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zimovet [89]
3 years ago
12

If state law mandates that elevators cannot accelerate more than 4.80 m/s2 or travel faster than 19.8 m/s , what is the minimum

time in which an elevator can travel the 373 m from the ground floor to the observatory floor?
Physics
1 answer:
Rudik [331]3 years ago
8 0

Answer:

23.0 s

Explanation:

Given:

v₀ = 0 m/s

v = 19.8 m/s

a = 4.80 m/s²

Find: Δx and t

v² = v₀² + 2aΔx

(19.8 m/s)² = (0 m/s)² + 2 (4.80 m/s²) Δx

Δx = 40.84 m

v = at + v₀

19.8 m/s = (4.80 m/s²) t + 0 m/s

t = 4.125 s

The elevator takes 40.84 m and 4.125 s to accelerate, and therefore also 40.84 m and 4.125 s to decelerate.

That leaves 291.3 m to travel at top speed.  The time it takes is:

291.3 m / (19.8 m/s) = 14.71 s

The total time is 4.125 s + 14.71 s + 4.125 s = 23.0 s.

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H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)

We already know the last quantity: \frac{m}{2} *(vel_1^2-vel_2^2)=-Δ (\frac{mvel^2}{2})=23400W

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The exit state is a liquid-vapor mixture, so its enthalpy is:

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C) For the area, consider the equation of mass flow:

m=p*vel*A where p is the density, and A the area. The density is the inverse of the specific volume, so m=\frac{vel*A}{v}

The specific volume of the inlet steam can be read also from the steam tables, and its value is: 0.08643\frac{m^3}{kg}, so:

A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2

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