Answer:
v_2 = 2*v
Explanation:
Given:
- Mass of both charges = m
- Charge 1 = Q_1
- Speed of particle 1 = v
- Charge 2 = 4*Q_1
- Potential difference p.d = 10 V
Find:
What speed does particle #2 attain?
Solution:
- The force on a charged particle in an electric field is given by:
F = Q*V / r
Where, r is the distance from one end to another.
- The Net force acting on a charge accelerates it according to the Newton's second equation of motion:
F_net = m*a
- Equate the two expressions:
a = Q*V / m*r
- The speed of the particle in an electric field is given by third kinetic equation of motion.
v_f^2 - v_i^2 = 2*a*r
Where, v_f is the final velocity,
v_i is the initial velocity = 0
v_f^2 - 0 = 2*a*r
Substitute the expression for acceleration in equation of motion:
v_f^2 = 2*(Q*V / m*r)*r
v_f^2 = 2*Q*V / m
v_f = sqrt (2*Q*V / m)
- The velocity of first particle is v:
v = sqrt (20*Q / m)
- The velocity of second particle Q = 4Q
v_2 = sqrt (20*4*Q / m)
v_2 = 2*sqrt (20*Q / m)
v_2 = 2*v
The gravitational force on the woman is A) 500 N
Explanation:
There are two forces acting on the woman during her fall:
- The force of gravity,
, acting downward - The air resistance,
, acting upward
According to Newton's second law, the net force acting on the woman is equal to the product between the woman's mass and her acceleration:

where m is the mass of the woman and a her acceleration.
The net force can be written as

Also, we know that the woman falls at a constant velocity (5 m/s), this means that her acceleration is zero:

Combining the equations together, we get:

which means that the magnitude of the gravitational force is equal to the magnitude of the air resistance:

Learn more about forces and Newton's second law:
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Sorry to say but I know that t(e introduction is first and the coda is last
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