Answer:
8.08 x 10^-5 m
Explanation:
A = 2 mm^2 = 2 x 10^-6 m^2
Total number of electrons, N = 9.4 x 10^18
time, t = 3 s
n = 5.8 x 10^28 electrons/ m^3
Current, i = Q / t = N x e / t = (9.4 x 10^18 x 1.6 x 10^-19) / 3 = 0.5 A
Let vd be the drift velocity.
i = n e A vd
0.5 = 5.8 x 10^28 x 1.6 x 10^-19 x 2 x 10^-6 x vd
vd = 2.7 x 10^-5 m/s
Distance traveled by the electrons = velocity x time
= vd x t = 2.7 x 10^-5 x 3 = 8.08 x 10^-5 m
Answer:
0.20kg-m^2
Explanation:
Let the linear velocity of the rope(=of pulley) is v m/s
Using kinematic equation
=> v = u + at
=>v = 0 + 4.9a
=>v = 4.9a ------------ eq1
By v^2 = u^2 + 2as
=>v^2 = 0 + 2 x v/4.9 x 1.2
=>4.9v^2 - 2.4v = 0
=>v(4.9v - 2.4) = 0
=>v = 2.4/4.9 = 0.49 m/s
Thus by v = r x omega
=>omega = v/r = 0.49/0.02 = 24.49 rad/sec
BY W = F x s = 50 x 1.2 = 60 J
=>KE(rotational) = W = 1/2 x I x omega^2
=>60 = 1/2 x I x (24.49)^2
=>I = 0.20 kg-m^2
Answer:
The electric field value is 240 N/C
Explanation:
Given that,
Distance = 5.0 mm
Potential difference = 1.2 V
We need to calculate the electric field value
Using formula of potential difference
Where, E = electric field
V = potential difference
d = distance
Put the value into the formula
Hence, The electric field value is 240 N/C
Answer:
Explanation:
Check the attachment for solution
