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GREYUIT [131]
2 years ago
11

A 35.0 g bullet strikes a 5.3 kg stationary wooden block and embeds itself in the block. The block and bullet fly off together a

t 7.1 m/s. What was the original speed of the bullet? (WILL GIVE BRAINLIEST)​
Physics
1 answer:
Alisiya [41]2 years ago
3 0

Answer:

= 1200m/s or 1.2 x 10^{3} m/s

Explanation:

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Answer:

ΔK = -6 10⁴ J

Explanation:

This is a crash problem, let's start by defining a system formed by the two trucks, so that the forces during the crash have been internal and the moment is preserved

initial instant. Before the crash

        p₀ = m v₁ + M 0

final instant. Right after the crash

        p_f = (m + M) v

        p₀ = p_f

        mv₁ = (m + M) v

        v = \frac{m}{m+M} \  v_1

     

we substitute

        v = \frac{20}{20+40}   3

        v = 1.0 m / s

having the initial and final velocities, let's find the kinetic energy

        K₀ = ½ m v₁² + 0

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        K_f = ½ (m + M) v²

        K_f = ½ (20 +40) 10³  1²

        K_f = 3 10⁴ J

the change in energy is

       ΔK = K_f - K₀

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The negative sign indicates that the energy is ranked in another type of energy

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