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Alex777 [14]
2 years ago
8

A 7000-kg plane is launched from an aircraft carrier in 2.0 seconds

Physics
1 answer:
ExtremeBDS [4]2 years ago
5 0

The acceleration and velocity of the plane is 78.57 m/s² and 157.14 m/s respectively

To calculate the acceleration of the plane, we use the formula below.

<h3>Formula:</h3>
  • a = F/m..................... Equation 1

Where:

  • a = Acceleration of the plane
  • F = Force applied to the plane
  • m = mass of the plane.

From the question,

Given:

  • F = 550000 N
  • m = 7000 kg

Substitute these values into equation 1

  • a = 550000/7000
  • a = 78.57 m/s²

To calculate the velocity, we use the formula below.

  • v = u+at............. Equation 2

Where:

  • v = Final velocity
  • u = initial velocity
  • a = acceleration
  • t = time.

From the question,

Given:

  • u = 0 m/s
  • a = 78.57 m/s
  • t = 2.0 seconds

Substitute these values into equation 2

  • v = 0+2(78.57)
  • v = 157.14 m/s

Hence, The acceleration and velocity of the plane is 78.57 m/s² and 157.14 m/s respectively.
Learn more about acceleration here: brainly.com/question/460763

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Answer:

a) 1.20227 seconds

b) 0.98674 m

c) 7.3942875 m/s

Explanation:

t = Time taken

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v = Final velocity

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v=u+at\\\Rightarrow 0=4.4-9.81\times t\\\Rightarrow \frac{-4.4}{-9.81}=t\\\Rightarrow t=0.44852\ s

s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.4\times 0.44852+\frac{1}{2}\times -9.81\times 0.44852^2\\\Rightarrow s=0.98674\ m

b) Her highest height above the board is 0.98674 m

Total height she would fall is 0.98674+1.8 = 2.78674 m

s=ut+\frac{1}{2}at^2\\\Rightarrow 2.78674=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.78674\times 2}{9.81}}\\\Rightarrow t=0.75375\ s

a) Her feet are in the air for 0.75375+0.44852 = 1.20227 seconds

v=u+at\\\Rightarrow v=0+9.81\times 0.75375\\\Rightarrow v=7.3942875\ m/s

c) Her velocity when her feet hit the water is 7.3942875 m/s

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