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Nimfa-mama [501]

2 years ago

8

A Ping-Pong ball has a mass of 2.3 g and a terminal speed of 9.2 m/s. The drag force is of the form bv2 What is the value of b?

1 answer:

Bingel [31]2 years ago

8 0

Drag force is a force that resist the motion of a certain body through a fluid like air or water. This can be calculated through the equation,
D = 0.5Cρv²
where D is drag, C is the drag coefficient, ρ is the density and v is the velocity. The value therefore of b in the item above is,
b = 0.5Cρ

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At a given moment, a plane passes directly above a radar station at an altitude of 6 km. Let θ be the angle that the line throug

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<u>Explanation:</u>

Let the distance x and angle θ be defined as in the figure below. Then

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Now, differentiate with respect to t, we get

$\sec ^{2} \theta \frac{d \theta}{d t}=-\left(\frac{6}{x^{2}}\right) \frac{d x}{d t}$

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Substituting ‘x’ value, we get

$\tan \theta=\frac{6}{160}=\frac{3}{80}$

Find the rate of change of theta after 12 min,

$\frac{d \theta}{d t}=-\frac{1}{\sec ^{2} \theta} \times \frac{6}{x^{2}} \times \frac{d x}{d t}$

We know, the formula for,

$\sec ^{2} \theta=1+\tan ^{2} \theta=1+\frac{3^{2}}{80^{2}}$

So, then, $\frac{d x}{d t}=800 \mathrm{km} / \mathrm{hr}(\text { let assume })$

$\frac{d \theta}{d t}=-\frac{1}{\left(1+\frac{3^{2}}{80^{2}}\right)} \times \frac{6}{160^{2}} \times 800$

$=-\frac{1}{\left(1+\frac{3^{2}}{80^{2}}\right)} \times \frac{6}{160^{2}} \times 800$

$=-\frac{1}{1.0014} \times \frac{6}{25600} \times 800$

$=-\frac{4800}{25635.84}=-0.187237 \mathrm{rad} / \mathrm{hr}$

When express the value in km/he, we get, the change in theta as

$=-187.237 \mathrm{km} / \mathrm{hr}$

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