Answer:
a)Current will flow perpendicularly.
b)Magnitude of flux will be 2.987 N m2 C−1
Answer:
Radius r = 20.34 cm
The radius that can produces such a disk is 20.34 cm
Explanation:
Area of a circle;
A = πr^2
A = area
r = radius
Making r the subject of formula;
r = √(A/π) ........1
Given;
A = 1300 cm^2
Substituting into the equation 1;
r = √(1300/π)
r = 20.34214472564 cm
r = 20.34 cm
The radius that can produces such a disk is 20.34 cm
Answer:
α = 0.0135 rad/s²
Explanation:
given,
t = 133 min = 133 x 60 = 7980 s
angular speed varies from 570 rpm to 1600 rpm
now,
570 rpm = 
= 59.69 rad/s
1600 rpm = = 
= 167.6 rad/s
using equation of rotational motion
ωf = ωi + αt
167.6 = 59.7 + α x 7980
α x 7980 = 107.9
α = 0.0135 rad/s²
Answer:
W = 100000 J = 100 KJ
Explanation:
Here we will use the most basic and general formula of work, which is as follows:

where,
W = Work Done = ?
F = Force Required = 200 N
d = Length of Track = 500 m
Therefore,

<u>W = 100000 J = 100 KJ</u>
Answer:
Vi = 8.28 m/s
Explanation:
This problem is related to the projectile motion.
As we know there are two components of motion associated with this, the horizontal component and vertical component.
The horizontal distance covered by the ball is
Vx*t = x
Vx*t = 5.3
Vx = 5.3/t eq. 1
Also we know that
Vx = Vicos(60)
Vx = Vi*0.5 eq. 2
equate eq. 1 and eq. 2
5.3/t = Vi*0.5
5.3/0.5 = Vi*t
Vi*t = 10.6 eq. 3
The vertical distance is
Vy = y1 + Vyi*t - 0.5gt²
also we know that
Vyi = Visin(60)
Vyi = Vi*0.866
It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance
3 = 1.9 + Vi*0.866*t - 0.5gt²
3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
1.1 = 0.866(Vi*t) - 4.9t²
0.866(Vi*t) = 4.9t² + 1.1
substitute Vi*t = 10.6 in above equation
0.866(10.6) = 4.9t² + 1.1
9.18 = 4.9t² + 1.1
4.9t² = 8.08
t² = 8.08/4.9
t² = 1.648
t = 1.28 sec
Finally, initial speed can be found by substituting the value of t into eq. 3
Vi*t = 10.6
Vi = 10.6/t
Vi = 10.6/1.28
Vi = 8.28 m/s