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)
Answer:
- A hypothesis is a tentative statement about the relationship between two or more variables. It is a specific, testable prediction about what you expect to happen in a study.
A hypothesis is a tentative statement about the relationship between two or more variables. is a specific, testable prediction about what you expect to happen in a study.
Explanation:
This is answer
Answer:
the kinetic energy of clown A is 0.444 times the kinetic energy of clown B.
Explanation:
Let the spring constant of the spring is k.
For clown A:
m = 40 kg
let the extension in the spring is y.
So, the spring force, F = k y
m g = k y
40 x g = k x y
y = 40 x g / k ..... (1)
For clown B:
m' = 60 kg
Let the extension in the spring is y'.
So, the spring force, F' = k y'
m' g = k y'
y' = 60 x g / k .....(2)
Kinetic energy for A, K = 1/2 ky^2
Kinetic energy for B, K' = 1/2 ky'^2
So, K/K' = y^2/y'^2 K / K' = (40 x 40) / (60 x 60) (from equation (1) and (2))
K / K' = 0.444
K = 0.444 K'
So the kinetic energy of clown A is 0.444 times the kinetic energy of clown B.
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Let's see what variables we've got first. Hmmm. We have:
Displacement, d = 28 m
Time taken, t = 11 s
Initial velocity, u = 0 m/s (at rest)
And now we need to find the final velocity, v. Among the 4 (or 5) equations of motions, there's no equation that will let us simply plug in the values and give an answer sigh. But fear not! We'll do it in steps.
I'm going to pick one of the motion equation to find more information:
![d = ut + \frac{1}{2} a {t}^{2}](https://tex.z-dn.net/?f=d%20%3D%20ut%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20a%20%7Bt%7D%5E%7B2%7D%20)
I know everything except for a in this one, so I I'll use this! After plugging in values, I get a = 0.4628 m/s^2.
Now I'm going to use another motion equation that has v in it because that needs to be solved!
![v = u + at](https://tex.z-dn.net/?f=v%20%3D%20u%20%2B%20at)
Now I know everything except dial velocity v. Nice!
v = 0 + (0.4628)(11)