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garri49 [273]
2 years ago
9

A point charge 20.0 nC is located at the center of a cube whose sides are of length 10.0 cm. If there is no other charges in thi

s system, what is the electric flux through one face of the cube?
Physics
1 answer:
stealth61 [152]2 years ago
3 0

Answer:

ϕ_Net = 2258.87 Nm²/C

Explanation:

We are given;

Charge; q = 20 nC = 20 × 10^(-9) C

Length of each side; L = 10 cm = 0.1 m

The net electric flux through one face of the cube is given by the formula;

ϕ_Net = q/ε_o

Where ε_o is a constant known as vacuum of Permittivity and has a value of 8.854 × 10^(−12) C²/N.m²

Thus;

ϕ_Net = (20 × 10^(-9))/(8.854 × 10^(−12))

ϕ_Net = 2258.87 Nm²/C

You might be interested in
How many significant figures are there in : (a) 0.000054 (b) 3.001 x 10^5 (c) 5.600
melomori [17]

Answer:

(a) 2 (b) 4 (c) 4

Explanation:

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

  • Digits from 1 to 9 are always significant and have infinite number of significant figures.
  • All non-zero numbers are always significant. For example: 654, 6.54 and 65.4 all have three significant figures.
  • All zero’s between integers are always significant. For example: 5005, 5.005 and 50.05 all have four significant figures.
  • All zero’s preceding the first integers are never significant. For example: 0.0078 has two significant figures.
  • All zero’s after the decimal point are always significant. For example: 4.500, 45.00 and 450.0 all have four significant figures.
  • All zeroes used solely for spacing the decimal point are not significant. For example : 8000 has one significant figure.

As per question,

0.000054 has 2 significant figures.

3.001 x 10⁵ has 4 significant figures.

5.600 has 4 significant figures.

4 0
3 years ago
A car starts from rest and accelerates uniformly over a time of 7.25 seconds for a distance of 210 m. Determine the acceleration
lara31 [8.8K]

Answer:

Acceleration is 7.990487515m/s²

Initial velocity is 0m.s

Explanation:

s=ut+(1/2)at²

210=0(7.25)+(1/2)a(7.25²)

210=26.28125a

∴a=7.990487515m/s²

'Vi' or 'u' is the inital speed. Since it starts from rest, this equals 0.

8 0
2 years ago
6
choli [55]

Answer:

B. Acceleration in the direction of motion speeds you up

Explanation:

Acceleration is defined as when something gains speed. For example, when a car speeds up. Deceleration is when the car slows down, and looses speed. When defining these terms, think of a car going faster, then slowing down at a red light.

3 0
3 years ago
If you ran 15 km/h for 20 min, how much distance would you cover?
nikitadnepr [17]
Since we have 15 kilometers per hour, and we're looking for 20 minutes, let's set up proportions.
20/60 minutes = x/15
20/60 = 1/3, so let's leave that simplified.
1/3 = x/15
Look at the denominators, 3 to 15 is a factor of 5, so multiply the numerator by 5.
1 • 5 = 5, so you will cover 5 kilometers in 20 minutes.

I hope this helps!
4 0
2 years ago
A proton orbits a long charged wire, making 1.80 ×106 revolutions per second. The radius of the orbit is 1.20 cm What is the wir
Fantom [35]

Answer:

linear charge density = -9.495 × 10^{-34} C/m

Explanation:

given data

revolutions per second = 1.80 × 10^{6}

radius = 1.20 cm

solution

we know that when proton to revolve around charge wire then centripetal force is require to be in orbit of radius around provide by electric force

so

- q × E = m × w² × r     ..................1

- 9 × 10^{9}  × \frac{2*linear\ charge\ density}r} q =  m × w² × r   ............2

and w = \frac{2*\pi}{T}  

w = \frac{d\theta }{dt}

w = 1.80 × 10^{6} × \frac{2*\pi}{1}

w = 11304000 rad/s

so here from equation 2

- 9 × 10^{9}  × \frac{2*linear\ charge\ density}{0.012} 1.80 × 10^{6} =  1.672 × 10^{-27} × 11304000² × 0.0120  

linear charge density = -9.495 × 10^{-34} C/m

8 0
2 years ago
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