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gizmo_the_mogwai [7]
4 years ago
7

Which of the following statements are true? A solid can diffuse into a liquid, but a solid cannot diffuse into another solid. A

solid can diffuse into both a liquid and another solid. A liquid can diffuse into another liquid. A gas can diffuse into another gas.
Physics
2 answers:
Papessa [141]4 years ago
5 0

A solid can diffuse into a liquid, but a solid cannot diffuse into another solid.

A solid can diffuse into both a liquid and another solid.

skad [1K]4 years ago
3 0

Explanation:

The statements which are true are as follows.

  • A solid can diffuse into both a liquid and another solid.
  • A liquid can diffuse into another liquid.
  • A gas can diffuse into another gas.

When we heat a solid it starts to melt as a result it diffuses into liquid. Whereas when we write with a chalk on a blackboard then the solid particles of chalk disperse on the solid surface of blackboard.

Therefore, a solid can diffuse into solid as well but the rate of diffusion is slow.

When we mix two liquids they tend to mix with each other. As a result, they diffuse into each other.

Similarly, a gas can diffuse into another gas as there is weak force of attraction between gas molecules and high kinetic energy. As a result, rate of diffusion is fast in gases.

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We can compare these two interactions on the basis of impulse (see above), but sometimes, we are more interested in the forces (
kiruha [24]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

 I =  476 \ N \cdot s

b

 I_1 =  14.21 \  N\cdot s

c

    F  = 20300 \  N

Explanation:

Considering the first question

From the question we are told that

   The force produced is F  =  3400 \ N

   The duration of the punch is  t =  0.14 \  s

Generally the impulse delivered is mathematically represented as

    I =  F  *  t

=>    I =  3400  *  0.14

=>    I =  476 \ N \cdot s

Considering the second  question

   The approaching velocity of the ball is  v_b  =  45 \ m/s

    The leaving  velocity of the ball is  v_l  =  -53 \ m/s

     The mass of the ball is  m_b  =  0.145 \  kg

Generally the magnitude of the impulse delivered is mathematically represented as

     I_1 =  m*  v_b  - m *  v_l

=>     I_1 =  [0.145 *  45]  - [0.145 * -53]

=>     I_1 =  14.21 \  N\cdot s

Considering the third  question

     The  duration of the impact of the bat is  t _1 =  0.7 \ ms  =  0.7 *10^{-3} \  s

      Generally the average force exerted by the bat is mathematically represented as  

       F  =  \frac{I_1}{t_1}

=>     F  =  \frac{14.21 }{0.7 *10^{-3}}

=>       F  = 20300 \  N

 

7 0
4 years ago
Four friends each took a different path walking from the drinking fountain to the cypress tree the table shows the distance that
lorasvet [3.4K]

Answer:

:)

Explanation:

5 0
3 years ago
Which is best used with a bar magnet to produce an electric current?
Charra [1.4K]
A complete loop/ring of wire
3 0
4 years ago
If the angle of incidence is 30°, degrees what is the angle of reflection? ​
algol13

Answer:

Angle of reflection = 30°

Explanation:

One of the laws of reflection is that the angle of incidence is equal to the angle of reflection.

The angle at which the light strikes the mirror is called the angle of incidence while the angle at which the light gets reflected back is called angle of reflection.

In this case, the angle of incidence is 30 degrees. So, we can say that the angle of reflection is also equal to 30 degrees.

3 0
4 years ago
Two large, parallel, nonconducting sheets of positive charge face each other. What is at points (a) to the left of the sheets, (
alexira [117]

Answer:

a)The electric Field will be zero at the point between the sheets

b)E_1=\dfrac{\sigma}{\epsilon_0}

c)E_2=\dfrac{\sigma}{\epsilon_0}

Explanation:

Let \sigma be the surface charge density of the of the non conducting parallel sheet.Let consider a Gaussian surface in the form of of cylinder such that its cross-sectional is A . Then there will be flux only due to cross sectional area as the curved sectional is perpendicular to the the electric field  so the Electric Flux due to it is zero.

Now using Gauss law we have, E be the electric Field at the distance r from the sheet then

E\times 2A=\dfrac{\sigma A}{\epsilon_0}\\E=\dfrac{\sigma}{2\epsilon_0}

The Field will be away from the sheet and perpendicular to it.

a) The Electric Field between them

E_1=\dfrac{\sigma}{2\epsilon_0}-\dfrac{\sigma}{2\epsilon_0}\\=0

b)The Electric Field to the right of the sheets

E_1=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}

c)The Electric Field to the left of the sheets

E_2=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}

3 0
3 years ago
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