The speed
of the elevator at the beginning of the 8 m descent is nearly 4 m/s. Hence, option A is the correct answer.
We are given that-
the mass of the elevator (m) = 1000 kg ;
the distance the elevator decelerated to be y = 8m ;
the tension is T = 11000 N;
let us determine the acceleration 'a' by using Newton's second law of motion.
∑Fy = ma
W - T = ma
(1000kg x 9.8 m/s² ) - 11000N = 1000 kg x a
9800 - 11000 = 1000
a = - 1.2 m/s²
Using the equation of kinematics to determine the initial velocity.
² =
² + 2ay
= √ ( 2 x 1.2m/s² x 8 m )
= √19.2 m²/s²
= 4.38 m/s ≈ 4 m/s
Hence, the initial velocity of the elevator is 4m/s.
Read more about the Equation of kinematics:
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Answer:
I think it might be A: Alec but tell me if I'm wrong
Explanation:
A load of 20N is lifted through a height of 8m by a machine.If the effort of 80N moves a distance of 6m.Calculate the efficiency percentage of the machine
Explanation:
A load of 20N is lifted through a height of 8m by a machine.If the effort of 80N moves a distance of 6m.Calculate the efficiency percentage of the machine
Answer:
P1 = 0 gage
P2 = 87.9 lb/ft³
Explanation:
Given data
Airplane flying = 200 mph = 293.33 ft/s
altitude height = 5000-ft
air velocity relative to the airplane = 273 mph = 400.4 ft/s
Solution
we know density at height 5000-ft is 2.04 ×
slug/ft³
so here P1 +
= P2 +
and here
P1 = 0 gage
because P1 = atmospheric pressure
and so here put here value and we get
P1 +
= P2 +
0 +
solve it we get
P2 = 87.9 lb/ft³