Please help on this one?

A cord 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.

disa [49]

Answer:

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 = $\frac{1}{2}MR^2$

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

I = $\frac{1}{2}(8 kg)(0.3 m)^2$

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

6 0

3 years ago

Who serves as the presiding officer of the Senate?

Cerrena [4.2K]

Who serves as the presiding officer of the Senate?

${ \bf{ \underbrace{Answer :}}}$

The $\sf\purple{Vice\:President}$ of the United States serves as the presiding officer of the Senate.

At present, $\sf\red{Kamala\:Harris}$ is the president of the Senate.

$\bold{ \green{ \star{ \orange{Hope\:it\:helps.}}}}⋆$

3 0

3 years ago

In a charming 19th-century hotel, an old-style elevator is connected to a counterweight by a cable that passes over a rotating d

melisa1 [442]

Answer:

2.38732 rpm

1.22625 rad/s²

163.292°

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

a = Acceleration = $\frac{1}{8}g$

d = Diameter of wheel = 2 m

r = Radius of wheel = $\frac{d}{2}=\frac{2}{2}=1\ m$

v = Speed of elevator = 25 cm/s

Angular speed is given by

$\omega=\frac{v}{r}\\\Rightarrow \omega=\frac{0.25}{1}\\\Rightarrow \omega=0.25\ rad/s=0.25\times \frac{60}{2\pi}\\\Rightarrow \omega=2.38732\ rpm$

The angular speed of the wheel is 2.38732 rpm

Angular acceleration is given by

$\alpha=\frac{a}{r}\\\Rightarrow \alpha=\frac{\frac{1}{8}g}{r}\\\Rightarrow \alpha=\frac{\frac{1}{8}\times 9.81}{1}\\\Rightarrow \alpha=1.22625\ rad/s^2$

The angular acceleration of the wheel is 1.22625 rad/s²

Angular displacement is given by

$\theta=\frac{s}{r}\\\Rightarrow \theta=\frac{2.85}{1}\\\Rightarrow \theta=2.85\ rad=2.85\times \frac{360}{2\pi}\\ =163.292\ ^{\circ}$

The angle the disk turned when it has raised the elevator is 163.292°

4 0

3 years ago

A boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s .

zimovet [89]

Answer:

Force exerted, F = 1.5 N

Explanation:

It is given that, a boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s.

i.e. u = 0

v = 30 m/s

Time taken, t = 0.06 s

Mass of the paper, m = 0.003 kg

We need to find the force the boxer exert on it. The force can be calculated using second law of motion as :

$F=m\times a$

$F=m\times (\dfrac{v-u}{t})$

$F=0.003\times (\dfrac{30}{0.06})$

F = 1.5 N

So, the force the boxer exert on the paper is 1.5 N. Hence, this is the required solution.

6 0

3 years ago

!!! HELP PLEASE !!!

maksim [4K]

Answer:

A. you would want to use the 12 ound ball and the 18 mph speed in order to transfer the most kinetic energy

B. KE= 1/2 MV^2 so you would want to maximize both your Mass(weight of the ball) and Velocity(speed of the roll) in order to achive the greatest KE.

C. see image

Explanation:

Full disclosure not totally sure what you are being asked in the last one but I put togethere a rough sketch that is my best guess.

3 0

3 years ago