<span>Data:
mass =
110-g bullet
d = 0.636 m
Force =
13500 + 11000x - 25750x^2, newtons.
a) Work, W
W = ∫( F* )(dx) =∫[13500+ 11000x - 25750x^2] (dx) =
W = 13500x + 5500x^2 - 8583.33 x^3 ] from 0 to 0.636 =
W = 8602.6 joule
b) x= 1.02 m
</span><span><span>W = 13500x + 5500x^2 - 8583.33 x^3 ] from</span> 0 to 1.02
W = 10383.5
c) %
[W in b / W in a] = 10383.5 / 8602.6 = 1.21 => W in b is 21% more than work in a.
</span>
The correct answer is 1.25 because it is 1/2 of 1 1/2 and that is 1.25.
Answer:
54.67 N
Explanation:
The total energy produced is the product of power and time duration:
E = Pt = 82 * 1 = 82 J
Which is converted from work, product of forced extended over a displacement
W = E = Fs = F*1.5 = 82
F = 82 / 1.5 = 54.67 N
So the magnitude of the force exerting on the handle is 54.67 N
Answer:
e. Only(a) and (b) above are correct
Explanation:
Impulse
= Fx t = m ( v-u )
v-u = change in velocity
F x t = mass x change in velocity
change in velocity = F t / mass
=a t
change in velocity ∝ t ( time ) , if a is constant
dv = a_avg dt
∫dv = a_avg ∫dt
v-u = a_avg t
change in velocity ∝ t ( time )
So both (a) and (b) are correct.
