327,360 feet
most aviation experts agree that the point where space begins is approximately 62 miles up. (62x5280=327,360).
Answer:
1.68m/s^2
Explanation:
using V^2=U^2+2×a×s formula.
If,
(Final Velocity)V=45
(Initial Velocity)U=0 because it starts from rest
(Distance)S=600m
(acceleration)a= ?
now,
using formula,
45^2=0^2+2×a×600
2025= 1200a
a=2025÷1200
a = 1.68m/s^2(The unit of acceleration is m/s^2)
Therefore the acceleration is 1.68m/s^2
Hope it works !!!
Answer:
Explanation:
The question relates to motion on a circular path .
Let the radius of the circular path be R .
The centripetal force for circular motion is provided by frictional force
frictional force is equal to μmg , where μ is coefficient of friction and mg is weight
Equating cenrtipetal force and frictionl force in the case of car A
mv² / R = μmg
R = v² /μg
= 26.8 x 26.8 / .335 x 9.8
= 218.77 m
In case of moton of car B
mv² / R = μmg
v² = μRg
= .683 x 218.77x 9.8
= 1464.35
v = 38.26 m /s .
Answer:
a) W=85.225 kW
b) 
Explanation:
First, consider the energy balance for the compressor: The energy that enters to the system (W and enthalpy of the feed flow) is equal to the energy that goes out from it (Heat Q and enthalpy of the exit flow):
Consider the enthalpy data from van Wylen 6th edition, Table B.2.2. According to that, 
, 
So, the power input to the compressor is:
b) The differential entropy change dS for a reversible heat transfer dQ at a temperature T is:
This equation can be integrated if the heat transfer surface temperature remains constant, which is the case, giving as a result:
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