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3 years ago

What is the weight in pounds of a 7.0 kilogram bowling ball on earth surface

Physics

1 answer:

alisha [4.7K]3 years ago

5 0

To convert a quantity in kilograms to pounds, you simply multiply by 2.2 lb/kg.

(7 kg)(2.2 lb/kg) = 15.4 lb.

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A student has the following equipment - copper wire, nail made of iron,a battery, paperclips.

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Taking the copper wire, he has to wind it around the nail made of iron. After which, he then connect both ends of the copper wire to the battery, so an electric charge travels through the wire. This is the basic electromagnet. Since a current is now flowing through the wire, a magnetic field is produced. Placing the electromagnet near the mixture of copper and iron, the magnet should attract the pieces of iron, as iron is more magnetic compared to copper. This is done over a period of time, so that only the copper pieces are left in the mixture.

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4 years ago

8. Define resistance and resistivity and also give the relation between them. Explain the

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3 years ago

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

Lilit [14]

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