(1)The initial kinetic energy imparted to a 0.0210 kg bullet is 1298 J. (a) Assu

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Question

(1)The initial kinetic energy imparted to a 0.0210 kg bullet is 1298 J.

(a) Assuming it accelerated down a 1.00 m long rifle barrel, estimate the average power delivered to it during the firing.
___________kW

(b) Neglecting air resistance, find the range of this projectile when it is fired at an angle such that the range equals the maximum height attained.

______________km

(a) How much work does the force of the pole do on the boat during the displacement?

(b) Does the boat

move at constant speed?slow down? speed up?

(b) What is the angle (in degrees) between the force and the displacement?
°

the kinetic energy of the bullet is

K = 1/2 * m v^2

v = root 2 K/m = root 2 (1298 J)/ 0.0210 kg) = 351.59 m/s

power is

P = Fv = (W/s) v =(1298 J/1 m) ( 351.59)=456370.3 W

(b)

the maximum height reached by projectile is

H = u^2 sin ^2 theta/ 2g

range R = u^2 sin 2thea/g

u^2 sin ^2 theta/ 2g= u^2 sin 2thea/g

2 sin(theta) cos(theta) = sin(theta) * sin ( theta) / 2
4 cos ( theta ) = sin (theta)
Tan( theta ) = 4
theta = 75.96 degrees

Therefore range of the projectile

R = V ^ 2 sin 2( Theta) / g
R = 351.59^ 2 * sin 2(75.96) / 9.8
= 5937.3 m or 5.93 km

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