A 2.3kg bicycle wheel has a diameter of 50cm. What torque must you apply to take the wheel from 0rpm to 120rpm in 5.5s?

τ = 0.26 N.m

Explanation:

First we find the moment of inertia of the wheel, by using the following formula:

I= mr²

where,

I = Moment of Inertia = ?

m = mass of wheel = 2.3 kg

r = radius of wheel = 50 cm/2 = 25 cm = 0.25 m

Therefore,

I = (2.3 kg)(0.25 m)²

I = 0.115 kg.m²

Now, we find the angular acceleration of the wheel:

α = (ωf - ωi)/t

where,

α = angular acceleration = ?

ωf = final angular velocity = (120 rpm)(2π rad/1 rev)(1 m/60 s) = 12.56 rad/s

ωi = Initial Angular Velocity = 0 rad/s

t = time = 5.5 s

Therefore,

α = (12.56 rad/s - 0 rad/s)/(5.5 s)

α = 2.28 rad/s²

Now, the torque is given as:

Torque = τ = Iα

τ = (0.115 kg.m²)(2.28 rad/s²)

τ = 0.26 N.m

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j