While riding a multispeed bicycle, the rider can select the radius of the rear sprocket that is fixed to the rear axle. the front sprocket of a bicycle has radius 12.0 cm. if the angular speed of the front sprocket is 0.600 rev/s, what is the radius of the rear sprocket for which the tangential speed of a point on the rim of the rear wheel will be 5.00 m/s? the rear wheel has radius 0.310 m.

2.8 cm

Explanation:

radius of front sprocket (R1) = 12 cm = 0.12 m

angular speed of the front sprocket (ω1) = 0.6 rev/s = 3.77 rad/s

radius of the rear wheel (R°2) = 0.31 m

tangential speed of the rear wheel (V°2) = 5 m/s

(take note that the tangential speed of the rear wheel is the same as that of the rear sprocket, V°2 = V2 (rear sprocket))

speed at any point on the front sprocket = speed at any point on the rear sprocket

V1 = V2 ....equation 1

        V1 = 0.12 x 3.77 = 0.45 m/s

        take note that the speed of the rear wheel is the same as the speed of                

        the rear sprocket, which means ω2 is the same for both the rear  

        wheel and the rear sprocket. This also means we can use the

        parameters for the rear wheel (V°2 and R2°) to find ω2

         ω2 =

         ω2 =   = 16.1 rad/s

        therefore V2 = R1 x ω2 = 16.1 x R1

recall that  V1 = V2 ....equation 1

now we can put the values of V1 and V2 into equation 1

0.45 = 16.1 x R1

R1 = 0.45 / 16.1 = 0.028 m = 2.8 cm

4.725 kgm/s to the right

468.6 kgm/s southwest

1.782\times 10^{49}\ kgm/s1.782×1049 kgm/s forward

Explanation:

When the mass of an object is multiplied by its velocity we get momentum

m = Mass

v = Velocity

Proton

\begin{gathered}p=mv\\\Rightarrow p=1.67\times 10^{-27}\times 5.04\times 10^6\\\Rightarrow p=8.4168\times 10^{-21}\ kgm/s\end{gathered}p=mv⇒p=1.67×10−27×5.04×106⇒p=8.4168×10−21 kgm/s

The momentum of the proton is 8.4168\times 10^{-21}\ kgm/s8.4168×10−21 kgm/s upward

Bullet

\begin{gathered}p=mv\\\Rightarrow p=0.015\times 315\\\Rightarrow p=4.725\ kgm/s\end{gathered}p=mv⇒p=0.015×315⇒p=4.725 kgm/s

The momentum of the bullet is 4.725 kgm/s to the right

Sprinter

\begin{gathered}p=mv\\\Rightarrow p=71\times 6.6\\\Rightarrow p=468.6\ kgm/s\end{gathered}p=mv⇒p=71×6.6⇒p=468.6 kgm/s

The momentum of the sprinter is 468.6 kgm/s southwest

Earth

\begin{gathered}p=mv\\\Rightarrow p=5.98\times 10^{24}\times 2.98\times 10^4\\\Rightarrow p=1.782\times 10^{29}\ kgm/s\end{gathered}p=mv⇒p=5.98×1024×2.98×104⇒p=1.782×1029 kgm/s

The momentum of the sprinter is 1.782\times 10^{29}\ kgm/s1.782×1029 kgm/s forward