Answer:
μ = 0.535
Explanation:
On a level floor, normal force = weight.
N = W
Friction force = normal force × coefficient of friction.
F = Nμ
Substitute:
F = Wμ
83 = 155μ
μ = 0.535
Round as needed.
Answer:
<em>0.45 mm</em>
Explanation:
The complete question is
a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?
A) 0.45 mm
B) 0.63 mm
C.) 0.68 mm
D) 0.91 mm
Current in the fuse is 1.0 A
Current density of the fuse when it melts is 620 A/cm^2
Area of the wire in the fuse = I/ρ
Where I is the current through the fuse
ρ is the current density of the fuse
Area = 1/620 = 1.613 x 10^-3 cm^2
We know that 10000 cm^2 = 1 m^2, therefore,
1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2
Recall that this area of this wire is gotten as
A = 
where d is the diameter of the wire
1.613 x 10^-7 = 
6.448 x 10^-7 = 3.142 x 
=
d = 4.5 x 10^-4 m = <em>0.45 mm</em>
As the external magnetic field decreases, an induced current flows in the coil. The direction of the induced magnetic field would be pointing to the screen. The flux through the coil is said to decrease. In order to counter this change, the coil would generate or produce a magnetic field that is induced that would be pointing to the same direction as the external field that is flowing which is into the the screen. This is according to Lenz's law or the right hand rule. It states that an induced current in a circuit that is due to the change or motion in magnetic field should be directed opposing to the change in the flux.