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Marianna [84]
3 years ago
13

It is well known that bullets and other missiles fired at Superman simply bounce off his chest. Suppose that a gangster sprays S

uperman's chest with 6.5 g bullets at the rate of 170 bullets/min, and the speed of each bullet is 540 m/s. Suppose too that the bullets rebound straight back with no change in speed. What is the magnitude of the average force on Superman's chest from the stream of bullets
Physics
1 answer:
valkas [14]3 years ago
6 0

Answer:

F = 76.05\,N

Explanation:

The phenomenon can be described and analyzed by means of the Principle of Momentum Conservation and the Impact Theorem:

(6.5\times 10^{-3}\,kg)\cdot (650\,\frac{bullets}{min} )\cdot (\frac{1\,min}{60\,s} )\cdot (540\,\frac{m}{s} ) - F = (6.5\times 10^{-3}\,kg)\cdot (650\,\frac{bullets}{min} )\cdot (\frac{1\,min}{60\,s} )\cdot (-540\,\frac{m}{s} )

The magnitude of the average force on Superman's chest is:

F = 76.05\,N

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Answer:

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Initial momentum = final momentum

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⇒ 730 ×v = (4054.9 - 2081.2) =1973.7

⇒v=2.7 m/s

Thus, the resulting speed of the block is 2.7 m/s.

(b) since, the momentum is conserved, the speed of the bullet-block center of mass would be constant.

V_{COM} = \frac{m_b}{m_b+m_{bl}}v_{bi}=\frac{4.30}{4.30+730}\times 943 m/s = 5.52 m/s

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3 years ago
3. A coil of 100 turns encloses an area of 100 cm2. It is placed at an angle of 700 with a
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Explanation:

Given that,

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Magnetic flux is given by :

\phi =BA\cos\theta\\\\\text{For N turns},\\\phi =NBA\cos\theta \\\\\phi=100\times 0.1\times 0.01\times \cos(70)\\\\=0.034\ Wb

So, the magnetic flux through the coil is 0.1 Wb.

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\epsilon=\dfrac{-d\phi}{dt}\\\\=\dfrac{0.034}{10^{-3}}\\\\=34\ V

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Answer:

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