<span>F* t = (m x v_final) - (m x v_initial)
200 x t = 50x8 - 50x0
t= 50 x 8 /200
t= 2s </span>
The resistance of a conductor is given by
where L is the length of the wire,
the resistivity of the material and A the cross-sectional area.
We can see that if all the other quantities do not change, if the new length of the conductor is 4 times the original length:
, then the new resistance is also 4 times the original value:
Answer:
★The second law of refraction
The ratio of sine of angle of incidence to the sine of angle of refraction is a constant for a light of given colour and for a given pair of media. This law is also called Snell's law of refraction. If 'i' is the angle of incidence and 'r' is the angle of refraction then, Sin i/Sin r = constant
This constant value is called the refractive index of the second medium with respect to the first.
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
