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liberstina [14]
3 years ago
10

What is its time interval between the release

Physics
1 answer:
azamat3 years ago
3 0
There are several many equations that are available to relate the distance,
speed, and time of a body moving vertically in gravity.  Happily, the only one
I can always remember without looking it up happens to be the right one to
use for this question !

             Distance  =  (1/2) x (gravity) x (time)²

                   3.8 m  =  (1/2) x (9.8 m/s²) x (time)²

Divide each side
by 4.9 m/s² :

                   (3.8 m) / (4.9 m/sec²)  =  (time)²

                     0.7755  sec²  =  time²

Square root
of each side:

                       0.88 second  =  time
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Answer:

The value is  KE_b =0.710 \ J

Explanation:

From the question we are told that

   The mass  of the bullet is m_b  = 5.00 \ g  = 0.005 \  kg

  The mass  of the wood is  m_w =  900 \  g  =  0.90\  kg

   The height attained by the combined mass is  h =  8.0 \ cm  =  0.08 \ m

Generally according to the law of energy conservation    

    KE _b  =  PE_c

Here KE_b is the kinetic energy of the bullet before collision.

and PE_c is the  potential  energy of the combined mass of bullet and wood at the height h which is mathematically represented as

      PE_m  =  [m_b  + m_w] *  g *  h

So

   KE_b =PE_c   = [0.005  + 0.90] * 9.8 *0.08

=> KE_b =0.710 \ J

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What do you think would happen to the force of attraction of two interacting charges if their distance apart is halved?
sweet [91]

Answer:

The new force becomes 4 times the initial force.

Explanation:

The force of attraction or repulsion is given by the relation as follows :

F=k\dfrac{q_1q_2}{d^2}

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F is inversely proportional to the distance between charges.

If the distance is halved, d'=(d/2), new force is given by :

F'=k\dfrac{q_1q_2}{d'^2}\\\\=k\dfrac{q_1q_2}{(\dfrac{d}{2})^2}\\\\=k\dfrac{q_1q_2}{\dfrac{d^2}{4}}\\\\=4\times \dfrac{kq_1q_2}{d^2}\\\\F'=4F

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