It is called a photon i believe
Answer:
a) q_1=q_2= 7.42*10^-7 C
b) q_2= 3.7102*10^-7 C , q_1 = 14.8*10^-7 C
Explanation:
Given:
F_e = 0.220 N
separation between spheres r = 0.15 m
Electrostatic constant k = 8.99*10^9
Find: charge on each sphere
part a
q_1 = q_2
Using coulomb's law:
F_e = k*q_1*q_2 / r^2
q_1^2 = F_e*r^2/k
q_1=q_2= sqrt (F_e*r^2/k)
Plug in the values and evaluate:
q_1=q_2= sqrt (0.22*0.15^2/8.99*10^9)
q_1=q_2= 7.42*10^-7 C
part b
q_1 = 4*q_2
Using coulomb's law:
F_e = k*q_1*q_2 / r^2
q_2^2 = F_e*r^2/4*k
q_2= sqrt (F_e*r^2/4*k)
Plug in the values and evaluate:
q_2= sqrt (0.22*0.15^2/4*8.99*10^9)
q_2= 3.7102*10^-7 C
q_1 = 14.8*10^-7 C
Answer:31.62 m/s
Explanation:
Given
mass of body 
Pull on chain is 
Pull get smaller at the rate of 
Net Upward Force 
net acceleration 
but g is acting downward
using 
here initial velocity is zero
Answer:
132 N
Explanation:
Given that a 1.1 kg hammer strikes a nail. Before the impact, the hammer is moving at 4.5 m/s; after the impact it is moving at 1.5 m/s in the opposite direction. If the hammer is in contact with the nail for 0.025 s, what is the magnitude of the average force exerted by the hammer on the nail
From Newton 2nd law of motion,
Change in momentum = impulse.
Change in momentum = m( V - U )
Substitute all the parameters into the formula
Change in momentum = 1.1 ( 4.5 - 1.5 )
Change in momentum = 1.1 × 3
Change in momentum = 3.3 kgm/s
Impulse = Ft
That is,
Ft = 3.3
Substitute time t into the formula above
F × 0.025 = 3.3
F = 3.3 / 0.025
F = 132 N
Therefore, the magnitude of the average force exerted by the hammer on the nail is 132 N.
