More water evaporates than falls as rain
did u know 3 million tonnes of water gets evaporated from oceans per day
Displacement equals (Velocity times Time) plus half times the (acceleration times time squared). =. (48 * 4) + 1/2 * (12 *12^2) = 288meters
Hey there, the answer is .............................. About 0.7 m/sec^2
<span>
Acceleration is the change in speed / time </span>
<span>Change in speed is 60 m/sec </span>
<span>Time is 1 minute 25 second. Convert that to seconds. </span>
<span>Divide the change in speed by the time in seconds.
About 0.7 m/sec^2
</span><span>So the acceleration is - 60 / 85 = - 0.71 m/s^2
HOPE I HELPED!!!!!!!!!!</span>
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
