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Allushta [10]
3 years ago
10

The front and rear sprockets on a bicycle have radii of 8.40 and 4.91 cm, respectively. The angular speed of the front sprocket

is 12.3 rad/s. Determine (a) the linear speed (in cm/s) of the chain as it moves between the sprockets and (b) the centripetal acceleration (in cm/s2) of the chain as it passes around the rear sprocket.
Physics
2 answers:
Rudik [331]3 years ago
8 0

Explanation:

It is given that,

Radius of the front sprockets, r_f=8.4\ cm=0.084\ m

Radius of the rear sprockets, r_r=4.91\ cm=0.0491\ m

The angular speed of the front sprocket is 12.3 rad/s, \omega_f=12.3\ rad/s

(a) Linear speed of the front sprockets, v_f=r_f\times \omega

v_f=0.084\times 12.3    

v_f=1.0332\ m/s

v_f=103.32\ cm/s

Linear speed of the rear sprockets, v_r=r_r\times \omega

v_r=0.0491\times 12.3    

v_r=0.60393\ m/s

v_r=60.393\ cm/s

(b) Let a_r is the centripetal acceleration of the chain as it passes around the rear sprocket.

a_r=\dfrac{v_r^2}{r_r}

a_r=\dfrac{(60.393)^2}{0.0491}

a_r=74283.39\ m/s^2

a_r=7428339\ cm/s^2

`Hence, this is the required solution.

steposvetlana [31]3 years ago
3 0

Answer:

a) the linear speed is 103.32 cm/s

b) the centripetal acceleration is 2174.139 cm/s²

Explanation:

the solution is in the attached Word file

Download docx
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Describe the earth's rotation on it axis
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It's 3 deep

Under 129, comes 125 and 685;

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Under 52, comes 46 : Under 511, is 601.

This is illustrated below.

129

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A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is
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Answer:

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0

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The athlete only interacts with the gravitational potential energy. Air resistance is neglected.

At height y = 0

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At height y = 0 the potential energy is 0 as

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As the the potential and kinetic energy are conserved

The general equation

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