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ohaa [14]
3 years ago
6

There is no blank in time between the action and reaction

Physics
1 answer:
Anna11 [10]3 years ago
4 0
I'm not sure what you mean by this...
Please rephrase the question. Thanks!
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Its often used for monuments because of its beauty and its variety of colors
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An object in motion is usually seen moving in relation to an object that appears to
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The answer is Stay In Place
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What are Base Quantities ?​
Norma-Jean [14]

Answer:

<em>physical quantities that cannot be defined in terms of other quantities.</em>

<em>Examples</em><em>:</em><em> </em><em>Length - meter (m)Time - second (s)Amount of substance - mole (mole)Electric current - ampere (A)Temperature - kelvin (K)Luminous intensity - candela (cd)Mass - kilogram (kg)</em>

6 0
3 years ago
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Select the statements that correctly interpret the data in this heating curve.
tekilochka [14]

The consistent temperature indicated by section 2 represents the flow of energy lost as the water freezes.

Section 2 indicates that energy is being utilized to bring all molecules to the melting point.

Section 4 indicates that the temperature is stable until all molecules reach the boiling point.

The increasing temperature in section 5 indicates that molecular motion is increasing.

Explanation:

The figure is missing: find it in attachment.

The graph represents the temperature of a block of ice as heat is continuously supplied to it.

At the beginning (section 1), the temperature increases: in this phase, the heat supplied to the ice is used to increase the kinetic energy of vibration of the molecules of ice.

When the ice reach point A (melting point), the ice starts to melt. In this phase (section 2), the temperature of the substance remains constant, because all the heat supplied is used to break the bonds between the molecules, converting the ice into liquid water.

At point B, the ice has completely melted, and therefore we now have liquid water.

In sector 3 (between B and C), the temperature of the water increases again as we supply more and more heat: this heat, in fact, is used to increase the kinetic energy of the molecules, which move faster and faster.

Then, the water reaches the boiling point (point C): after that, the water starts to boil, (sector 4) and the heat supplied is entirely used to remove the intermolecular forces between the molecules of water; as a result, the temperature remains constant during the process.

At poind D, the water has completely boiled, and converted into gas state (steam).

After that, the temperature of the steam starts to increase again (sector 5) as more and more heat is supplied, because it is used to increase the kinetic energy of the molecules.

Therefore, the correct statements are:

The consistent temperature indicated by section 2 represents the flow of energy lost as the water freezes.

Section 2 indicates that energy is being utilized to bring all molecules to the melting point.

Section 4 indicates that the temperature is stable until all molecules reach the boiling point.

The increasing temperature in section 5 indicates that molecular motion is increasing.

3 0
3 years ago
Read 2 more answers
The rotational inertia of a thin rod about one end is 1/3 ML2. What is the rotational inertia of the same rod about a point loca
zlopas [31]

Answer:

The value is  I = 0.0932 ML ^2  

Explanation:

From the question we are told that

  The rotational inertia about one end is I_R =  \frac{1}{3} ML^2

   The location of the axis of rotation considered is d =  0.4 L

Generally the mass of the portion of the rod from the axis of rotation considered to the end of the rod is  0.4 M

Generally the length of the rod from the its beginning to the axis of rotation consider is

      k = 1 - 0.4 L = 0.6L

Generally the mass of the portion  of the rod from the its beginning to the axis of rotation consider is

    m  =  1- 0.4 M = 0.6 M

Generally the rotational inertia about the axis of rotation consider for the first portion of the rod is

     I_{R1} =  \frac{1}{3} (0.6 M )(0.6L)^2

    I_{R1} =  \frac{1}{3} (0.6 M )L^2 0.6^2

Generally the rotational inertia about the axis of rotation consider for the second  portion of the rod is

     I_{R2} =  \frac{1}{3} (0.6 M )(0.6L)^2

=> I_{R2} =  \frac{1}{3} (0.4 M )(0.4L)^2

=>  I_{R2} =  \frac{1}{3} (0.4 M )L^2 0.4^2

Generally by the principle of superposition that rotational inertia of the rod at the considered axis of rotation is

  I =   \frac{1}{3} (0.6 M )L^2 0.6^2 +   \frac{1}{3} (0.4 M )L^2 0.4^2

=>   I =  \frac{1}{3} ML ^2  [0.6 * (0.6)^2 + 0.4 * (0.4)^2 ]

=>   I = 0.0932 ML ^2  

8 0
3 years ago
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