Respuesta :
Answer:
The angular momentum is 22.5 kgm²/s.
Explanation:
Given that,
Force = 300 N
Time = 0.250 s
Distance = 0.300 m
We need to calculate the torque
The torque is the product of the force and distance.
Using formula of torque
[tex]\vec{\tau}=\vec{F}\times \vec{r}[/tex]
Where, F = force
r = distance
Put the value into the formula
[tex]\vec{\tau}=300\times0.300[/tex]
[tex]\vec{\tau}=90\ N-m[/tex]
We need to calculate the angular momentum
The torque is equal to the rate of change of angular momentum of the particle.
[tex]\vec{\tau}=\dfrac{\Delta L}{\Delta t}[/tex]
[tex]\Delta L=\vec{\tau}\Delta t[/tex]
We know,
[tex]\Delta L=L_{f}-L_{i}[/tex]
Initial angular momentum is zero.
So, [tex]\Delta L=L_{f}-0[/tex]
Put the value into the formula
[tex]L_{f}=0.250\times90[/tex]
[tex]L_{f}=22.5\ Kgm^2/s[/tex]
Hence, The angular momentum is 22.5 kgm²/s.
The angular momentum given to the engine is; 22.5 kg.m/s²
What is the angular momentum?
We are given;
Force; F = 300 N
Time; t = 0.250 s
Distance; r = 0.300 m
Formula for torque is;
τ = F * r
τ = 300 * 0.3
τ = 90 N.m
Now, torque is equal to rate of change in angular momentum with time. Thus;
τ = (L_f - L_i)/Δt
90 = (L_f - 0)/0.25
L_f = 0.25 * 90
L_f = 22.5 kg.m/s²
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