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lana [24]
3 years ago
13

A wind-tunnel experimentis performed on a 1/25scale model of a supersonic aircraft. The prototype aircraft flies at 450 m/s in c

onditions where the speed of sound is 250 m/s and the air density is 0.89kg/m3. The model aircraft is tested in a wind tunnel in which the speed of sound is 280m/s and the air density is 0.56kg/m3. The drag force on the model is 100N. What speed must the flow in the wind tunnel be for dynamic similitude, and what is the drag force on the prototype?

Physics
1 answer:
KengaRu [80]3 years ago
4 0

Answer: V = 504m/a

F = 4N

Explanation: please find the attached file for the solution

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Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr
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Two magnetic poles that repel one another are held three inches apart. What happens when the poles are moved closer together, to
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PLEASE PROVIDE EXPLANATION.<br><br> THANK YOU!!
Ksju [112]

Answer:

11,000 kg

(a) 11.2 m/s

(b) 1.6 m/s

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(2200 kg) (60.0 km/h) + m (0 km/h) = (2200 kg) (10 km/h) + m (10 km/h)

132,000 kg km/h = 22,000 kg km/h + m (10 km/h)

110,000 kg km/h = m (10 km/h)

m = 11,000 kg

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(m) (-v) + (2m) (5v) = (m) (v₁) + (2m) (v₂)

-mv + 10mv = m v₁ + 2m v₂

9mv = m (v₁ + 2 v₂)

9v = v₁ + 2 v₂

Since the collision is elastic, kinetic energy is also conserved.

½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²

m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²

(m) (-v)² + (2m) (5v)² = m v₁² + (2m) v₂²

mv² + 50mv² = m v₁² + 2m v₂²

51mv² = m (v₁² + 2 v₂²)

51v² = v₁² + 2 v₂²

We know v = 1.60 m/s.  So the two equations are:

14.4 = v₁ + 2 v₂

130.56 = v₁² + 2 v₂²

Solve the system of equations using substitution.

130.56 = (14.4 − 2 v₂)² + 2 v₂²

130.56 = 207.36 − 57.6 v₂ + 4 v₂² + 2 v₂²

0 = 6 v₂² − 57.6 v₂ + 76.8

0 = v₂² − 9.6 v₂ + 12.8

v₂ = [ 9.6 ± √(9.6² − 4(1)(12.8)) ] / 2(1)

v₂ = 1.6 or 8

If v₂ = 1.6 m/s, then v₁ = 14.4 − 2 v₂ = 11.2 m/s.

If v₂ = 8 m/s, then v₁ = 14.4 − 2 v₂ = -1.6 m/s.

We know v₁ can't be -1.6 m/s, since that would mean puck A didn't change speeds after the collision.  Therefore, v₁ = 11.2 m/s and v₂ = 1.6 m/s.

8 0
3 years ago
Addition of a lubricant such as oil to a surface will __ friction ?
Mashcka [7]

Answer:

decrease

Explanation:

Addition of a lubricant such as oil to a surface will decrease friction. This makes the surface more greasy or slippery. This is very much efficient in machines, for example.

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