-- loud sounds
-- bright lights
-- strong radio signals
-- Slinkies that can pinch you painfully
-- a tsunami in the ocean
-- earthquakes above Richter 5 or 6
Answer:
a) P = 1240 lb/ft^2
b) P = 1040 lb/ft^2
c) P = 1270 lb/ft^2
Explanation:
Given:
- P_a = 2216.2 lb/ft^2
- β = 0.00357 R/ft
- g = 32.174 ft/s^2
- T_a = 518.7 R
- R = 1716 ft-lb / slug-R
- γ = 0.07647 lb/ft^3
- h = 14,110 ft
Find:
(a) Determine the pressure at this elevation using the standard atmosphere equation.
(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.
(c) Determine the pressure if the air is assumed to have a constant temperature of 59 oF.
Solution:
- The standard atmospheric equation is expressed as:
P = P_a* ( 1 - βh/T_a)^(g / R*β)
(g / R*β) = 32.174 / 1716*0.0035 = 5.252
P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252
P = 1240 lb/ft^2
- The air density method which is expressed as:
P = P_a - γ*h
P = 2116.2 - 0.07647*14,110
P = 1040 lb/ft^2
- Using constant temperature ideal gas approximation:
P = P_a* e^ ( -g*h / R*T_a )
P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )
P = 1270 lb/ft^2
Answer: Both A and B
Explanation:
The objects are in contact so it would depend on the amount of force that is pressing them as to the amount of friction and it would depend on the smoothness as too the amount of friction that is added
Answer:
2 m/s²
Explanation:
Given:
v₀ = 0 m/s
v = 20 m/s
t = 10 s
Find: a
a = (v − v₀) / t
a = (20 m/s − 0 m/s) / 10 s
a = 2 m/s²
<span> static friction :) your very welcome
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