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Mandarinka [93]
4 years ago
15

an ice cube is placed in the sun later there is a puddle of water later still the.puddle is gone what kind of change has occured

​
Physics
1 answer:
TiliK225 [7]4 years ago
7 0

Clear question is;

An ice cube is placed in the sun. Later there is a puddle of water. Later still the puddle is gone. What kind of change has occurred?

Answer:

Physical change.

Explanation:

There are two type of changes called physical change and chemical change.

In Physical changes, there is only change in the appearance of the substance and not its chemical composition. Whereas, in Chemical changes the substance changes entirely into another substance with a new chemical formula.

Now, for the ice cube, it has melted from a solid state to a liquid state. Thus, there is only a change in appearance and not its chemical composition. Thus, it's a physical change that has occurred.

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Two children hang by their hands from the same tree branch. the branch is straight, and grows out from the tree trunk at an angl
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The net torque exerted by the children on the branch of the tree is 1382 N-m.

The torque exert by the kids is calculated as

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4 years ago
A block of mass m is pushed up against a spring with spring constant k until the spring has been compressed a distance x from eq
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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
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