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1 year ago

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an object is acted on by a drag force with a magnitude that is proportional to the speed. the object accelerates downward at 3.0

0 m/s2 when it is falling with a speed 20.0 m/s. what is the terminal speed of the object as it is falling?

1 answer:

Nutka1998 [239]1 year ago

7 0

The terminal speed of the object falling down is 66.67 m/s.

The terminal speed acquired by the body when,

Weight of the body = Drag force of the body

It is given,

Drag force is directly proportional to the speed,

So,

F = CV

Where F is drag force,

V is the speed,

C is the constant,

So, it can be written as C = F/V.

The weight of the body = mg

The weight of the body = 10m

M is the mass and g is the acceleration due to gravity,

The drag force when the speed is 20m/s.

Drag force = ma

a is the acceleration during the drag force which is given to be 3m/s²,

Drag force = 3m

Now we can write,

F₁/V₁ = F₂/V₂

F₁ is the drag force at 20m/s speed.

F₂ is the weight of the body and V₂ is the terminal speed,

Now, it can be written,

3m/20 = 10m/V₂

V₂ = 66.67 m/s.

So, the terminal speed is 66.67m/s.

To know more about terminal speed, visit,

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3 years ago

Read 2 more answers

The pressure in a traveling sound wave is given by the equation ΔP = (1.78 Pa) sin [ (0.888 m-1)x - (500 s-1)t] Find (a) the pre

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Answer:

a) $P_m=1.78\ Pa$

b) $f=79.5775\ Hz$

c) $\lambda=7.076\ m$

d) $v=563.06\ m.s^{-1}$

Explanation:

<u>Given equation of pressure variation:</u>

$\Delta P= (1.78\ Pa)\ sin\ [(0.888\ m^{-1})x-(500\ s^{-1})t]$

We have the standard equation of periodic oscillations:

$\Delta P=P_m\ sin\ (kx-\omega.t)$

<em>By comparing, we deduce:</em>

(a)

amplitude:

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(b)

angular frequency:

$\omega=2\pi.f$

$2\pi.f=500$

∴Frequency of oscillations:

$f=\frac{500}{2\pi}$

$f=79.5775\ Hz$

(c)

wavelength is given by:

$\lambda=\frac{2\pi}{k}$

$\lambda=\frac{2\pi}{0.888}$

$\lambda=7.076\ m$

(d)

Speed of the wave is gives by:

$v=\frac{\omega}{k}$

$v=\frac{500}{0.888}$

$v=563.06\ m.s^{-1}$

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3 years ago