges of magnitude 2.0 ✕ 10^−9 C. The plates are 1.8 mm apart
Romashka [77]

Answer:

E   = 5291.00 N/C

Explanation:

Expression for capacitance is

C = \frac{\epsilon  A}{d}

where

A is area of square plate

D = DISTANCE BETWEEN THE PLATE

C = \frac{\epsilon\times(19\times 10^{-2})^2}{1.5\times 10^{-3}}

C = 24.06 \epsilon

C = 24.06\times 8.854\times 10^{-12} F

C =2.1\times 10^{-10} F

We know that capacitrnce and charge is related as

V = \frac{Q}{C}

 = \frac{2\tiimes 10^{-9}}{2.\times 10^{-10}}

v = 9.523 V

Electric field is given as

E = \frac{V}{d}

   = \frac{9.52}{1.8*10^{-3}}

E   = 5291.00 V/m

E   = 5291.00 N/C

5 0
3 years ago
I need help with #40 need answer and show me what you multiplied please
ale4655 [162]
(0.5)×(0squared)×(3)=(1.5j)
3 0
3 years ago
18 points! Brainliest! Physics!
Naddik [55]
You will use the Pythagorean Theorem to solve it.
c^2 = a^2 + b^2
c^2 = (1.5)^2 + (2)^2
c^2 = 6.25
c = square root of 6.25
c = 2.5
I hope this helps!
3 0
4 years ago
Why is the scientific method described as cyclic?
Troyanec [42]
Because the scientific method can go around in a circle as many times as neccisary to get the results you need
8 0
3 years ago
The space shuttle usually orbited Earth at altitudes of around 300.0 km. 1) Determine the time for one orbit of the shuttle abou
777dan777 [17]

Answer:

Explanation:

distance of shuttle from centre of the earth = radius of the orbit

= 6300 + 300 = 6600 km

= 6600 x 10³

Formula of time period of the satellite

T = 2π R /v₀ , v₀ is orbital velocity

v₀ = √gR , ( if height  is small with respect to radius )

T = 2π R /√gR

= 2π√ R /√g

= 2 x 3.14 x √ 6600 x 10³ / √9.8

= 2 x 3.14 x 256.9 x 10 / 3.13

= 5154.41 s

= 5154.41 / 60 minutes

= 85.91 m

85.9 minutes.

2 ) No of sunrise per day = no of rotation per day

= 24 x 60 / 85.9

= 16.76

or 17 sunrises.

3 0
4 years ago
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