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olya-2409 [2.1K]
3 years ago
15

Upon reaching a velocity of 100fps, the pilot of the airplane decides to abort the take off and applies brakes and stops the air

plane in 1000ft. What is the airplane’s deceleration?
Physics
1 answer:
masha68 [24]3 years ago
8 0

Answer:

5 ft/s²

Explanation:

u = Initial velocity = 100 ft/s

v = Final velocity = 0

s = Displacement = 1000 ft

a = Acceleration

From equation of motion

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-100^2}{2\times 1000}\\\Rightarrow a=-5\ ft/s^2

1\ ft/s^2=0.3048\ m/s^2\\\Rightarrow 5\ ft/s^2=5\times 0.3048\ m/s^2=1.524\ m/s^2

The airplane's deceleration is 5 ft/s² or 1.524 m/s²

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A 3.45-kg centrifuge takes 100 s to spin up from rest to its final angular speed with constant angular acceleration. A point loc
Dafna11 [192]

Answer:

(a) 18.75 rad/s²

(b) 14920.78 rev

Explanation:

(a)

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Where v = final velocity, u = initial velocity, t = time.

Given: v = 150 m/s,  u = 0 m/s ( from rest), t = 100 s

Substitute into equation 1

a = (150-0)/100

a = 1.5 m/s²

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α = a/r..................... Equation 2

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Given: a = 1.5 m/s², r = 8 cm = 0.08 m

substitute into equation 2

α = 1.5/0.08

α = 18.75 rad/s²

(b)

Using,

Ф = (ω'+ω).t/2........................... Equation 3

Where Ф = number of revolution of the centrifuge, ω' = initial angular velocity, ω = Final angular velocity.

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ω = v/r and ω' = u/r

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