Answers
The moment of inertia of Pulley is 8.89 kgm².
Angular acceleration of the pulleyThe angular acceleration of the pulley is calculated as follows;
[tex]\alpha = \frac{\omega_f - \omega _i}{t}[/tex]
where;
ωi is the initial angular velocity = 0 ωf is the final angular velocity = 17 rev/mint is the time of motionFinal angular velocity in radian per second is calculated as
[tex]\omega _f = 17 \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 \ s} = 1.78 \ rad/s[/tex]
Now, solve for angular acceleration
[tex]\alpha = \frac{1.78-0}{5} \\\\\alpha = 0.36 \ rad/s^2[/tex]
Moment of inertia of the pulleyThe the moment of inertia of Pulley is determined using the formula for torque.
Iα = Fr
I = Fr/α
I = (16 x 0.2)/(0.36)
I = 8.89 kgm²
Thus, the moment of inertia of Pulley is 8.89 kgm².
Learn more about moment of inertia here: https://brainly.com/question/25803184
Related Questions
Tarnish is produced by a redox reaction that occurs when a metal reacts with
a nonmetallic compound. The green tarnish on a copper penny might be
produced by a reaction between copper and hydrogen sulfide. What occurs
during this reaction?
A. Hydrogen sulfide acts as an acid, and copper acts as a base.
B. Copper atoms gain electrons from sulfur atoms.
C. A double-replacement reaction takes place.
D. Copper atoms lose electrons to sulfur atoms.
Answers
Answer:
Copper atoms lose electrons to sulfur atoms
Explanation:
a p e x (:
In the process of changing a flat tire, a motorist uses a hydraulic jack. She begins by applying a force of 58 N to the input piston, which has a radius r1. As a result, the output plunger, which has a radius r2, applies a force to the car. The ratio r2/r1 has a value of 8.2. Ignore the height difference between the input piston and output plunger and determine the force that the output plunger applies to the car.
Answers
Answer:
The correct answer is "3899.92 N".
Explanation:
The given values are:
Force,
[tex]F_{app}=58 N[/tex]
Ratio,
[tex]\frac{R_2}{R_1}=8.2[/tex]
As we know,
Area, [tex]A=\pi r^2[/tex]
or,
⇒ [tex]\frac{F_2}{F_1} =\frac{A_2}{A_1}[/tex]
On substituting the value of "A", we get
⇒ [tex]\frac{F_2}{F_1} =\frac{\pi r_2^2}{\pi r_1^2}[/tex]
⇒ [tex]\frac{F_2}{F_1} =\frac{r_2^2}{r_1^2}[/tex]
On applying cross-multiplication, we get
⇒ [tex]F_2=F_1\times (\frac{r_2}{r_1} )^2[/tex]
On substituting the given values, we get
⇒ [tex]=58\times (8.2)^2[/tex]
⇒ [tex]=58\times 67.2[/tex]
⇒ [tex]=3899.92 \ N[/tex]