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Lynna [10]
3 years ago
9

An 8.0g bullet, moving at 400 m/s, goes through a stationary block of wood in 4.0 x 10^-4s, emerging at a speed of 100 m/s. (a)

what average force did the wood exert on the bullet? (b) how thick is the wood?
Physics
1 answer:
Ierofanga [76]3 years ago
7 0

Answer:

Explanation:

Initial velocity (u) of the Bullet = 400 m/sec

Final velocity (v) of the Bullet = 100 m/sec

Bullet passed through the block in (t) = 0.0004 sec  

Using 1st Equation of motion :

400 m/s = 100 m/s - a (0.0004)

<em>Deceleration of Bullet = 750,000 m/sec^2 </em>

<em />

(a) F (force exerted by the wooden block on the bullet) = F (force exerted by the bullet on the wooden block)

F = m * a = 0.008 * 750,000<em> = </em>6000 N

(b) Using 3rd Equation of motion :

v^{2} = u^{2} - 2aS

10000 = 160000 - 2 * 750,000 * S

Thickness of wood (S) = 0.1 m

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39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th
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<u>Answer:</u> The final temperature of the solution is 80.14^oC

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, \text{heat}_{absorbed}=\text{heat}_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

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m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]    ..........(1)

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q = heat absorbed or released

m_1 = mass of aluminium = 39 g

m_2 = mass of coffee = 166 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 83^oC

c_1 = specific heat of aluminium = 0.904J/g^oC

c_2 = specific heat of coffee= 4.1801J/g^oC

Putting all the values in equation 1, we get:

39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]

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