Ask question

Ask question

All categories

3 years ago

9

Consider a motor that exerts a constant torque of 25.0 nâ‹…m to a horizontal platform whose moment of inertia is 50.0 kgâ‹…m2. A

ssume that the platform is initially at rest and the torque is applied for 12.0 rotations. Neglect friction.

1 answer:

Crazy boy [7]3 years ago

3 0

Answer:

W = 1884J

Explanation:

This question is incomplete. The original question was:

<em>Consider a motor that exerts a constant torque of 25.0 N.m to a horizontal platform whose moment of inertia is 50.0kg.m^2 . Assume that the platform is initially at rest and the torque is applied for 12.0rotations . Neglect friction. </em>

<em> How much work W does the motor do on the platform during this process? Enter your answer in joules to four significant figures.</em>

The amount of work done by the motor is given by:

$W=\Delta K$

$W= 1/2*I*\omega f^2-1/2*I*\omega o^2$

Where I = 50kg.m^2 and ωo = rad/s. We need to calculate ωf.

By using kinematics:

$\omega f^2=\omega o^2+2*\alpha*\theta$

But we don't have the acceleration yet. So, we have to calculate it by making a sum of torque:

$\tau=I*\alpha$

$\alpha=\tau/I$ => $\alpha = 0.5rad/s^2$

Now we can calculate the final velocity:

$\omega f = 8.68rad/s$

Finally, we calculate the total work:

$W= 1/2*I*\omega f^2 = 1883.56J$

Since the question asked to "<em>Enter your answer in joules to four significant figures.</em>":

W = 1884J

You might be interested in

A lighthouse is located on a small island, 3 km away from the nearest point on a straight shoreline, and its light makes four re

lbvjy [14]

Answer:

The beam of light is moving at the peed of:

$\frac{dy}{dt} = \frac{80\pi}{3}$ km/min

Given:

Distance from the isalnd, d = 3 km

No. of revolutions per minute, n = 4

Solution:

Angular velocity, $\omega = \frac{d\theta'}{dt} = 2\pi n = 2\pi \times 4 = 8\pi$ (1)

Now, in the right angle in the given fig.:

$tan\theta' = \frac{y}{3}$

Now, differentiating both the sides w.r.t t:

$\frac{dtan\theta'}{dt} = \frac{dy}{3dt}$

Applying chain rule:

$\frac{dtan\theta'}{d\theta'}.\frac{d\theta'}{dt} = \frac{dy}{3dt}$

$sec^{2}\theta'\frac{d\theta'}{dt} = \frac{dy}{3dt} = (1 + tan^{2}\theta')\frac{d\theta'}{dt}$

Now, using $tan\theta = \frac{1}{m}$ and y = 1 in the above eqn, we get:

$(1 + (\frac{1}{3})^{2})\frac{d\theta'}{dt} = \frac{dy}{3dt}$

Also, using eqn (1),

$8\pi\frac{10}{9})\theta' = \frac{dy}{3dt}$

$\frac{dy}{dt} = \frac{80\pi}{3}$

7 0

2 years ago

6. Draw a velocity-time graph for an object originally traveling at -3 m/s. The object

faltersainse [42]

See the graph in attachment

Explanation:

In this problem we have to draw a velocity-time graph for an object travelling initially at -3 m/s, then slowing down and turning around.

In the graph, we see that the initial velocity at time t = 0 is

$v_0 = -3 m/s$

and it is negative, so below the x-axis.

Later, the object slows down: this means that the magnitude of its velocity increases, therefore (since the velocity is negative) the curve must go upward, approaching and reaching the x-axis (which corresponds to zero velocity).

After that, the object's velocity keep increasing, but now it is positive: this means that the object is travelling in a direction opposite to the initial direction, so it has turned around.

Learn more about velocity:

brainly.com/question/5248528

#LearnwithBrainly

4 0

3 years ago

4.39 moles of gas in a box has a pressure of 2.25 atm at temperature of 385K. What is the volume of the box?

bazaltina [42]

Answer:

0.0619 m^3

Explanation

number of moles = n = 4.39 mol

pressure = P = 2.25 atm =2.25×1.01×10^5 Pa= 2.27×10^5 Pa

Molar gas constant =R = 8.31 J/(mol K)

Temperature T= 385K

volume of gas = V =?

BY GENERAL GAS LAW WE HAVE

PV = nRT

or V = nRT/P

or V = (4.39×8.31×385)/(2.27×10^5)

V = 0.0618728

V = 0.0619 m^3

5 0

3 years ago

Read 2 more answers

A physics major is working to pay her college tuition by performing in a traveling carnival. She rides a motorcycle inside a hol

il63 [147K]

Answer:

v = 12.1 m/s

Explanation:

When at the top of the circle, there are two forces acting on the combined mass of the rider and the motorcycle.
These are the force of gravity (downward) and the normal force, which is directed from the surface away from it, perpendicular to the surface.
In this case, as the motorcycle runs in the interior of the circle, at the top point this force is completely vertical, and is also downward.
Since the motorcycle is moving in a vertical circle, there must be a force, keeping the object moving around a circle.
This force is the centripetal force, aims towards the center of the circle, and is just the net force aiming in this direction at any point.
At the top point, this force is just the sum of the normal force and the weight of the mass of the rider and the motorcycle combined, as follows (we take the direction towards the center as positive):

$F_{c} = N + m*g (1)$

Now, we know that the centripetal force is related with the tangential speed at this point and the radius of the circle as follows:

$F_{c} = m*\frac{v^{2}}{r} (2)$

Since the normal force takes any value as needed to make (1) equal to (2), if the speed diminishes, it will be needed less force to keep the equality valid.
In the limit, when the motorcyvle tires barely touch the surface, this normal force becomes zero.
In this condition, from (1) and (2), we can find the minimum possible value of the speed that still keeps the motorcycle touching the surface, as follows:
$v_{min} =\sqrt{r*g} =\sqrt{15.0m*9.8m/s2} = 12.1 m/s (3)$

6 0

3 years ago

What is the law of conservationl

Alik [6]

Conservation law, also called law of conservation, in physics, several principles that state that certain physical properties (i.e., measurable quantities) do not change in the course of time within an isolated physical system

6 0

3 years ago