Answer:
v_{f} = 74 m/s, F = 230 N
Explanation:
We can work on this exercise using the relationship between momentum and moment
I = ∫ F dt = Δp
bold indicates vectors
we can write this equations in its components
X axis
Fₓ t = m ( -v_{xo})
Y axis
t = m (v_{yf} - v_{yo})
in this case with the ball it travels horizontally v_{yo} = 0
Let's use trigonometry to write the final velocities and the force
sin 30 = v_{yf} / vf
cos 30 = v_{xf} / vf
v_{yf} = vf sin 30
v_{xf} = vf cos 30
sin40 = F_{y} / F
F_{y} = F sin 40
cos 40 = Fₓ / F
Fₓ = F cos 40
let's substitute
F cos 40 t = m ( cos 30 - vₓ₀)
F sin 40 t = m (v_{f} sin 30-0)
we have two equations and two unknowns, so the system can be solved
F cos 40 0.1 = 0.4 (v_{f} cos 30 - 20)
F sin 40 0.1 = 0.4 v_{f} sin 30
we clear fen the second equation and subtitles in the first
F = 4 sin30 /sin40 v_{f}
F = 3.111 v_{f}
(3,111 v_{f}) cos 40 = 4 v_{f} cos 30 - 80
v_{f} (3,111 cos 40 -4 cos30) = - 80
v_{f} (- 1.0812) = - 80
v_{f} = 73.99
v_{f} = 74 m/s
now we can calculate the force
F = 3.111 73.99
F = 230 N
Answer:
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
<u>ΔP/Δt = 42.6 N</u>
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b.
According to Newton's Second Law, the force is equal to rate of change of linear momentum:
Average Force = F = ΔP/Δt
<u>F = 42.6 N</u>
<u></u>
c.
The pressure is given by:
Average Pressure = P = Average Force/Area
P = 42.6 N/ 3 x 10⁻⁴ m²
<u>P = 142042.4 Pa = 1.42 KPa</u>
Explanation :
Work is done when a force is applied to create a displacement on an object.
Thus, the work done depends on the two factors i.e.
(1) Applied force (F)
(2) Distance or displacement (d)
Mathematically, work done is 
It also depends on the angle between the force and the displacement.
For example,
A person carries a weight of 20 kg and lifts it on his head 1.5 m above the surface. So, the work done by him on the luggage will be:
or
So, 
Hence, the work done by him on the luggage is 294 Joules.
