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yaroslaw [1]
2 years ago
12

What is the electric potential of a 2.2 µC charge at a distance of 6.3 m from the charge? Recall that Coulomb’s constant is k =

8.99 × 109 N • ___ V What is the electric potential at a distance of 99 m from the charge?___ V
Physics
2 answers:
Doss [256]2 years ago
6 0
The first part of the question is 3,100 V.

The second part of the question is 200 V.
const2013 [10]2 years ago
3 0

Electric potential due to a point charge is given by

V = \frac{kQ}{r}

here we know that

Q = charge

r = distance from the charge

V = \frac{9*10^9* 2.2* 10^{-6}}{6.3}

V = 3142.8 Volts

now again by the same equation

V = \frac{kQ}{r}

V = \frac{9*10^9* 2.2* 10^{-6}}{99}

V = 200 Volts


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A charming friend of yours who has been reading a little bit about astronomy accompanies you to the campus observatory and asks
lidiya [134]

Answer:

b

Explanation:

3 0
2 years ago
An air-track cart with mass m1=0.28kg and initial speed v0=0.75m/s collides with and sticks to a second cart that is at rest ini
arsen [322]
Kinetic energy is calculated through the equation,

   KE = 0.5mv²

At initial conditions,

  m₁:  KE = 0.5(0.28 kg)(0.75 m/s)² = 0.07875 J

  m₂ : KE = 0.5(0.45 kg)(0 m/s)² = 0 J

Due to the momentum balance,

   m₁v₁ + m₂v₂ = (m₁ + m₂)(V)

Substituting the known values,

   (0.29 kg)(0.75 m/s) + (0.43 kg)(0 m/s) = (0.28 kg + 0.43 kg)(V)

   V = 0.2977 m/s

The kinetic energy is,
   KE = (0.5)(0.28 kg + 0.43 kg)(0.2977 m/s)²
   KE = 0.03146 J

The difference between the kinetic energies is 0.0473 J. 
7 0
2 years ago
A certain part of a flat screen TV has a thickness of 150 nanometers. How<br> many meters is this?
Bess [88]

Answer:

1.5e-7 meters

.00000015 meters

Explanation:

.000000001 meters = 1 nanometer. Multiply that by 150 and an answer is there.

5 0
2 years ago
A large solar panel on a spacecraft in Earth orbit produces 1.0 kW of power when the panel is turned toward the sun. What power
Mandarinka [93]

Answer:

e*P_s = 11 W

Explanation:

Given:

- e*P = 1.0 KW

- r_s = 9.5*r_e

- e is the efficiency of the panels

Find:

What power would the solar cell produce if the spacecraft were in orbit around Saturn

Solution:

- We use the relation between the intensity I and distance of light:

                                  I_1 / I_2 = ( r_2 / r_1 ) ^2

- The intensity of sun light at Saturn's orbit can be expressed as:

                                  I_s = I_e * ( r_e / r_s ) ^2

                                  I_s = ( 1.0 KW / e*a) * ( 1 / 9.5 )^2

                                  I_s = 11 W / e*a

- We know that P = I*a, hence we have:

                                  P_s = I_s*a

                                  P_s = 11 W / e

Hence,                       e*P_s = 11 W

3 0
2 years ago
A girl pushes an 18.15 kg wagon with a force of 3.63 N. what is the acceleration?
Anton [14]

Divide the force given by mass and you will find the acceleration of the object :-

F = m × a

3.63 = 18.15 × a

3.63 = 18.15a

a = 3.63/18.15

a = 0.2 m/s^2

hope it helps!

3 0
2 years ago
Read 2 more answers
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