Answer:
a) - 72.5°c
b) pressure = 3625.13 Pa
c) density = 0.063 kg/m^3
d) it is a subsonic aircraft
Explanation:
a) Determine Temperature
Temperature at 19.5 km ( 19500 m )
T = -131 + ( 0.003 * altitude in meters )
= -131 + ( 0.003 * 19500 ) = - 72.5°c
b) Determine pressure and density at 19.5 km altitude
Given :
Po (atmospheric pressure at sea level ) = 101kpa
R ( gas constant of air ) = 0.287 KJ/Kgk
T = -72.5°c ≈ 200.5 k
pressure = 3625.13 Pa
hence density = 0.063 kg/m^3
attached below is the remaining part of the solution
C) determine if the aircraft is subsonic or super sonic
Velocity ( v ) = 
= 
= 283.8 m/s
hence it is a subsonic aircraft
When the spring is extended by 44.5 cm - 34.0 cm = 10.5 cm = 0.105 m, it exerts a restoring force with magnitude R such that the net force on the mass is
∑ F = R - mg = 0
where mg = weight of the mass = (7.00 kg) g = 68.6 N.
It follows that R = 68.6 N, and by Hooke's law, the spring constant is k such that
k (0.105 m) = 68.6 N ⇒ k = (68.6 N) / (0.105 m) ≈ 653 N/m
3260÷4=815 which is you average seed
Answer:
2.28
Explanation:
From mirror formula,
1/f = 1/u+1/v .......... Equation 1
Where f = focal length of the mirror, v = image distance, u = object distance.
Note: The focal length mirror is positive.
make v the subject of the equation,
v = fu/(u-f)............ Equation 2
Given: f = 2.5 cm, u = 1.4 cm
Substitute into equation 2
v = 2.5(1.4)/(1.4-2.5)
v = 3.5/-1.1
v = -3.2 cm.
Note: v is negative because it is a virtual image.
But,
Magnification = image distance/object distance
M = v/u
Where M = magnification.
Given: v = 3.2 cm, u = 1.4 cm
M = 3.2/1.4
M = 2.28.
Thus the magnification of the tooth = 2.28.
