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Fynjy0 [20]
3 years ago
10

A torque of 0.77 N⋅m is applied to a bicycle wheel of radius 30 cm and mass 0.70 kg

Physics
1 answer:
Naddik [55]3 years ago
4 0

Answer:

α = τ/I = 0.77 / (0.70(0.30²)) = 12.22222... = 12 rad/s²

Explanation:

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A circular saw blade with radius 0.175 m starts from rest and turns in a vertical plane with a constant angular acceleration of
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Answer:

The distance the piece travel in horizontally axis is

L=3.55m

Explanation:

a=2 \frac{rev}{s^{2}} \\h=0.820m\\r = 0.125 m&#10;\\d=150rev

d= 155 rev = 155(2\pi ) = 310\pi rad

a= 2.0 \frac{rev}{s^{2} } = 2.0(2\pi )  = 4.0\pi \frac{rev}{s^{2} }

d=d_{i}+vo*t+\frac{1}{2}*a*t^{2} \\ di=0\\vo=0\\d=\frac{1}{2}*a*t^{2}\\t=\sqrt{\frac{2*d}{a}}\\t=\sqrt{\frac{2*310 rad}{4\frac{rad}{s^{2}}}} \\t=12.449

w=a*t\\w=4\frac{rad}{s^{2}}*12.449s\\ w=49.79 \frac{rad}{s}

Now the angular velocity is the blade speed so:

V=w*r\\V=49.79 \frac{rad}{s}*0.175m\\V=8.7 \frac{m}{s}

assuming no air friction effects affect blade piece:

time for blade piece to fall to floor

t=\sqrt{\frac{2*h}{g}}\\t=\sqrt{\frac{2*0.820m}{9.8\frac{m}{s^{2} } }}\\t=0.409s

Now is the same time the piece travel horizontally

L=t*V\\L=0.409s*8.7\frac{m}{s}\\L=3.55m

blade piece travels  HORIZONTALLY = (24.5)(0.397) = 9.73 m  ANS

6 0
4 years ago
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