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romanna [79]
3 years ago
6

The initial kinetic energy imparted to a 0.020 kg bullet is 1200 J. (a) Assuming it

Physics
1 answer:
Lilit [14]3 years ago
5 0

Answer:

(a) Power= 207.97 kW

(b) Range= 5768.6 meter

Explanation:

Given,

Mass of bullet, m=0.02 kg

Kinetic energy imparted, K=1200 J

Length of rifle barrel, d=1 m

(a)

Let the speed of bullet when it leaves the barrel is v.

Kinetic energy, K=\frac{1}{2} mv^{2}

v=\sqrt{\frac{2K}{m} }

=\sqrt{\frac{2\times1200}{0.02} }

=346.4m/s

Initial speed of bullet, u=0

The average speed in the barrel, v_a_v_g=\frac{u+v}{2}

=\frac{0+346.4}{2} \\=173.2 m/s

Time taken by bullet to cross the barrel, t=\frac{d}{v}

=\frac{1}{173.2}\\ =0.00577 second

Power, P_a_v_g=\frac{W}{t}

=\frac{1200}{0.00577} \\=207.97kW

(b)

In projectile motion,

Maximum height, H_m=\frac{v^2\sin^2\theta}{2g} \\

Range, R=\frac{v^2\sin2\theta}{g}

given that, H_m=R

then, \frac{v^2\sin^2\theta}{2g}=\frac{v^2\sin2\theta}{g}\\\sin^2\theta=2\sin\theta\cos\theta\\\\\tan\theta=4\\\theta=\tan^-^14\\\theta=75.96^0\\R=\frac{v^2\sin2\theta}{g}\\=\frac{346.4^2\times\sin(2\times75.96)}{9.8}\\5768.6 meter

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Answer:

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Explanation:

a.

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<u>E = 86.36 x 10⁻³ V/m = 86.36 mV/m</u>

<u></u>

b.

Now, from Ohm's Law:

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Therefore,

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R = 1.27 Ω

Now, the resistance of a wire can be given as:

R = ρL/A

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Therefore,

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A point charge q = +4.50 nC moves through a potential difference ΔV = Vf − Vi = +27.0 V. What is the change in the electric pote
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For a charged particle moving in an electric field, the change in electric potential energy of the particle is given by

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For the point charge in this problem, we have:

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Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

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