Answer:
The 10-meter long rod of an SR-71 airplane expands 0.02 meters (2 centimeters) when plane flies at 3 times the speed of sound.
Explanation:
From Physics we get that expansion of the rod portion is found by this formula:
[tex]\Delta l = \alpha\cdot l_{o}\cdot (T_{f}-T_{o})[/tex] (Eq. 1)
Where:
[tex]\Delta l[/tex] - Expansion of the rod portion, measured in meters.
[tex]\alpha[/tex] - Linear coefficient of expansion for titanium, measured in [tex]\frac{1}{^{\circ}C}[/tex].
[tex]l_{o}[/tex] - Initial length of the rod portion, measured in meters.
[tex]T_{o}[/tex] - Initial temperature of the rod portion, measured in Celsius.
[tex]T_{f}[/tex] - Final temperature of the rod portion, measured in Celsius.
If we know that [tex]\alpha = 5\times 10^{-6}\,\frac{1}{^{\circ}C}[/tex], [tex]l_{o} = 10\,m[/tex], [tex]T_{o} = 0\,^{\circ}C[/tex] and [tex]T_{f} = 400\,^{\circ}C[/tex], the expansion experimented by the rod portion is:
[tex]\Delta l = \left(5\times 10^{-6}\,\frac{1}{^{\circ}C} \right)\cdot (10\,m)\cdot (400\,^{\circ}C-0\,^{\circ}C)[/tex]
[tex]\Delta l = 0.02\,m[/tex]
The 10-meter long rod of an SR-71 airplane expands 0.02 meters (2 centimeters) when plane flies at 3 times the speed of sound.
