Answer:
a. 0.21 rad/s2
b. 2.205 N
Explanation:
We convert from rpm to rad/s knowing that each revolution has 2π radians and each minute is 60 seconds
200 rpm = 200 * 2π / 60 = 21 rad/s
180 rpm = 180 * 2π / 60 = 18.85 rad/s
r = d/2 = 30cm / 2 = 15 cm = 0.15 m
a)So if the angular speed decreases steadily (at a constant rate) from 21 rad/s to 18.85 rad/s within 10s then the angular acceleration is
![\alpha = \frac{\Delta \omega}{\Delta t} = \frac{21 - 18.85}{10} = 0.21 rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5CDelta%20%5Comega%7D%7B%5CDelta%20t%7D%20%3D%20%5Cfrac%7B21%20-%2018.85%7D%7B10%7D%20%3D%200.21%20rad%2Fs%5E2)
b) Assume the grind stone is a solid disk, its moment of inertia is
![I = mR^2/2](https://tex.z-dn.net/?f=I%20%3D%20mR%5E2%2F2)
Where m = 28 kg is the disk mass and R = 0.15 m is the radius of the disk.
![I = 28*0.15^2/2 = 0.315 kgm^2](https://tex.z-dn.net/?f=%20I%20%3D%2028%2A0.15%5E2%2F2%20%3D%200.315%20kgm%5E2)
So the friction torque is
![T_f = I\alpha = 0.315*0.21 = 0.06615 Nm](https://tex.z-dn.net/?f=T_f%20%3D%20I%5Calpha%20%3D%200.315%2A0.21%20%3D%200.06615%20Nm)
The friction force is
![F_f = T_f/R = 0.06615 / 0.15 = 0.441 N](https://tex.z-dn.net/?f=F_f%20%3D%20T_f%2FR%20%3D%200.06615%20%2F%200.15%20%3D%200.441%20N)
Since the friction coefficient is 0.2, we can calculate the normal force that is used to press the knife against the stone
![N = F_f/\mu = 0.441/0.2 = 2.205 N](https://tex.z-dn.net/?f=N%20%3D%20F_f%2F%5Cmu%20%3D%200.441%2F0.2%20%3D%202.205%20N)
<span>movement of energy through space or a medium Hope this works! ;3</span>
Answer:
while all of the world can experience a solar eclipse it is not normally at the same time so one part of the earth might have it but not the other and even in one area if there is one in north america like the last one in texas we did not have a full eclipse
Answer:
D-Driving the car faster down the road.