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Grace [21]
3 years ago
9

A. Consider a charge Q fixed in place at the origin that creates an electric field in the surrounding region. We call a charge w

hich gives rise to the electric field the source charge. You bring a small charge q from infinite distance to a distance r from the source charge adiabatically, that is, infinitely slowly such that the electric field due to the source charge is not disturbed. This small charge q is called a test charge. Now answer the following questions. All of your answers will be in terms of q, Q, r and the dielectric constant εo, as appropriate and be careful about your signs. 1. What is the electric potential energy of the charge q as a function of the distance r away from
Physics
1 answer:
zvonat [6]3 years ago
6 0

Answer:

U = k Q q / r

Explanation:

The potential electric energy for a point charge is

       U = k q₁ q₂ / r

Where k is the Coulomb constant, q are the electric charges and r is the distance between them

Let's replace our values

     U = k Q q / r

Note that the positive power is because the two charges have the same sign, so we have to do a job to move the test load and also a scalar

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Jane, looking for tarzan, is running at top speed 5.3 m/s and grabs a vine hanging vertically from a tall tree in the jungle. ho
lutik1710 [3]

v₀ = initial speed as tarzan grabs the vine = 5.3 m/s

v = final speed as the tarzan reach the maximum height = 0 m/s

h = maximum height gained by the tarzan

m = mass of tarzan

using conservation of energy

initial kinetic energy = final kinetic energy + potential energy

(0.5) m v²₀ = (0.5) m v² + m g h

(0.5) v²₀ = (0.5) v² + g h

(0.5) (5.3)² = (0.5) (0)² + (9.8) h

h = 1.43 m


3 0
3 years ago
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Sedaia [141]

Answer:

V = 20.5 m/s

Explanation:

Given,

The mass of the cart, m = 6 Kg

The initial speed of the cart, u = 4 m/s

The acceleration of the cart, a = 0.5 m/s²

The time interval of the cart, t = 30 s

The final velocity of the cart is given by the first equation of motion

                              v = u + at

                                  = 4 + (0.5 x 30)

                                = 19 m/s

Hence the final velocity of cart at 30 seconds is, v = 19 m/s

The speed of the cart at the end of  3 seconds

                                    V = 19 + (0.5 x 3)

                                       = 20.5 m/s

Hence, the final velocity of the cart at the end of this 3.0 second interval is, V = 20.5 m/s

6 0
3 years ago
Faiyez wrote the problem below
Ilya [14]

445/100 - 5/4 = 445/100 - 125/100 = 320/100 = 16/5 = 3 1/5.

6 0
3 years ago
Someone help me with these two questions ASAP!!!
Sonbull [250]

For the first question it is the fourth option. Cryosphere is a term for the portions of earth that are covered in water when the water is in solid form. this includes both snow and ice.

For the second question the answer is a delta is formed at the mouth of the river a sediment is carried down stream. The hydrosphere refers to all water on earth.

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3 years ago
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An electron is released from rest in a weak electric field given by E =-2.30 x 10-10 N/Cj. After the electron has traveled a ver
Goshia [24]

Explanation:

It is given that,

An electron is released from rest in a weak electric field of, E=2.3\times 10^{-10}\ N/C

Vertical distance covered, s=1\ \mu m=10^{-6}\ m

We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :

v^2-u^2=2as

v^2=2as.............(1)

Electric force is F_e and force of gravity is F_g. As both forces are acting in downward direction. So, total force is:

F=mg+qE

F=9.1\times 10^{-31}\times 9.8+1.6\times 10^{-19}\times 2.3\times 10^{-10}

F=4.57\times 10^{-29}\ N

Acceleration of the electron, a=\dfrac{F}{m}

a=\dfrac{4.57\times 10^{-29}\ N}{9.1\times 10^{-31}\ kg}

a=50.21\ m/s^2

Put the value of a in equation (1) as :

v=\sqrt{2as}

v=\sqrt{2\times 50.21\times 10^{-6}}

v = 0.010 m/s

So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.

6 0
3 years ago
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