The magnetic field strength of a very long current-carrying wire is proportional to the inverse of the distance from the wire. The farther you go from the wire, the weaker the magnetic field becomes.
B ∝ 1/d
B = magnetic field strength, d = distance from wire
Calculate the scaling factor for d required to change B from 25μT to 2.8μT:
2.8μT/25μT = 1/k
k = 8.9
You must go to a distance of 8.9d to observe a magnetic field strength of 2.8μT
Answer:
a)15 N
b)12.6 N
Explanation:
Given that
Weight of block (wt)= 21 N
μs = 0.80 and μk = 0.60
We know that
Maximum value of static friction given as
Frs = μs m g = μs .wt
by putting the values
Frs= 0.8 x 21 = 16.8 N
Value of kinetic friction
Frk= μk m g = μk .wt
By putting the values
Frk= 0.6 x 21 = 12.6 N
a)
When T = 15 N
Static friction Frs= 16.8 N
Here the value of static friction is more than tension T .It means that block will not move and the value of friction force will be equal to the tension force.
Friction force = 15 N
b)
When T= 35 N
Here value of tension force is more than maximum value of static friction that is why block will move .We know that when body is in motion then kinetic friction will act on the body.so the value of friction force in this case will be 12.6 N
Friction force = 12.6 N
Answer:
1.71 km
Explanation:
Convert 30 minutes to seconds:
30 min × (60 s / min) = 1800 s
Find the displacement:
0.95 m/s × 1800 s = 1710 m
Convert to kilometers:
1710 m × (1 km / 1000 m) = 1.71 km
Answer:
1.27 m
Explanation:
Distance = 192 m
number of rotations = 48
Distance traveled in one rotation = 2 x π x r
Where, r be the radius of wheel.
so, distance traveled in 48 rotations = 48 x 2 x 3.14 x r
It is equal to the distance traveled.
192 = 48 x 2 x 3.14 x r
r = 0.637 m
diameter of wheel = 2 x radius of wheel = 2 x 0.637 = 1.27 m