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bagirrra123 [75]
3 years ago
11

Passengers on an airplane move from rest to 86 meters per second before the airplane takes off. If the airplane takes 100 second

s to take off what is the acceleration, in m/s^2, of the airplane? Show your answer to one decimal point. Make sure you include the proper unit. Do not put a space between the number and the unit!
Physics
1 answer:
VikaD [51]3 years ago
4 0

Utilize the formula:  V _{f} = V _{i} + a\Delta t

V _{f} = Final Velocity (86 m/s)

V _{f} = Initial Velocity (0 m/s)

a = acceleration (m/s²)

\Deltat = Time (100 seconds)

As a result,

86 m/s = 0 + (a)(100 seconds)

Using algebra, divide 86 m/s by 100 seconds:

86 m/s = 100a

a = 0.86 m/s²

Rounded to one decimal place: 0.9 m/s²

Let me know if you have any questions!

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Answer:

The second projectile was 1.41 times faster than the first.

Explanation:

In the ballistic pendulum experiment, the speed (v) of the projectile is given by:  

v = \frac{m + M}{m} \cdot \sqrt{2gh}

<em>where m: is the mass of the projectile, M: is the mass of the pendulum, g: is the gravitational constant and h: is the maximum height of the pendulum.   </em>

To know how many times faster was the second projectile than the first, we need to take the ratio for the velocities for the projectiles 2 and 1:    

\frac{v_{2}}{v_{1}} = \frac{\frac{m_{2} + M}{m_{2}} \cdot \sqrt{2gh_{2}}}{\frac{m_{1} + M}{m_{1}} \cdot \sqrt{2gh_{1}}}           (1)

<em>where m₁ and m₂ are the masses of the projectiles 1 and 2, respectively, and h₁ and h₂ are the maximum height reached by the pendulum by the projectiles 1 and 2, respectively.  </em>

Since the projectile 1 has the same mass that the projectile 2, we can simplify equation (1):

\frac{v_{2}}{v_{1}} = \frac{\sqrt{h_{2}}}{\sqrt{h_{1}}}  

\frac{v_{2}}{v_{1}} = \frac{\sqrt{5.2 cm}}{\sqrt{2.6 cm}}

\frac{v_{2}}{v_{1}} = 1.41  

Therefore, the second projectile was 1.41 times faster than the first.  

I hope it helps you!

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