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3 years ago

10

A stone is dropped from rest from the top of a cliff into a pond below. If its initial height is 10 m, what is its speed when it

hits the water?

1 answer:

Brut [27]3 years ago

8 0

Answer:

14 m/s

Explanation:

The motion of the stone is a free fall motion, so an accelerated motion with constant acceleration g = 9.8 m/s^2 towards the ground. So, we can use the following SUVAT equation:

$v^2 -u^2 = 2gh$

where

v is the final speed of the stone as it reaches the water

u = 0 is the initial speed

g = 9.8 m/s^2 is the acceleration

h = 10 m is the distance covered by the stone

Solving for v, we find

$v=\sqrt{u^2+2gh}=\sqrt{0+2(9.8 m/s^2)(10 m)}=14 m/s$

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Sidana [21]

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

7 0

3 years ago

In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the following expression,

eduard

(a) 4.06 cm

In a simple harmonic motion, the displacement is written as

$x(t) = A cos (\omega t + \phi)$ (1)

where

A is the amplitude

$\omega$ is the angular frequency

$\phi$ is the phase

t is the time

The displacement of the piston in the problem is given by

$x(t) = (5.00 cm) cos (5t+\frac{\pi}{5})$ (2)

By putting t=0 in the formula, we find the position of the piston at t=0:

$x(0) = (5.00 cm) cos (0+\frac{\pi}{5})=4.06 cm$

(b) -14.69 cm/s

In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:

$v(t) = x'(t) = -\omega A sin (\omega t + \phi)$ (3)

Differentiating eq.(2), we find

$v(t) = x'(t) = -(5 rad/s)(5.00 cm) sin (5t+\frac{\pi}{5})=-(25.0 cm/s) sin (5t+\frac{\pi}{5})$

And substituting t=0, we find the velocity at time t=0:

$v(0)=-(25.00 cm/s) sin (0+\frac{\pi}{5})=-14.69 cm/s$

(c) -101.13 cm/s^2

In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:

$a(t) = v'(t) = -\omega^2 A cos (\omega t + \phi)$

Differentiating eq.(3), we find

$a(t) = v'(t) = -(5 rad/s)(25.00 cm/s) cos (5t+\frac{\pi}{5})=-(125.0 cm/s^2) cos (5t+\frac{\pi}{5})$

And substituting t=0, we find the acceleration at time t=0:

$a(0)=-(125.00 cm/s) cos (0+\frac{\pi}{5})=-101.13 cm/s^2$

(d) 5.00 cm, 1.26 s

By comparing eq.(1) and (2), we notice immediately that the amplitude is

A = 5.00 cm

For the period, we have to start from the relationship between angular frequency and period T:

$\omega=\frac{2\pi}{T}$

Using $\omega = 5.0 rad/s$ and solving for T, we find

$T=\frac{2\pi}{5 rad/s}=1.26 s$

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rosijanka [135]

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Hope this helps :)

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4 years ago