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2 years ago

How long does it take the baton to complete one spin

Physics

1 answer:

sweet [91]2 years ago

8 0

The time it takes the baton to complete one spin will be 0.56 s. Option B is correct.

<h3>What is centripetal acceleration?</h3>

The acceleration needed to move a body in a curved way is understood as centripetal acceleration.

The direction of centripetal acceleration is always in the path of the center of the course. The total acceleration is the result of tangential and centripetal acceleration.

The entire question is;

"How long does it take the baton to complete one spin?

A twirler’s baton is 0.76 m long and spins around its center. The end of the baton has a centripetal acceleration of 47.8 m/s2.

a.0.31 s

b.0.56 s

c.4.3 s

d.70 s"

The given data in the problem;

Length of baton,L = 0.76 m

Centripetal acceleration, $\rm F_c=47.8 \ m/s^2$

The centripetal acceleration is found by;

$a= \frac{4 \pi^2R}{T^2}$

Substitute the given value:

$\rm 47.82= \frac{4 \pi^2\times 0.38}{T^2} \\\\ 47.82\times T^2 = 4(3.14)^2 \\\\T = 0.56 s$

The time it takes the baton to complete one spin will be 0.56 s.

Hence option B is correct.

To learn more about centripetal acceleration, refer to the link;

brainly.com/question/17689540

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4 years ago

The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de

spin [16.1K]

Answer:

the spring constant k = $5.409*10^4 \ N/m$

the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

Explanation:

From Hooke's Law

$F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m$

Thus; the spring constant k = $5.409*10^4 \ N/m$

The amplitude is decreasing 37% during one period of the motion

$e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}$

$b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) } }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s$

Therefore; the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

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3 years ago