A 4.2-g bullet leaves the muzzle of a rifle with a speed of 338 m/s. What force (assumed constant) is exerted on the bullet while it is traveling down the 0.83-m-long barrel of the rifle

289.05N

Explanation:

The mass of the bullet is 4.2-g bullet, if we convert it to "kg" we have

4.2-g bullet/1000

= 0.0042kg

The speed of the bullet leaving the missle is 338 m/s.

The Lenght of the barrel= 0.83-m-long

From kinematics, v²= u² +2al

V= final speed when it leaves the barell=

338 m/s

U= intial speed=0

L= length of the barrel travelled by the bullet

V²= 2al( initial speed=0)

If we make "a" subject of the formula we have

a=v²/2l

a= (338)²/(2×0.83)

=68821.687 m/s²

We can now calculate the force acting on the body using the Expression below

F=ma

Where F= force acting on the body

m= mass

a= Acceleration

F= 0.0042kg×68821.687

= 289.05N

Therefore, the force exerted on the bullet while it is traveling down the 0.83-m-long barrel of the rifle is 289.05N

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

im doing peak rn so sry if its wrong

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)