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bekas [8.4K]
3 years ago
12

The fastest propeller-driven aircraft is the Russian TU-95/142, which canreach a maximum speed of 925 km/h. For this speed, calc

ulate the plane'sresultant displacement if it travels east for 1.50 h, then turns 135° north-west and travels for 2.00 h

Physics
1 answer:
valentinak56 [21]3 years ago
7 0

Answer:

The resultant displacement of plane = 1310.76 Km

Explanation:

Given that,

Speed of the plane = 925 Km/h

Therefore, in 1 hour plane travels 925 Km

From figure,

Displacement if it travels east for 1.5 h, AB = 1387.5 Km

Displacement if it travels north-west for 2 h, BC = 1850 Km

Angle formed between two displacement = 135°

The resultant displacement is given by the relation

          AC² = AB² + BC² + 2 AB BC Cos θ

Substituting the values in the equations,

           AC²  = 1387.5² + 1850² + 2 x 1387.5 x 1850 x Cos 135°

                    = 1718095

           AC =  1310.76

Therefore, the resultant displacement of plane is 1310.76 Km

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Answer:

h >5/2r

Explanation:

This problem involves the application of the concepts of force and the work-energy theorem.

The roller coaster undergoes circular motion when going round the loop. For the rider to stay in contact with the cart at all times, the roller coaster must be moving with a minimum velocity v such that at the top the rider is in a uniform circular motion and does not fall out of the cart. The rider moves around the circle with an acceleration a = v²/r. Where r = radius of the circle.

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The roller coaster starts from height h. Its potential energy changes as it travels on its course. The potential energy decreases from a value mgh at the height h to mg×2r at the top of the loop. No other force is acting on the roller coaster except the force of gravity which is a conservative force so, energy is conserved. Because energy is conserved the total change in the potential energy of the rider must be at least equal to or greater than the kinetic energy of the rider at the top of the loop

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The height at the roller coaster starts is usually higher than the top of the loop by design. So

ΔPE =mgh - mg×2r = mg(h – 2r)

2r is the vertical distance from the base of the loop to the top of the loop, basically the diameter of the loop.

In order for the roller coaster to move smoothly and not come to a halt at the top of the loop, the ΔPE must be greater than the ΔKE at the top.

So ΔPE > ΔKE at the top. The extra energy moves the rider the loop from the top.

ΔPE > ΔKE

mg(h–2r) > 1/2mv²

g(h–2r) > 1/2(√(rg))²

g(h–2r) > 1/2×rg

h–2r > 1/2×r

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3 years ago
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