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krok68 [10]
3 years ago
7

A 0.035 bullet strikes a 5kg wooden block that is stationary and gets stuck inside the block. The block and bullet move together

at 8.6 m/s. What is the initial velocity of the bullet?
Physics
1 answer:
hjlf3 years ago
7 0

Answer:

1240 m/s

Explanation:

[Do you mean 0.035 kg bullet?]

Since the momentum of the system is conserved,

Let u m/s be the initial velocity of the bullet

0.035u = 8.6(5+0.035)

u ~= 1237.1714 m/s

u = 1240 m/s (3 sig. fig.)

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Explanation:

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A 97.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.63 r
sammy [17]

Answer:

the final angular velocity of the platform with its load is 1.0356 rad/s

Explanation:

Given that;

mass of circular platform m = 97.1 kg

Initial angular velocity of platform ω₀ = 1.63 rad/s

mass of banana m_{b} = 8.97 kg

at distance r = 4/5  { radius of platform }

mass of monkey m_{m} = 22.1 kg

at edge = R

R = 1.73 m

now since there is No external Torque

Angular momentum will be conserved, so;

mR²/2 × ω₀ = [ mR²/2 + m_{b} (\frac{4}{5} R)² + m_{m}R² ]w

m/2 × ω₀ = [ m/2 + m_{b} (\frac{4}{5} )² + m_{m} ]w

we substitute

w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1

w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )

w = 48.55 × [ 1.63 / ( 76.3908 ) ]

w = 48.55 × 0.02133

w = 1.0356 rad/s

Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s

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