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xxTIMURxx [149]
2 years ago
14

As a 15000 kg jet plane lands on an aircraft carrier, its tail hook snags a cable to slow it down. The cable is attached to a sp

ring with spring constant 60000 N/m.If the spring stretches 30 m to stop the plane, what was the plane's landing speed?
Physics
1 answer:
RideAnS [48]2 years ago
5 0

Answer:

60 m/s

Explanation:

From the law of conservation of energy,

The kinetic energy of the plane = Energy of store in the spring when the plane lands.

1/2mv²  = 1/2ke²

making v the subject of the equation.

v = √(ke²/m).................... Equation 1

Where  v = the plane landing speed, k = spring constant, e = extension. m = mass of the plane.

Given: m = 15000 kg, k = 60000 N/m, e = 30 m.

Substitute into equation 1

v = √(60000×30²/15000)

v = √(4×900)

v = √(3600)

v = 60 m/s.

Hence the plane's landing speed = 60 m/s

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A pilot in a small plane encounters shifting winds. He flies 26.0 km northeast, then 45.0 km due north. From this point, he flie
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Answer:

a) v₃ = 19.54 km, b)  70.2º north-west

Explanation:

This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition

vector 1 moves 26 km northeast

let's use trigonometry to find its components

         cos 45 = x₁ / V₁

         sin 45 = y₁ / V₁

         x₁ = v₁ cos 45

         y₁ = v₁ sin 45

         x₁ = 26 cos 45

         y₁ = 26 sin 45

         x₁ = 18.38 km

         y₁ = 18.38 km

Vector 2 moves 45 km north

        y₂ = 45 km

Unknown 3 vector

          x3 =?

          y3 =?

Vector Resulting 70 km north of the starting point

           R_y = 70 km

we make the sum on each axis

X axis

      Rₓ = x₁ + x₃

       x₃ = Rₓ -x₁

       x₃ = 0 - 18.38

       x₃ = -18.38 km

Y Axis

      R_y = y₁ + y₂ + y₃

       y₃ = R_y - y₁ -y₂

       y₃ = 70 -18.38 - 45

       y₃ = 6.62 km

the vector of the third leg of the journey is

         v₃ = (-18.38 i ^ +6.62 j^ ) km

let's use the Pythagorean theorem to find the length

         v₃ = √ (18.38² + 6.62²)

         v₃ = 19.54 km

to find the angle let's use trigonometry

           tan θ = y₃ / x₃

           θ = tan⁻¹ (y₃ / x₃)

           θ = tan⁻¹ (6.62 / (- 18.38))

           θ = -19.8º

with respect to the x axis, if we measure this angle from the positive side of the x axis it is

          θ’= 180 -19.8

          θ’= 160.19º

I mean the address is

          θ’’ = 90-19.8

          θ = 70.2º

70.2º north-west

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