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xxMikexx [17]
3 years ago
15

Consider the following four objects: a hoop, a flat disk, a solid sphere, and a hollow sphere. Each of the objects has mass M an

d radius R. The axis of rotation passes through the center of each object, and is perpendicular to the plane of the hoop and the plane of the flat disk. Which of these objects requires the largest torque to give it the same angular acceleration?
Physics
1 answer:
Mumz [18]3 years ago
4 0

Answer:

The hoop

Explanation:

We need to define the moment of inertia of the different objects, that is,

DISK:

I_{disk} = \frac{1}{2} mR^2

HOOP:

I_{hoop} = mR^2

SOLID SPHERE:

I_{ss} = \frac{2}{5}mR^2

HOLLOW SPHERE

I_{hs} = \frac{2}{3}mR^2

If we have the same acceleration for a Torque applied, then

mR^2>\frac{2}{3}mR^2>\frac{1}{2} mR^2>\frac{2}{5}mR^2

I_{hoop}>I_{hs} >I_{disk}>I_{ss}

The greatest momement of inertia is for the hoop, therefore will require the largest torque to give the same acceleration

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irakobra [83]

The answer is true about the cabins in commercial airliners that require pressurization.

<h3>Why are the cabins of commercial airplanes pressurized?</h3>

Airplanes are pressurized because the air is very thin at the high altitude where they fly. The passenger jet has a cruising altitude of about 30,000 - 40,000 feet. At this altitude or height, humans can't breathe very well and our body gets less amount of oxygen. Most aircraft cabins are pressurized to an altitude about 8,000 feet. This is called cabin altitude. Aircraft pilots have access to the control's mode of a cabin pressure control system and if needed it can command the cabin to depressurize.

So we can conclude that cabins in commercial airliners require pressurization because of the greater pressure of the surrounding environment.

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4 0
1 year ago
A 13.0-g wad of sticky clay is hurled horizontally at a 110-g wooden block initially at rest on a horizontal surface. The clay s
Alisiya [41]

Answer:

v_{ic}=92.53 m/s

Explanation:

We need to apply conservation of momentum and energy to solve this problem.

<u>Conservation of momentum</u>

p_{i}=p_{f}

m_{c}v_{ic}=(m_{c}+m_{w})V (1)

  • m(c) is the mass of stick clay
  • m(w) is the mass of the wooden block
  • v(ic) is the initial velocity of clay
  • V is the final velocity of the system clay plus wood.

<u>Conservation of total energy</u>

The change in kinetic energy is equal to the change in internal energy, in our case it would be the energy loss due to the friction force. Let's recall the definition of work, it is the dot product between force and displacement, Therefore:

\Delta E=W

\frac{1}{2}(m_{c}+m_{w})V^{2}=F_{friction}*d

\frac{1}{2}(m_{c}+m_{w})V^{2}=\mu (m_{c}+m_{w})gd

We can find V from this equation:

V=\sqrt{2\mu gd}=\sqrt{2*0.65*9.81*7.5}=9.78 m/s

Now, let's put V into the equation (1) and find v(ic)

v_{ic}=\frac{(m_{c}+m_{w})V}{m_{c}}=\frac{123*9.78}{13}=92.53 m/s

I hope it helps you!  

<u />

8 0
3 years ago
A support beam needs to be placed at a 28° angle of elevation so that the top meets a vertical beam 1.6 meters above the horizon
k0ka [10]

The distance  from the vertical beam that the lower end of the support beam be placed along the horizontal floor is 3.00 m

<h3>Angle of elevation</h3>

The angle of elevation is the angle to which the eye must be raised in order to see an elevated obect at a height.

Now, we know that a right angle triangle is formed; we can obtain the distance  from the vertical beam that the lower end of the support beam be placed along the horizontal floor from;

tan 28 = 1.6/x

x = 1.6 /tan 28

x = 3.0 meters

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7 0
2 years ago
If Sherry took her pulse for 6 seconds and felt 14 beats, What would her heart rate be?
Schach [20]

Answer:

140 beats per minute

Explanation:

There are 60 seconds in a minute and we know that Sherry felt 14 beats in 6 seconds if you multiply 6 by 10 you get 60 seconds which can be 1 minute. Then you multiply 14 by 10 since you 14 by 10 since you multiplied 6 by 10 and you get 140 beats per minute.

5 0
3 years ago
How much kinetic energy does a 4kg cat have while running at 9 m/s
ludmilkaskok [199]

Answer:

How much kinetic energy does a 4 Kg cat have while running at 9 m/s?

its 5 J of kinetic energy.

Explanation:

3 0
2 years ago
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