Answer:
Honey how can i draw you cant draw here
Explanation:
I wish i can help you
Answer:
1 C
Explanation:
The intensity of electric current is defined as

where
I is the current
q is the amount of charge transferred
t is the time interval during which the charge is transferred
For the lightning in this problem, we have
is the current
is the time interval
Solving the formula for q, we find the amount of charge transferred:

To explain, I will use the equations for kinetic and potential energy:

<h3>Potential energy </h3>
Potential energy is the potential an object has to move due to gravity. An object can only have potential energy if 1) <u>gravity is present</u> and 2) <u>it is above the ground at height h</u>. If gravity = 0 or height = 0, there is no potential energy. Example:
An object of 5 kg is sitting on a table 5 meters above the ground on earth (g = 9.8 m/s^2). What is the object's gravitational potential energy? <u>(answer: 5*5*9.8 = 245 J</u>)
(gravitational potential energy is potential energy)
<h3>Kinetic energy</h3>
Kinetic energy is the energy of an object has while in motion. An object can only have kinetic energy if the object has a non-zero velocity (it is moving and not stationary). An example:
An object of 5 kg is moving at 5 m/s. What is the object's kinetic energy? (<u>answer: 5*5 = 25 J</u>)
<h3>Kinetic and Potential Energy</h3>
Sometimes, an object can have both kinetic and potential energy. If an object is moving (kinetic energy) and is above the ground (potential), it will have both. To find the total (mechanical) energy, you can add the kinetic and potential energies together. An example:
An object of 5 kg is moving on a 5 meter table at 10 m/s. What is the objects mechanical (total) energy? (<u>answer: KE = .5(5)(10^2) = 250 J; PE = (5)(9.8)(5) = 245 J; total: 245 + 250 = 495 J</u>)
Answer:
The unknown quantities are:
E and F
The final velocity of the proton is:
√(8/3) k e^2/(m*r)
Explanation:
Hello!
We can solve this problem using conservation of energy and momentum.
Since both particles are at rest at the beginning, the initial energy and momentum are:
Ei = k (q1q2)/r
pi = 0
where k is the coulomb constant (= 8.987×10⁹ N·m²/C²)
and q1 = e and q2 = 2e
When the distance between the particles doubles, the energy and momentum are:
Ef = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2
pf = m1v1 + m2v2
with m1 = m, m2 = 4m, v1=vf_p, v2 = vf_alpha
The conservation momentum states that:
pi = pf
Therefore:
m1v1 + m2v2 = 0
That is:
v2 = (1/4) v1
The conservation of energy states that:
Ei = Ef
Therefore:
k (q1q2)/r = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2
Replacing
m1 = m, m2 = 4m, q1 = e, q2 = 2e
and v2 = (1/4)v1
We get:
(1/2)mv1^2 = k e^2/r + (1/2)4m(v1/4)^2 = k e^2/r + (1/8)mv1^2
(3/8) mv1^2 = k e^2/r
v1^2 = (8/3) k e^2/(m*r)