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Vanyuwa [196]
3 years ago
8

Which of the following statement(s) about energy and phase is/are correct? Select all that apply. Choose one or more: A. While o

nly one phase is present, adding or removing energy changes PE but not KE. B. While only one phase is present, adding or removing energy changes KE but not PE. C. During a phase change, adding or removing energy changes KE but not PE. D. During a phase change, adding or removing energy changes PE but not KE.
Physics
1 answer:
Free_Kalibri [48]3 years ago
3 0

Answer: Option (A) and (D) are the correct statements.

Explanation:

Potential energy is defined as the energy obtained by an object due to its position. Whereas kinetic energy is defined as the energy obtained by an object due to  the motion of its molecules.

When there will be only one phase present then addition or removal of energy will lead to change in kinetic energy of the substance but no change in potential energy will take place.

Whereas if change in phase is occurring then adding or removing any energy will lead to change in potential energy of the substance while kinetic energy will remain the same.

Thus, we can conclude that correct statements are as follows.

  • While only one phase is present, adding or removing energy changes PE but not KE.
  • During a phase change, adding or removing energy changes PE but not KE.
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A 5 kg ball is sitting on top of a hill. It has a Potential Energy of 6000J. What is the height of the ball?​
natali 33 [55]

Explanation:

mass=5 kg

potential energy=6000j

height=?

Now

potential energy =m.g.h

or 6000=5*9.8*h

or 6000=49h

or 6000÷49=h

or h= 122.45m

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Despite a very strong wind, a tennis player
Gnoma [55]

Answer:

Option 5. 1 and 3

Solution:

The only forces acting on the tennis ball after it has left contact with the racquet and the instant before it touches the ground are the force of gravity in the downward direction and the force by the air exerted on the ball.

The ball after it left follows the path of trajectory and as it moves forward in the horizontal direction the force of the air acts on it.

In the whole projectile motion of the ball, the acceleration due to gravity acts on the ball thus the force of gravity acts on the ball in the downward direction  before it hits the ground.

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3 years ago
Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi
tigry1 [53]

Answer:

a

The  radial acceleration is  a_c  = 0.9574 m/s^2

b

The horizontal Tension is  T_x  = 0.3294 i  \ N

The vertical Tension is  T_y  =3.3712 j   \ N

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

   The length of the string is  L =  10.7 \ cm  =  0.107 \ m

     The mass of the bob is  m = 0.344 \  kg

     The angle made  by the string is  \theta  =  5.58^o

The centripetal force acting on the bob is mathematically represented as

         F  =  \frac{mv^2}{r}

Now From the diagram we see that this force is equivalent to

     F  =  Tsin \theta where T is the tension on the rope  and v is the linear velocity  

     So

          Tsin \theta  =   \frac{mv^2}{r}

Now the downward normal force acting on the bob is  mathematically represented as

          Tcos \theta = mg

So

       \frac{Tsin \ttheta }{Tcos \theta }  =  \frac{\frac{mv^2}{r} }{mg}

=>    tan \theta  =  \frac{v^2}{rg}

=>   g tan \theta  = \frac{v^2}{r}

The centripetal acceleration which the same as the radial acceleration  of the bob is mathematically represented as

      a_c  =  \frac{v^2}{r}

=>  a_c  = gtan \theta

substituting values

     a_c  =  9.8  *  tan (5.58)

     a_c  = 0.9574 m/s^2

The horizontal component is mathematically represented as

     T_x  = Tsin \theta = ma_c

substituting value

   T_x  = 0.344 *  0.9574

    T_x  = 0.3294 \ N

The vertical component of  tension is  

    T_y  =  T \ cos \theta  = mg

substituting value

     T_ y  =  0.344 * 9.8

      T_ y  = 3.2712 \ N

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is  

         

       T  = T_x i  + T_y  j

substituting value  

      T  = [(0.3294) i  + (3.3712)j ] \  N

         

3 0
3 years ago
A charge of 5.67 x 10^-18 C is placed 3.5 x 10^-6 m away from another charge of -
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+ 1.58 e -15

Please hit thanks button! :)
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A 24 cm radius aluminum ball is immersed in water. Calculate the thrust you suffer and the force. Knowing that the density of al
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Answer:

W =1562.53 N

Explanation:

It is given that,

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The density of Aluminium, d=2698.4\ kg/m^3

We need to find the thrust and the force. The mass of the liquid displaced is given by :

m=dV

V is volume

Weight of the displaced liquid

W = mg

W=dVg

So,

W=dg\times \dfrac{4}{3}\pi r^3\\\\W=2698.4\times 10\times \dfrac{4}{3}\times \pi \times (0.24)^3\\\\W=1562.53\ N

So, the thrust and the force is 1562.53 N.

7 0
3 years ago
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