Ahorizontal rifle is fired at a bull's-eye. the muzzle speed of the bullet is 660 m/s. the gun is pointed directly at the center of the bull's-eye, but the bullet strikes the target 0.027 m below the center. what is the horizontal distance between the end of the rifle and the bull's-eye?

The answer is 48.5 m

Explanation:

First we need to know that in the y-axis there is acceleration, in specific the acceleration caused by the gravity. I consider this value as a 10 m/s2.

Then, we define the next equation for the y-axis:

Where:

Vi = 0 (because at the beginning there is no vertical velocity)

d = 0.027

Finally we have:

Vf = 0.7348 m/s

Also, we define another equation for the y-axis:

we replace the variables and we have:

t = 0.07348 s

Finally, to calculate the horizontal distance, we need the next equation:

Replacing the variables we have as a result:

D = 48.5 m

the answer is b i think

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6