Answer:
Explanation:
Given
car A had a head start of 
and it starts at x=0 and t=0
Car B has to travel a distance of 
where 
is the distance travel by car A in time t
distance travel by car A is
For car B with speed 
Heat used by electric heater :
Q = m • c • ∆T
Q = (75 kg)(4200 J/kg°C)(43°C - 15°C)
Q = 8.82 × 10⁶ J
Cost of electrical energy :
Cost = (8.82 × 10⁶ J)/(3.6 × 10⁶ J) • ($ 0.15)
Cost = $ 0.3675
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
Taking ratio of W & w. ≈ 6 . w = 1/6 W. Therefore , Weight of an object on the moon is 1/6 of its weight on the earth.
Sexual reproduction. thats the answer i think
