A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (assume a uniform acceleration). a. 1.62*105 m/s2.b. 1000 m/s2.c. 0.62*105 m/s2.d. 3.62*105 m/s2.

Explanation:

Given

velocity of the bullet = 521m/s

distance = 0.840m

required

acceleration of the bullet a

Using the equation of motion:

v² = u²+2aS where:

v is the final velocity

u is the initial velocity

a is the acceleration

S is the distance

Substitute the given parameters into the equation to get a:

521² = 0²+2a(0.840)

271,441 = 0+1.68a

271,441 = 1.68a

a = 271,441/1.68

a = 161,572.02

a = 1.62×10⁵m/s²

Hence the acceleration of the bullet is 1.62×10⁵m/s²

a charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a third charge so it would experience no net electric force? enter an expression for the exact answer, which will involve a square-root sign. (do not enter an approximate decimal answer.)

ill do it for 100, also this isnt 80 its 40