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stepan [7]
3 years ago
7

15 POINTS

Physics
1 answer:
zhenek [66]3 years ago
3 0

Answer:

Acceleration of the bullet while moving into the clay is -312500 ms⁻²

Explanation:

Given data

Initial velocity (v₁) = 250 m/s                 Final velocity (v₂) = 0 m/s

Distance (s)= 0.1 m                            acceleration (a) = ?

Using 3rd equation of motion

                  2as = v₂²- v₁²

       2 × a × 0.1 = 0² - 250²

            0.2 × a = -62500

                      a = -62500/0.2

                     a = -312500 m/s²

Here negative sign indicate that bullet slow down and it is deceleration i.e negative acceleration.

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\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)

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Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m

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\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\

Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C

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