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3 years ago

Can you guys help me to solve first question???? Torque question

Physics

1 answer:

Kobotan [32]3 years ago

5 0

A uniform spherical shell of mass M = 11.0 kg and radius R = 0.480 m can rotate about a vertical axis on frictionless bearings (see the figure). A massless cord passes around the equator of the shell, over a pulley of rotational inertia I = 0.160 kg m2 and radius r = 0.110 m, and is attached to a small object of mass m = 1.60 kg. There is no friction on the pulley's axle; the cord does not slip on the pulley. What is the speed of the object when it has fallen a distance 0.700 m after being released from rest? Use energy considerations.

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The following acceleration vs time plots show

fenix001 [56]

The data set that is most likely from the car equipped with ABS is :

The First ( upper ) plot

Although the plots related to your question is missing attached below is the missing data.

The First plot ( acceleration vs time ) plot is the data that represents the application of brakes in a car fitted with ABS ( antilock braking system ) because of the absence of skidding effect between 6.5 and 9 seconds in the graph.

The skidding effect is present in the second plot around the same time because the car is not fitted with ABS

Hence we can conclude that the graph that shows the car fitted with ABS is the first plot .

Learn more : brainly.com/question/4360615

7 0

3 years ago

URGENT!!!!!!! PLEASE HELP WITH THIS PHYSICS PROBLEM

Levart [38]

Explanation:

Let

$x_1$ = distance traveled while accelerating

$x_2$ = distance traveled while decelerating

The distance traveled while accelerating is given by

$x_1 = v_0t + \frac{1}{2}at^2 = \frac{1}{2}at^2$

$\:\:\:\:\:= \frac{1}{2}(2.5\:\text{m/s}^2)(30\:\text{s})^2$

$\:\:\:\:\:= 1125\:\text{m}$

We need the velocity of the rocket after 30 seconds and we can calculate it as follows:

$v = at = (2.5\:\text{m/s}^2)(30\:\text{s}) = 75\:\text{m/s}$

This will be the initial velocity when start calculating for the distance it traveled while decelerating.

$v^2 = v_0^2 + 2ax_2$

$0 = (75\:\text{m/s})^2 + 2(-0.65\:\text{m/s}^2)x_2$

Solving for $x_2,$ we get

$x_2 = \dfrac{(75\:\text{m/s})^2}{2(0.65\:\text{m/s}^2)}$

$\:\:\:\:\:= 4327\:\text{m}$

Therefore, the total distance x is

$x = x_1 + x_2 = 1125\:\text{m} + 4327\:\text{m}$

$\:\:\:\:= 5452\:\text{m}$

3 0

3 years ago

A toy cart at the end of a string 0.70 m long moves in a circle on a table. The cart has a mass of 2.0 kg and the string has a b

luda_lava [24]

Given that the mass of the toy cart is 2.0 kg and and the acceleration is unknown, the normal formula would be a=f/m where a is acceleration, f is force and m is mass but the string's breaking strength is 40n so I think the formula in this case will be f is greater than m*a
40 is greater than 2a
40 is greater than 2a
40/2 is greater than 2a/2
20m/s² is greater than a
Therefore the maximum speed the toy cart should have should be less than 20m/s²

8 0

4 years ago

A 900-N lawn roller is to be pulled over a 5-cm (high) curb. Radius of roller is 25 cm. What minimum pulling force is needed if

Feliz [49]

I believe b 30 degrees

8 0

4 years ago

........................

mezya [45]

Answer:

uh what are you saying

6 0

3 years ago