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2 years ago

Which statements describe a situation in which work is being done? Check all that apply.

Physics

2 answers:

slava [35]2 years ago

5 0

The work is being done in following cases:

A mover carries a box up a flight of stairs.

A mover carries a box across a room.

Wind blows a pool chair across the yard.

A weightlifter lifts a barbell off the ground.

These are the only cases in which work is being done. The Formula of Work is:

Work= Force applied x Distance moved.

In all of these cases force is being applied and the object on which the force is being applied move in a horizontal or Vertical direction. Thus Work is done !

DanielleElmas [232]2 years ago

5 0

Answer:

A. A box is carried up a flight of stairs by a mover.

D. Wind blows a pool chair across the yard.

E. A weightlifter lifts a barbell off the ground.

Explanation:

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Ryan swings a pail of water in a vertical circle 1.0 m in radius at a constant speed. If the water is NOT to spill on him:

nexus9112 [7]

Part 1

If water does not spill at the top point of the circular motion then for the minimum speed condition we can say normal force will be zero at the top position

$F_g = ma$

$mg = m\frac{v^2}{R}$

$g = \frac{v^2}{R}$

$v = \sqrt{Rg}$

given that

R = 1 m

g = 9.8 m/s^2

now from above equation we have

$v = \sqrt{1(9.8)} = 3.13 m/s$

Part b)

for minimum value of angular speed we will have

$\omega = \frac{v}{R}$

$\omega = \frac{3.13}{1}$

$\omega = 3.13 rad/s$

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2 years ago

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Semmy [17]

Answer:

B. 1200

Explanation:

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The Moment of Inertia of the Disc is represented by $I = \frac{15}{32}\cdot M\cdot R^{2}$ . (Correct answer: A)

Let suppose that the Disk is a Rigid Body whose mass is uniformly distributed. The Moment of Inertia of the element is equal to the Moment of Inertia of the entire Disk minus the Moment of Inertia of the Hole, that is to say:

$I = I_{D} - I_{H}$ (1)

Where:

$I_{D}$ - Moment of inertia of the Disk.

$I_{H}$ - Moment of inertia of the Hole.

Then, this formula is expanded as follows:

$I = \frac{1}{2}\cdot M\cdot R^{2} - \frac{1}{2}\cdot m\cdot \left(\frac{1}{2}\cdot R^{2} \right)$ (1b)

Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole ( $m$ ):