A mass on the end of a spring undergoes simple harmonic motion. At the instant when the mass is at its maximum displacement from
1 answer:
You might be interested in
Answer:
The current in the coil is 60 Ampere.
Explanation:
Given:
Number of turns in the coil is N = 25
Dimension of the coil = 15cm X 12cm
magnitude of magnetic field = 0.20T
angle in the xy plane is θ = 0 degree
torque τ = 5.4 N-m
To find:
current in the coil is i = ?
Solution:
The torque acting on the coil is given by
=>
Converting cm to m
12 cm = 0.12 m
15 cm = 0.15 m
The area of the coil is
A = 0.12 X 0.15
A =
Substituting the values
=>
=>
=>
=>
=>
=>
=> i = 60 A
Answer:
vB' = 0.075[m/s]
Explanation:
We can solve this problem using the principle of linear momentum conservation, which tells us that momentum is preserved before and after the collision.
Now we have to come up with an equation that involves both bodies, before and after the collision. To the left of the equal sign are taken the bodies before the collision and to the right after the collision.
where:
mA = 0.355 [kg]
vA = 0.095 [m/s] before the collision
mB = 0.710 [kg]
vB = 0.045 [m/s] before the collision
vA' = 0.035 [m/s] after the collision
vB' [m/s] after the collison.
The signs in the equation remain positive since before and after the collision, both bodies continue to move in the same direction.