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Tems11 [23]
3 years ago
12

passengers on an airplane move from rest to 73 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.
Physics
1 answer:
Brrunno [24]3 years ago
3 0

Answer:

0.73m/s2

Explanation:

The speed at which the passengers move is the final speed of the plane

Hence acceleration = V-U/t; change in velocity and t is time taken. U is the initial velocity which is 0m/s

a = 73-0/100= 0.73m/s2

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Answer:

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Explanation:

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3 years ago
Is this true or false the kinetic energy of a freely falling object decreases as it begins to fall
Naddika [18.5K]
The kinetic energy of an object is the energy of the object as it moves. It has more kinetic energy the faster it moves. Because a falling object is increasing in speed, it would also increase in kinetic energy. Hope this helps! :)
8 0
3 years ago
Read 2 more answers
A person 1.8m tall stands 0.75m from a reflecting globe in a garden.
Maru [420]

Answer:

1. The image of the person is 1.41 m, virtual and formed at the back of the surface of the globe.

2. The person's image is 3.38 m tall.

Explanation:

From the given question, object distance, u = 0.75 m, object height = 1.8 m, radius of curvature of the reflecting globe, r = 8 cm = 0.08 m.

f = \frac{r}{2} = \frac{0.08}{2} = 0.04 m

1. The image distance, v, can be determined by applying mirror formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

\frac{1}{0.04} = \frac{1}{0.75} + \frac{1}{v}

\frac{4}{100} - \frac{75}{100} = \frac{1}{v}

\frac{1}{v} = \frac{4 - 75}{100}

  = - \frac{71}{100}

⇒ v = -\frac{100}{71}

      = - 1.41 m

The image of the person is 1.41 m, virtual and formed at the back of the surface of the globe.

2.  \frac{image distance}{object distance} = \frac{image height}{object height}

  \frac{1.41}{0.75} = \frac{v}{1.8}

v = \frac{2.538}{0.75}

  = 3.384

v = 3.38 m

The person's image is 3.38 m tall.

6 0
3 years ago
a race car accelerates uniformly from 18.5 m/s to 46.1m/s in 2.47 seconds detrrmine the acceleration of the car and distance tra
LenaWriter [7]

Because acceleration is constant, the acceleration of the car at any time is the same as its average acceleration over the duration. So

a=\dfrac{\Delta v}{\Delta t}=\dfrac{46.1\,\frac{\mathrm m}{\mathrm s}-18.5\,\frac{\mathrm m}{\mathrm s}}{2.47\,\mathrm s}=11.2\,\dfrac{\mathrm m}{\mathrm s^2}

Now, we have that

{v_f}^2-{v_0}^2=2a\Delta x

so we end up with a distance traveled of

\left(46.1\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(18.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(11.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)\Delta x

\implies\Delta x=79.6\,\mathrm m

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3 years ago
How far does a wave travel in three cycles?
eduard
It will be 3 wavelengths because 1 cycle = 1 wavelength. 
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4 years ago
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