Acord is wrapped around the rim of a solid uniform wheel 0.300 m in radius and of mass 8.00 kg. a steady horizontal pull of 34.0 n to the right is exerted on the cord, pulling it off tangentially from the wheel. the wheel is mounted on frictionless bearings on a horizontal axle through its center.
part acompute the angular acceleration of the wheelpart bcompute the acceleration of the part of the cord that has already been pulled off the wheelpart cfind the magnitude of the force that the axle exerts on the wheelpart dfind the direction of the force that the axle exerts on the wheelpart ewhich of the answers in parts a, b, c, and d would change if the pull were upward instead of horizontal

A. α = 94.4 rad/s

B. a = 28.32 m/s

C. N = 34N

D. α = 94.4 rad/s

    a = 28.32 m/s

     N =  44.4  N

Explanation:

part A:

using:

∑T = Iα

where T is the torque, I is the moment of inertia and α is the angular momentum.

firt we will find the moment of inertia I as:

I =

Where M is the mass and R is the radius of the wheel, then:

I =

I = 0.36 kg*m^2

Replacing on the initial equation and solving for α, we get::

∑T = Iα

Fr = Iα

34 N =  0.36α

α = 94.4 rad/s

part B

we need to use this equation :

a = αr

where a is the aceleration of the cord that has already been pulled off and r is the radius of the wheel, so replacing values, we get:

a = (94.4)(0.3 m)

a = 28.32 m/s

part C

Using the laws of newton, we know that:

N = T

where N is the force that the axle exerts on the wheel part and T is the tension of the cord

so:

N = 34N

part D

The anly answer that change is the answer of the part D, so, aplying laws of newton, it would be:

-Mg + N +T = 0

Then, solving for N, we get:

N = -T+Mg

N = -34 + (8 kg)(9.8)

N =  44.4  N

real, smaller and upside-down.   so the answer should be b.

the temperature in kelvin of 2.0 dm3 of an ideal gas is doubled and its pressure is increased by a factor of four. what is the final volume of the gas