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inessss [21]
3 years ago
5

A bee loaded with pollen flies in a circular path at a constant speed of 3.20 m/s. If the mass of the bee is 133 mg and the radi

us of its path is 11.0 m, what is the magnitude of the centripetal acceleration, what is the centripetal force? acceleration _ m/s^2 force _ N (3sigfigs)
Physics
2 answers:
BaLLatris [955]3 years ago
8 0
<h3><u>Answer;</u></h3>

Centripetal acceleration = 0.931 m/s²

Centripetal force = = 1.234 × 10^-4 Newtons

<h3><u>Explanation;</u></h3>

Centripetal acceleration is given by the formula v²/r, where r is the radius of the circular path and v is the velocity of a body;

Centripetal acc = 3.2²/11

                          = 0.931 m/s²

Centripetal force is a force that acts on an object or a body in circular path and is directed towards the center of the circular path.

Centripetal force is given by the formula;

mv²/r ; where m is the mass of the body, r is the radius of the circular path and v is the velocity of a body;

mass = 33 mg or 1.33 × 10^-6 kg, velocity = 3.20 m/s and r = 11 m

Therefore;

Centripetal force = (1.33 ×10^-6 × 3.1²)/ 11

                            = 1.234 × 10^-4 Newtons

                           

Valentin [98]3 years ago
6 0

Centripetal acceleration is is a caused by the constant change in the direction of the velocity of the particle describing the circular path.

Centripetal acceleration is given by a=v²/r where v is the velocity and r the radius of the circular path.

From the question, v=3.20m/s and r=11.0m

Therefore, a=3.20²/11.0

                     =0.931m/s²

from the newtons laws of motion, force= mass ×acceleration. therefore

centripetal acceleration=mv²/r since a=v²/r

to change 133mg to kg we divide by 1000000 since 100mg make a gram and 1000 grams make a kilogram, the SI unit for mass.

133÷1000000=1.33×10⁻⁴kg

thus centripetal force=1.33×10⁻⁴  kg×0.931 m/s²

                                    =1.24×10⁻⁴Newtons

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<h3>Angular Speed of the pulley </h3>

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K.E = P.E

\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\

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<h3>Linear speed of the block</h3>

The linear speed of the block after travelling 0.7 m;

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v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

Learn more about conservation of energy here: brainly.com/question/24772394

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