solution:
When an uncharged conducting object brought near to a charged insulating object there is a force on the conducting object to move the electrons within it to opposite sides of the conductor. That means there is a separation of charges in the conducting object in the presence of the charged insulating object near to it but the charge on the conducting object is neutral.
Thus, the conducting object is uncharged.
There is a force of attraction between the uncharged conducting object and the insulating object when it brought near to the insulating object.
Thus, there is a force on the conducting object.
The conductor remains uncharged and a force is exerted on it.
Answer:
for this problem, 2.5 = (5+2/2)-(5-2/2)erf (50×10-6m/2Dt)
It now becomes necessary to compute the diffusion coefficient at 750°C (1023 K) given that D0= 8.5 ×10-5m2/s and Qd= 202,100 J/mol.
we have D= D0exp( -Qd/RT)
=(8.5×105m2/s)exp(-202,100/8.31×1023)
= 4.03 ×10-15m2/s
Answer:
0.398
Explanation:
According to friction, the frictional force is directly proportional to the normal reaction
Ff = nR
Ff is the frictional force
n is the coefficient of friction
R is the reaction
Reaction is equal to the weight
R= W = 9.3N
Fm = Ff = 3.7N
Fm is the moving force
Get the coefficient of friction
n = Ff/R
n = 3.7/9.3
n = 0.398
Hence the coefficient of friction for the surface is 0.398
