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3 years ago

10

Estimate the force a person must exert on a massless string attached to a 0.15 kg ball to make the ball revolve in horizontal ci

rcle of radius 0.6 m. The ball makes 2 revolutions per second.

1 answer:

Mrrafil [7]3 years ago

6 0

Answer:

$F = 14.2 N$

Explanation:

From the question we are told that:

Mass $m=0.15kg$

Radius $r=0.6$

Angular Velocity $\omega=2rev/s$

$\omega= =2x2 \pi rad/s=>4 \pi rad/s$

Generally the equation for Force applied is mathematically given by

$F =mrw2$

$F=0.15*0.6* (4*x3.14^)2$

$F = 14.2 N$

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A cyclist is riding a bicycle at a speed of 22 mph on a horizontal road. The distance between the axles is 42 in., and the mass

stealth61 [152]

Answer:

The shortest distance is $S = 24.86 ft$

Explanation:

The free body diagram of this question is shown on the first uploaded image

From the question we are told that

The speed of the bicycle is $v_b = 22\ mph = 22 * \frac{5280}{3600} = 32.26 ft/s$

The distance between the axial is $d = 42 \ in$

The mass center of the cyclist and the bicycle is $m_c = 26 \ in$ behind the front axle

The mass center of the cyclist and the bicycle is $m_h = 40 \ in$ above the ground

For the bicycle not to be thrown over the

Momentum about the back wheel must be zero so

$\sum _B = 0$

=> $mg (26) = ma(40)$

=> $a = \frac{26}{40} g$

Here $g = 32.2 ft/s^2$

So $a = \frac{26}{40} (32.2)$

$a = 20.93 ft/s^2$

Apply the equation of motion to this motion we have

$v^2 = u^2 + 2as$

Where $u = 32.26 ft /s$

and $v = 0$ since the bicycle is coming to a stop

$v^2 = (32.26)^2 - 2(20.93) S$

=> $S = 24.86 ft$

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Answer:

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Explanation:

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