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elena55 [62]
3 years ago
12

An object goes from 10 m/s to 2 m/s in 2 seconds, what's its acceleration?

Physics
1 answer:
Mila [183]3 years ago
6 0

Answer:

a=(v-u)÷t

a= (2-10)÷2

a= -8÷2

a= -4 m/s²

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Which of the following represents an element?
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Megan was doing time trials on her bike around a 400m horizontal track. She took 32 seconds to travel 400m. What was her average
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A-10A twin-jet close-support airplane is approximately rectangular with a wingspan (the length perpendicular to the flow directi
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Solution :

Given :

Rectangular wingspan

Length,L = 17.5 m

Chord, c = 3 m

Free stream velocity of flow, $V_{\infty}$ = 200 m/s

Given that the flow is laminar.

$Re_L=\frac{\rho V L}{\mu _{\infty}}$

      $=\frac{1.225 \times 200 \times 3}{1.789 \times 10^{-5}}$

    $= 4.10 \times 10^7$

So boundary layer thickness,

$\delta_{L} = \frac{5.2 L}{\sqrt{Re_L}}$

$\delta_{L} = \frac{5.2 \times 3}{\sqrt{4.1 \times 10^7}}$

    = 0.0024 m

The dynamic pressure, $q_{\infty} =\frac{1}{2} \rho V^2_{\infty}$

                                           $ =\frac{1}{2} \times 1.225  \times 200^2$

                                          $=2.45 \times 10^4 \ N/m^2$

The skin friction drag co-efficient is given by

$C_f = \frac{1.328}{\sqrt{Re_L}}$

     $=\frac{1.328}{\sqrt{4.1 \times 10^7}}$

     = 0.00021

$D_{skinfriction} = \frac{1}{2} \rho V^2_{\infty}S C_f$

                  $=\frac{1}{2} \times 1.225 \times 200^2 \times 17.5 \times 3 \times 0.00021$

                  = 270 N

Therefore the net drag = 270 x 2

                                      = 540 N

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2 years ago
A rock has a mass of 8.0 kilograms falls 5 meters.
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Answer:

A. 392J

B. 392J

C. 9.899m/s

Explanation:

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