In 11.8 s, 151 bullets strike and embed themselves in a wall. The bullets strike the wall perpendicularly. Each bullet has a mass of 5 x 10^-3 kg and a speed of 1110 m/s. Required:
a. What is the average change in momentum per second for the bullets?
b. Determine the average force exerted on the wall.
c. Assuming the bullets are spread out over an area of 3.0×10^−4m^2 obtain the average pressure they exert on this region of the wall.

a. ΔP/Δt =  42.6 N

b. F = 42.6 N

c. P = 142042.4 Pa = 1.42 KPa

Explanation:

a.

First, we find the change in momentum of the bullets. For one bullet:

ΔP = m(Vf - Vi)

where,

ΔP = Change in Momentum = ?

m = mass of bullet = 5 x 10⁻³ kg

Vf = Final Speed = 1110 m/s

Vi = Initial Speed = 0 m/s (Since bullets are initially at rest)

Therefore,

ΔP = (3 x 10⁻³ kg)(1110 m/s - 0 m/s)

ΔP = 3.33 N.s

For 151 bullets:

ΔP = (151)(3.33 N.s)

ΔP = 502.83 N.s

Now, dividing this by time interval, Δt = 11.8 s

ΔP/Δt = 502.83 N.s/ 11.8 s

ΔP/Δt =  42.6 N

b.

According to Newton's Second Law, the force is equal to rate of change of linear momentum:

Average Force = F = ΔP/Δt

F = 42.6 N

c.

The pressure is given by:

Average Pressure = P = Average Force/Area

P = 42.6 N/ 3 x 10⁻⁴ m²

P = 142042.4 Pa = 1.42 KPa

positive reinforcement? ?

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.