Efficiency = useful energy out / total energy in x 100
= 100/400 x 100
=0.25 x 100
= 25%
25%
The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N
the slow masses that must be quicker are the pulley, ring, and the rolling sphere.
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m
The mass equal of the rolling sphere is computed by: the sphere revolves around the contact point with the table. So using the proposition of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for spin about the midpoint of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m
the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2
Answer:
I don't know why you are asking me?
Answer:
13.78 mT
Explanation:
The peak voltage ε = ωNAB where ω = angular speed of coil = 1500 rpm = 1500 × 2π/60 rad/s = 50π rad/s = 157.08 rad/s, N = number of turns of coil = 250, A = area of coil = πr² where r = radius of coil = 10 cm = 0.10 m,
A = π(0.1 m)² = 0.03142 m² and B = magnetic field strength
So,
B = ε/ωNA
substituting the values of the variables into the equation given that ε = 17 V
So, B = ε/ωNA
B = 17 V/(157.08 rad/s × 250 turns × 0.03142 m²)
B = 17 V/(1233.8634 rad-turns-m²/s)
B = 0.01378 T
B = 13.78 mT
Calculate the magnitude of the linear momen- tum for each of the following cases a) a proton with mass 1.67 × 10-27 kg mov- ing with a velocity of 6 × 106 m/s. Answer in units of kg · m/s.
