Question

A bullet of mass 30 grams passes between the two ends of a 40 m long road in 0.3 seconds the bullet travels with a constant speed<br /><br />What is the kinetic energy of the bullet?<br /><br />

Answer 1

Hi friend,

We know that,

K.E= 1/2 mv²

M= 0.03 kg

V= distance/time

V= 40/0.3

V=133 m/s

K.E= 1/2×{0.03)×{133}

K.E=265.3=265(app.) J

Hope this helped you a little!!!

Answer 2

$\maltese \: \red{\mid{\underline{\overline{\textbf{Answer}}}\mid}}$

Kinetic energy is the energy possessed by an object dye to it's motion. It is half of product of mass and square of velocity. Represented by :

$\maltese \: ke = \frac{1}{2} \times m {v}^{2}$

Now in this case :

• Mass of bullet = 30g=0.03kg
• Distance = 40m
• Time = 0.3s

Now velocity of this bullet = Distance / time

$\maltese \: v = \frac{d}{t}$

$\maltese \: v = \frac{40}{0.3} = \frac{400}{3}$

$\maltese \: v = 133.3 \frac{m}{s}$

Now the kinetic energy of bullet will be :

$\maltese \: ke = \frac{1}{2} \times m {v}^{2}$

After inputting the values we get :

$\maltese \: ke = \frac{1}{2} \times 0.03 \times {133}^{2}$

$\maltese \: ke = \frac{530}{2}$

$\maltese \: ke = 265j$

Therefore the kinetic energy of bullet is 265J.

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BY SCIVIBHANSHU

THANK YOU

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