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BabaBlast [244]
3 years ago
14

A bullet with a mass of 0.02 kg is fired horizontally into a block of wood hanging on a string. The bullet sricks in the wood an

d causes it to swing upward to a height of 0.2 m. If the mass of the wood block is 2 kg, what is the initial speed of the bullet?

Physics
1 answer:
djyliett [7]3 years ago
7 0

Answer:

u= 20.09 m/s

Explanation:

Given that

m = 0.02 kg

M= 2 kg

h= 0.2 m

Lets take initial speed of bullet = u m/s

The final speed of the system will be zero.

From energy conservation

1/2 m u²+ 0 = 0+ (m+M) g h

m u²=2 (m+M) g h

By putting the values

0.02 x u² = 2 (0.02+2) x 10 x 0.2       ( take g=10 m/s²)

u= 20.09 m/s

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a) T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}

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Ordering this periods from largest to smallest:

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b) Acceleration a is defined as the variation of velocity in time:

a=\frac{V}{T} (9)

Applying this equation to each satellite:

a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2} (10)

a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2} (11)

a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2} (12)

a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2} (13)

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a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2} (15)

Ordering this acceerations from largest to smallest:

a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}

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