Ahorizontal rifle is fired at a bull's-eye. the muzzle speed of the bullet is 670 m/s. the gun is pointed directly at the center of the bull's-eye, but the bullet strikes the target 0.025 m below the center. what is the horizontal distance between the end of the rifle and the bull's-eye?

47.86 m

Explanation:

Consider the motion along the vertical direction and assume the motion in downward direction as negative

y = vertical displacement = - 0.025 m

a = acceleration = - 9.8 m/s²

t = time taken to strike the target below the center

= initial velocity along the vertical direction = 0 m/s

Using the kinematics equation

y = t + (0.5) a t²

Inserting the values

- 0.025 = (0) t + (0.5) (- 9.8) t²

t = 0.0714 sec

Consider the motion along the horizontal direction

x = horizontal displacement = horizontal distance between the end of the rifle and the bull's-eye

v₀ = velocity along the horizontal direction = 670 m/s

Since there is no acceleration along the horizontal direction, we have

x = v₀ t

inserting the values

x = (670) (0.0714)

x = 47.86 m

coil wire, increase current, put iron inside the coil

electric field is a function of 1/r^2

yes, the force per unit charge for a point charge is inverse square law.