Explanation:
the inability to change the position from rest to motion or motion to rest by themselves is inertia.
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
This is a binary star system
784 Newtons or 176.37 lbs
Answer:
in the parallel connection the light bulbs shine less than in the series connection
Explanation:
In a series circuit the current through the whole circuit is the same, therefore the power (brightness) of each bulb is
P = i² R
where R is the resistance of each bulb and i the current of the circuit.
If we connect the light bulbs and the cells in parallel, the current in the circuit is the sum of the east that passes through each light bulb,
i = i₁ + i₂
if the two light bulbs are the same
i = 2 i₁
i₁ = i / 2
so the power of each bulb is is
P = i₁² R
P = R i² / 4
P = ¼ P_initial
Therefore we see that in the parallel connection the light bulbs shine less than in the series connection
