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viva [34]
3 years ago
9

Please help me:(

Physics
1 answer:
Lelechka [254]3 years ago
3 0

Answer:B

Explanation:

I’m doing the same thing you are!

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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
Mumz [18]

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

4 0
3 years ago
During a spring vacation, it rained on 13 days. When it rained in the morning, the afternoon was sunny. Every rainy afternoon wa
hammer [34]

Answer:

The holiday lasted for 18 days.

Explanation:

Let's assume the number of days are as following:

let the rain in the morning and lovely afternoon = X days

Clear morning and rain in the afternoon = Y days

No rain in the morning and in the afternoon = Z days

Now according to the question

Number of days with rain = X + Y = 13 days

Number of days with clear mornings = Y + Z = 11 days

Number of days with clear afternoons = X + Z = 12 days

Solving above 3 equations, we get

X = 7,  Y = 6 and  Z = 5

Now  total number of days on holiday = X+Y+Z = 7+6+5 =18

Hence, total number of days on holiday = 18 days.

8 0
4 years ago
An air-conditioning system is to be filled from a rigid container that initially contains 5 kg of liquid R-134a at 24°C. The val
OverLord2011 [107]

Answer:

Answer

The Final Quality of teh R-134a in the container  is  0.5056

The Total Heat transfer is Q_{in} = 22.62 KJ

Explanation:

Explanation is  in the following  attachments

3 0
3 years ago
Early one October, you go to a pumpkin patch to select your Halloween pumpkin. You lift the 2.9 kg pumpkin to a height of 1.5 m
Katen [24]

Answer:

Work done in lifting to a height of 1.5 m is 42.63 J.

Work done in carrying it to 50 m is 0 J

Explanation:

Given:

Mass of the pumpkin (m) = 2.9 kg

Vertical displacement of the pumpkin (y) = 1.5 m

Horizontal displacement of the pumpkin (x) = 50 m

Acceleration due to gravity (g) = 9.8 m/s²

Work done by a force is given by the formula:

W=FS\cos\theta

Where,

F\to force\ applied\\\\S\to displacement\ caused\\\\\theta\to angle\ between\ F\ and\ S

  • Now, the work done is maximum when both force and displacement is in same direction. (\theta=0^\circ)
  • Work done is minimum when both force and displacement are in opposite direction. (\theta=180^\circ)
  • Work done is zero when force and displacement are perpendicular to each other. (\theta=90^\circ)

Here, as the pumpkin is raised to a height 'h', there is force applied against gravity in the upward direction. So, force and displacement are in same direction and thus the angle is 0° between the force and displacement vectors.

So, work done is given as:

W=Force\times Vertical\ displacement

Force applied is equal to the gravitational force applied by the Earth but in the opposite direction.

So, Force = mg = 2.9\times 9.8 = 28.42\ N

So, work, W=Fy=28.42\times 1.5=42.63\ J

So, work done in lifting the pumpkin is 42.63 J.

Now, when the pumpkin is carried to the check-out stand, the displacement is horizontal while the force applied is still in the upward direction.

So, the force and displacement are perpendicular to each other. Thus, the work done is zero.

4 0
4 years ago
In what region of the em spectrum is each?
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