Answer:
The work done on the air is 4.699 x 10⁶ Joules
Explanation:
Given;
increase in air volume, ΔV = 0.227 m³
net pressure of the air, P = 2.07 x 10⁷ Pa
The work done on the air is given by;
W = PΔV
Where;
W is the work done on the air
P is the net pressure
ΔV is the increase in air volume
Substitute the given values and solve for work done;
W = (2.07 x 10⁷ Pa) (0.227 m³)
W = 4.699 x 10⁶ Joules
Therefore, the work done on the air is 4.699 x 10⁶ Joules
Answer:
(a) 5142.86 m
(b) 317.5 m/s
(c) 49.3 degree C
Explanation:
m = 100 kg, Q = 1200 kcal = 1200 x 1000 x 4.2 = 504 x 10^4 J
(a) Let the altitude be h
Q = m x g x h
504 x 10^4 = 100 x 9.8 x h
h = 5142.86 m
(b) Let v be the speed
Q = 1/2 m v^2
504 x 10^4 = 1/2 x 100 x v^2
v = 317.5 m/s
(c) The temperature of normal human body, T1 = 37 degree C
Let the final temperature is T2.
Q = m x c x (T2 - T1)
504 x 10^4 = 100 x 4.1 x 1000 x (T2 - 37)
T2 = 49.3 degree C
Answer:

Explanation:
7.0g is
the acceleration due to gravity.
Given that
, acceleration can be calculated as:

Since circular motion is involved and our radius of curvature is given as,
, <em>centripetal acceleration </em>is given by the formula:
.
Make
subject of the formula to find
:

The maximum speed of the pilot is 
Answer:
220.508 kPa
Explanation:
32 psi = 32 pound force per square inch = 32 lbf/in2 = 32 * (4.448 N/lbf) * (12 in/ft * 3.28 ft/m)^2
or 220508 Pa or 220.508 kPa
So the maximum safe air pressure of a tire is 220.508 kPa