Answer:
Magnetic force, F = 0.24 N
Explanation:
It is given that,
Current flowing in the wire, I = 4 A
Length of the wire, L = 20 cm = 0.2 m
Magnetic field, B = 0.6 T
Angle between force and the magnetic field, θ = 30°. The magnetic force is given by :
F = 0.24 N
So, the force on the wire at an angle of 30° with respect to the field is 0.24 N. Hence, this is the required solution.
Answer:
Angular acceleration, is
Explanation:
Given that,
Initial speed of the drill,
After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed,
We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.
So, the drill's angular acceleration is .
Answer:
Explanation:
a ) angular frequency ω =
k is spring constant and m is mass attached
ω =
= 3.6515 rad / s
frequency of oscillation n = 3.6515 / (2 x 3.14)
= .5814 s⁻¹
x = .1 mcos(ωt)
= .1 mcos(3.6515t)
b ) maximum speed = ωA , A is amplitude
= 3.6515 x .1
= .36515 m /s
36.515 cm /s
maximum acceleration = ω²A
= 3.6515² x .1
= 1.333 m / s²
c ) Kinetic energy at displacement x
= 1/2 m ω²( A²-x²)
potential energy =1/2 m ω²x²
so 1/2 m ω²( A²-x²) = 1/2 m ω²x²
A²-x² = x²
2x² = A²
x = A / √2