Answer:
0.52 Nm
Explanation:
A = 0.12 m^2, N = 200, i = 0.5 A, B = 0.050 T
Angle between the plane of loop and magnetic field = 30 Degree
Angle between the normal of loop and the magnetic field = 90 - 30 = 60 degree
θ = 60°
Torque = N i A B Sinθ
Torque = 200 x 0.5 x 0.12 x 0.050 x Sin 60
Torque = 0.52 Nm
Answer:
There isnt enough in your question to answer the question bro, like we need a picture or something bro.
Explanation:
Answer:
k = 9.6 x 10^5 N/m or 9.6 kN/m
Explanation:
First, we need to use the expression to calculate the spring constant which is:
w² = k/m
Solving for k:
k = w²*m
To get the angular velocity:
w = 2πf
The problem is giving the linear velocity of the car which is 5.7 m/s. With this we can calculate the frequency of the car:
f = V/x
f = 5.7 / 4.9 = 1.16 Hz
Now the angular velocity:
w = 2π*1.16
w = 7.29 rad/s
Finally, solving for k:
k = (7.29)² * 1800
k = 95,659.38 N/m
In two significant figures it'll ve 9.6 kN/m
This is true. let me know if im wrong.
