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2 years ago

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A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path wit

h a velocity of 103 m/s west. What is the final velocity of the block? Assume the block rests on a perfectly frictionless horizontal surface.

1 answer:

makvit [3.9K]2 years ago

7 0

Answer:

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

Initial velocity of block = 0 m/s

Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

Final velocity of bullet = -103 m/s

We need to find final velocity of the block( u )

Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u

We have

2.87 = -1.442 + 1.8 u

u = 2.40 m/s

Final velocity of the block = 2.40 m/s east.

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Answer:

The distance between knothole and the paint ball is 0.483 m.

Explanation:

Given that,

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The angle at which the forester aims his gun are,

$\tan\theta=\dfrac{4}{15}$

$\tan\theta=0.266$

$\cos\theta=\dfrac{15}{\sqrt{15^2+4^2}}$

$\cos\theta=0.966$

Using the equation of motion of the trajectory

The horizontal displacement of the paint ball is

$x=(u\cos\theta)t$

$t=\dfrac{x}{u\cos\theta}$

Using the equation of motion of the trajectory

The vertical displacement of the paint ball is

$y=u\sin\theta(t)-\dfrac{1}{2}gt^2$

$y=u\sin\theta(\dfrac{x}{u\cos\theta})-\dfrac{1}{2}g(\dfrac{x}{u\cos\theta})^2$

$y=x\tan\theta-\dfrac{gx^3}{2u^2(\cos\theta)^2}$

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$y=(15\times0.266)-(\dfrac{9.8\times(15)^2}{2\times(50)^2\times(0.966)^2})$

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A bag of potato chips contains 2.00 L of air when it is sealed at sea level at a pressure of 1.00 atm and a temperature of 20.0°

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Answer:

The volume at mountains is 2.766 L.

Explanation:

Given that,

Volume $V_{1} = 2.00\ L$

Pressure $P_{1}= 1.00\ atm$

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Temperature $T_{1}= 20.0°C = 293\ K$

Temperature $T_{2}= 7.00°C = 280\ K$

We need to calculate the volume at mountains

Using gas law

$\dfrac{PV}{T}=\ Constant$

For both temperature,

$\dfrac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}$

Put the value into the formula

$\dfrac{101.325\times2}{293}=\dfrac{70\times V_{2}}{280}$

$V_{2}=\dfrac{101.325\times2\times280}{293\times70}$

$V_{2}=2.766\ litre$

Hence, The volume at mountains is 2.766 L.

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