Find Displacement and Distance
displacement ...
north is 700+400+100 =1200m n
south=1200m
1200-1200=0
east is 300+300=600m
west is 600m
600-600=0
back at dtart. displ zero
distance is 700+ 300m + 400 m + 600m + 1200m + 300m + 100m = 3600m
Assuming north as positive direction, the initial and final velocities of the ball are:
(with negative sign since it is due south)
the time taken is
, so the average acceleration of the ball is given by
And the positive sign tells us the direction of the acceleration is north.
<span>v =Integral(F/m)dtv=Integral(F/m)dt
=â«50(6t2â’4t+3)/5dt=â«05(6t2â’4t+3)/5dt
=1/5â—(2t3â’2t2+3t)|(0,5)=1/5â—(2t3â’2t2+3t)|(0,5)
=1/5â—(250â’50+15)=1/5â—(250â’50+15)
=215/5=215/5
=43m/s.</span>
Mathematically it can be expressed as: I = FpΔt. Where F p is the average magnitude of the acting force and Δt = t 2 - t 1, the time lapse in the force acts.
The magnitude of the impulse applied to stop the cart is
I = FpΔt = (10N) * (5s) = 50 N.s
Answer:
14.4 Nm
Explanation:
Moment of Inertia, I = 0.9 kg m^2, w0 = 0, w = 8 rad/s, t = 0.5 second
Use first equation of motion for rotational motion
w = w0 + α t
where, α be the angular acceleration
8 = 0 + α x 0.5
α = 16 rad/s^2
Now Torque = Moment of inertia x angular acceleration
τ = I x α
τ = 0.9 x 16 = 14.4 Nm
