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g A frequently quoted rule of thumb in aircraft design is that wings should produce about 1000 N of lift per square meter of wing. (The fact that a wing has a top and bottom surface does not double its area.) a)At takeoff the aircraft travels at 61.1 m/s, so that the air speed relative to the bottom of the wing is 61.1 m/s. Given the sea level density of air to be 1.29 kg/m3, how fast (in m/s) must it move over the upper surface to create the ideal lift
Answer:
The value is [tex]u = 72.69 \ m/s[/tex]
Explanation:
From the question we are told that
The amount of force a square meter of an aircraft wing should produce is [tex]F = 1000 \ N[/tex]
The air speed relative to the bottom of the wing is [tex]v = 61.1 \ m/s[/tex]
The air level density of air is [tex]\rho_s = 1.29\ kg/m^3[/tex]
Gnerally this force per square meter of an aircraft wing is mathematically represented as
[tex]F = \frac{1}{2} * \rho_s * A * [ u^2 - v^2 ][/tex]
Here u is the speed air need to go over the top surface to create the ideal lift
A is the area of a square meter i.e [tex]A = 1 \ m^2[/tex]
So
[tex]1000 = \frac{1}{2} * 1.29 * 1 * [ u^2 - 61.1 ^2 ][/tex]
=> [tex]u = 72.69 \ m/s[/tex]
A 5-kg projectile is fired over level ground with a velocity of 200 m/s at an angle of 25
above the horizontal. Just before it hits the ground its speed is 150 m/s. Over the entire trip the
change in the thermal energy of the projectile and air is:
Answer: 43.8 kJ
Explanation:
Given;
mass of the object, m = 5kg
initial velocity of the projectile, v₁ = 200 m/s
final velocity of the projectile, v₂ = 150 m/s
To determine the the change in the thermal energy of the projectile and air, we consider change in potential and kinetic enrgy of the projectile. Since the projectile was fired over level ground, change in potential energy is zero.
Then, change in thermal energy of the projectile, KE = Δ¹/₂mv²
KE = Δ¹/₂mv² = ¹/₂m(v₁²-v₂²)
KE = ¹/₂ × 5(200²-150²) = 2.5(17500) = 43750 J = 43.8 kJ
Therefore, the change in thermal energy of the projectile is 43.8 kJ