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
-- Electric field lines DO never cross. <em>(A)
</em>
-- Electric field lines that are close together DO indicate a stronger electric field. <em>(B)
</em>
-- Electric field lines DO not affect the charge that created them. <em>(C)</em>
-- Electric field lines DON'T begin on north poles and end on south poles. North and South "poles" are the way we talk about magnets, not electric charges.
Answer:
4. It is the force of the road on the tires (an external force) that stops the car.
Explanation:
If there is no friction between the road and the tires, the car won't stop.
You can see this, for example, when there is ice on the road. You can still apply the brakes (internal force), but since there is no friction (external force) the car won't stop.
The force of the brakes on the wheels is not what makes the car stop, it is the friction of the road against still tires that makes it stop.
The answer is divergent boundaries.
I hope this helps you!
Thecorrect answer is B.
hope this helps brainliest would be cool if not its fine
