Answer:
200A
Explanation:
Given that
the distance between earth surface and power cable d = 8m
when the current is flowing through cable , the magnitude flux density at the surface is 15μT
when the current flow throught is zero the magnitude flux density at the surface is 20μT
The change in flux density due to the current flowing in the power cable is
B = 20μT - 15μT
B =5μT -----(1)
The expression of magnitude flux density produced by the current carrying cable is
-----(2)
Substitute the value of flux density
B from eqn 1 and eqn 2
Therefore, the magnitude of current I is 200A
Answer:
The answer is below
Explanation:
The length of the rope is equal to the radius of the circle formed by the complete rotation of the rope. Therefore the radius = 1.50 m.
a) The distance covered by the rope when completing one rotation is the same as the perimeter of the circle. Hence:
Distance covered in one rotation = 2π * radius = 2π * 1.5 = 3π meters
The velocity of the ball = Distance / time = 3π meters / 3.4 seconds = 2.77 m/s
b) The initial velocity (u) is 0 m/s, the final velocity is 2.77 m/s during time (t) = 3.4 s. Hence acceleration (a):
v = u + at
2.77 = 3.4a
a = 0.82 m/s²
c) Force on ball = mass * acceleration = 4 * 0.82 = 3.28 N
a) In this case the forces are the centrifugal force Fcp,
which is directed horizontally toward the wall; the force of static friction Ff
with the wall, directed upward; the normal force Fn by the wall, which is
directed away the wall; the force of gravity Fg, directed downwards. Then we
have that the horizontal forces are all equal in magnitude; similarly the
vertical forces are also all equal in magnitude.
b) The minimum coefficient s occurs when force of gravity
is equals the max friction force, that is
Fg = Ff,max
m g = s Fn
Also, the normal force has equal magnitude to the
centrifugal force:
m g = s Fcp
m g = s m w^2 r
g = s w^2 r
s = g / (r w^2)
With values: g = 9.81 m/s^2; r = 2.5 m; and w = 2pi *
0.60 = 3.77 rad/s; we find
s = 9.81 / (2.5 * 3.77^2) = 0.276
