Ask question

Ask question

All categories

4 years ago

14

A particular automotive wheel has an angular moment of inertia of 12 kg*m^2, and is decelerated from 135 rpm to 0 rpm in 8 secon

ds. a. How much torque is required to do this? b. How much work is done to accomplish this?

1 answer:

lozanna [386]4 years ago

8 0

Answer:

(A) Torque required is 21.205 N-m

(b) Wok done will be equal to 1199.1286 j

Explanation:

We have given moment of inertia $I=12kgm^2$

Wheel deaccelerate from 135 rpm to 0 rpm

135 rpm = $135\times \frac{2\pi }{60}=14.1371rad/sec$

Time t = 8 sec

So angular speed $\omega _i=135rpm$ and $\omega _f=0rpm$

Angular acceleration is given by $\alpha =\frac{\omega _f-\omega _i}{t}=\frac{0-14.1371}{8}=--1.7671rad/sec^2$

Torque is given by torque $\tau =I\alpha$

$=12\times 1.7671=21.205N-m$

Work done to accelerate the vehicle is

$\Delta w=K_I-K_F$

$\Delta W=\frac{1}{2}\times 12\times 14.137^2-\frac{1}{2}\times 12\times0^2=1199.1286J$

You might be interested in

Can anyone plz help me

juin [17]

Answer:

D,B,C,A,C

Explanation:

I believe that is the correct answers but it is unclear. I don't think the key for the second last question would let the current flowing so the bulb would be off.

8 0

3 years ago

Barium, a group 2 element, forms an ionic compound with sulfur, a group 16 element. What is the formula for barium sulfide?

WITCHER [35]

Answer:BaS

Explanation:

Ba- barium

S-sulfur

6 0

3 years ago

A 0.041-kg bullet has a kinetic energy of 600j. What is the velocity of the bullet?

Paladinen [302]

171.0798 M/S

In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s2.

Was this helpful

5 0

3 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

3 0

3 years ago

A pendulum has a mass of 3kg and is lifted to a height of .3m. What is the maximum speed of the pendulum

Kryger [21]

Given data

*The given mass of the pendulum is m = 3 kg

*The given height is h = 0.3 m

The formula for the maximum speed of the pendulum is given as

$v_{\max }=\sqrt[]{2gh}$

*Here g is the acceleration due to the gravity

Substitute the values in the above expression as

$\begin{gathered} v_{\max }=\sqrt[]{2\times9.8\times0.3} \\ =2.42\text{ m/s} \end{gathered}$

Hence, the maximum speed of the pendulum is 2.42 m/s

7 0

1 year ago