To solve this problem we will apply the concepts of equilibrium and Newton's second law.
According to the description given, it is under constant ascending acceleration, and the balance of the forces corresponding to the tension of the rope and the weight of the elevator must be equal to said acceleration. So


Here,
T = Tension
m = Mass
g = Gravitational Acceleration
a = Acceleration (upward)
Rearranging to find T,



Therefore the tension force in the cable is 10290.15N
False. reaction time would be lapse of time between external stimuli and muscular movement
Answer:
r = 2161.9 m
Explanation:
Aerodynamic lift(L) is perpendicular to the wing, which is tilted 40 degrees to the horizontal.
Since the plane is moving in a horizontal circle, the vertical component of the lift must cancel the weight W of the airplane, but the horizontal component is the centripetal force that keeps it in a circle.
L is perpendicular to wing at angle θ with respect to horizontal
Thus,
Vertical component of lift is:
L cosθ = W = mg
Thus, m = L cosθ / g - - - - (eq1)
Horizontal component of lift is:
L sinθ = centripetal force = mv² / r - - - - (eq2)
Combining equations 1 and 2,we have;
L sinθ = (L cosθ / g)(v² / r)
L cancels out on both sides to give;
tanθ = v²/ rg
r = v² / (g tanθ)
We are given;
velocity; v = 480 km/hr = 480 x 10/36 = 133.33 m/s
r = 133.33²/[(9.8) tan(40)] = 2161.9 m
Answer:
The angle formed of the rope with the surface = 40°
Force applied = 125Newtons
The displacement covered by the box =25metres
W= FDcos theta
[125×40×cos(40°) ] Joules
= [ (3125×0.76604444311)]Joules
= 2393.88888472 joules(ans)
Hope it helps
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