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3 years ago

15

The diagram shows a sealed container, which

2 answers:

Scrat [10]3 years ago

7 0

Answer:

The kinetic energy of the molecules would increase.

Explanation:

Drupady [299]3 years ago

5 0

Answer:

C. The kinetic energy of the molecules would increase.

Explanation:

On Edge 2021

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A horizontal pipeline is 660 mm in diameter and carries oil at a rate of 150 kg/s. this oil has a mass density of 850 kg/m3 and

timurjin [86]

A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs.
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6 0

3 years ago

Explain how the graphic organizer helped you formulate your decision and participate in the discussion.

nekit [7.7K]

Answer:

The graphic organizers help to keep track of the details. They are a visual representation of knowledge that rescue the important aspects of a concept using labels within a scheme. They also present information in a concise manner, highlighting the organization and the relationship of the concepts. Graphic organizers help students organize their thinking process and their writing.

Explanation:

6 0

3 years ago

Use the information and diagram to answer the following question.

Mumz [18]

Answer:b

Explanation:

Because I don’t know

8 0

3 years ago

A 22.0-kg child is riding a playground merrygo-round that is rotating at 40.0 rev/min. What centripetal force is exerted if he i

Tems11 [23]

Answer:

F=480.491 N

Explanation:

Given that

mass ,m = 22 kg

Angular speed ω = 40 rev/min

$\omega=\dfrac{2\pi \times 40}{60}\ rad/s$

ω =4.18 rad/s

The radius r= 1.25 m

We know that centripetal force is given as

F=m ω² r

Now by putting the values in the above equation we get

$F=22\times 4.18^2\times 1.25\ N$

F=480.491 N

Therefore the centripetal force on the child will be 480.491 N.

6 0

3 years ago

A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t

vladimir1956 [14]

a) $\omega=\frac{-mvR}{I+MR^2}$

b) $v=\frac{-mvR^2}{I+MR^2}$

Explanation:

a)

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

$L_1=0$

Later, after the girl throws the rock, the angular momentum will be:

$L_2=(I_M+I_g)\omega +L_r$

where:

$I$ is the moment of inertia of the merry-go-round

$I_g=MR^2$ is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

$\omega$ is the angular speed of the merry-go-round and the girl

$L_r=mvR$ is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

$L_1=L_2$

So we find:

$0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}$

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

b)

The linear speed of a body in rotational motion is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

$\omega=\frac{-mvR}{I+MR^2}$ is the angular speed

$r=R$ is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

$v=\omega R=\frac{-mvR^2}{I+MR^2}$

5 0

3 years ago