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: be alert for pedestrians near the bus.
Explanation: Due to road accidents many Governments around the world has adopted and put in place certain rules and regulations with regards to road safety, this is so to prevent the or reduce the chances of accidents happening.
Road safety rules are rules and guidelines put in place by Government in order to prevent road accidents and maintain a free flow of traffic. An example of such rules is 'be alert for pedestrians near the bus ' when approaching a local bus that is stopped.
Answer:
Explanation:
Let h be the height .
initial velocity in first case u = 0
final velocity v = 6 m /s
acceleration due to gravity g = 9.8 m /s²
v² = u² + 2 g h
6² = 0 + 2 x 9.8 x h
h = 1.837 m .
For second case u = 3 m /s
v² = u² + 2 gh
= 3² + 2 x 1.837 x 9.8
= 9 + 36
= 45 m
v = 6.7 m /s
The second and third laws of thermodynamics states that absolute zero cannot be reached. The correct option among all the options that are given in the question is the third option or option "C". Both the laws actually deal with the relations that exist between heat and other forms of energy. I hope the answer helps you.
Answer:
KE = 0.162 KJ
Explanation:
given,
mass of bullet (m)= 20 g = 0.02 Kg
speed of the bullet (u)= 1000 m/s
mass of block(M) = 1 Kg
velocity of bullet after collision (v)= 100 m/s
kinetic energy = ?
using conservation of momentum
m u = m v + M V
0.02 x 1000 = 0.02 x 100 + 1 x V
20 = 2 + V
V = 18 m/s
now,
Kinetic energy of the block
KE = 162 J
KE = 0.162 KJ
