So, the first question is: how many meters are 10 nm?
1nm =<span>0.000000001 m.
So 10 nanometers are </span><span>0.00000001 m!
Now, how many milimeter are those?
let's start with meters, 1 meter are 1000 milimeters.
so </span>
0.00000001*1000=0.<span><span>00001</span> m!
now, micrometers .1 micrometer are 1000 nanometers.
so 10 nanometers are 0.01 micrometers! (1 nanometer is 0.001 micrometers)
</span>
Answer:
A) ![\omega_f=17.503\ rad.s^{-1}](https://tex.z-dn.net/?f=%5Comega_f%3D17.503%5C%20rad.s%5E%7B-1%7D)
B) ![t=55.6822\ s](https://tex.z-dn.net/?f=t%3D55.6822%5C%20s)
C) ![\theta=1312\ rad](https://tex.z-dn.net/?f=%5Ctheta%3D1312%5C%20rad)
Explanation:
Given:
- mass of flywheel,
![m=40\ kg](https://tex.z-dn.net/?f=m%3D40%5C%20kg)
- diameter of flywheel,
![d=0.72\ m](https://tex.z-dn.net/?f=d%3D0.72%5C%20m)
- rotational speed of flywheel,
![N_i=450\ rpm \Rightarrow \omega_i=\frac{450\times 2\pi}{60} =15\pi\ rad.s^{-1}](https://tex.z-dn.net/?f=N_i%3D450%5C%20rpm%20%5CRightarrow%20%5Comega_i%3D%5Cfrac%7B450%5Ctimes%202%5Cpi%7D%7B60%7D%20%3D15%5Cpi%5C%20rad.s%5E%7B-1%7D)
- duration for which the power is off,
![t_0=35\ s](https://tex.z-dn.net/?f=t_0%3D35%5C%20s)
- no. of revolutions made during the power is off,
![\theta=180\times 2\pi=360\pi\ rad](https://tex.z-dn.net/?f=%5Ctheta%3D180%5Ctimes%202%5Cpi%3D360%5Cpi%5C%20rad)
<u>Using equation of motion:</u>
![\theta=\omega_i.t+\frac{1}{2} \alpha.t^2](https://tex.z-dn.net/?f=%5Ctheta%3D%5Comega_i.t%2B%5Cfrac%7B1%7D%7B2%7D%20%5Calpha.t%5E2)
![360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2](https://tex.z-dn.net/?f=360%5Cpi%3D15%5Cpi%5Ctimes%2035%2B%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%5Calpha%5Ctimes35%5E2)
![\alpha=-0.8463\ rad.s^{-2}](https://tex.z-dn.net/?f=%5Calpha%3D-0.8463%5C%20rad.s%5E%7B-2%7D)
Negative sign denotes deceleration.
A)
Now using the equation:
![\omega_f=\omega_i+\alpha.t](https://tex.z-dn.net/?f=%5Comega_f%3D%5Comega_i%2B%5Calpha.t)
![\omega_f=15\pi-0.8463\times 35](https://tex.z-dn.net/?f=%5Comega_f%3D15%5Cpi-0.8463%5Ctimes%2035)
is the angular velocity of the flywheel when the power comes back.
B)
Here:
![\omega_f=0\ rad.s^{-1}](https://tex.z-dn.net/?f=%5Comega_f%3D0%5C%20rad.s%5E%7B-1%7D)
Now using the equation:
![\omega_f=\omega_i+\alpha.t](https://tex.z-dn.net/?f=%5Comega_f%3D%5Comega_i%2B%5Calpha.t)
![0=15\pi-0.8463\times t](https://tex.z-dn.net/?f=0%3D15%5Cpi-0.8463%5Ctimes%20t)
is the time after which the flywheel stops.
C)
Using the equation of motion:
![\theta=\omega_i.t+\frac{1}{2} \alpha.t^2](https://tex.z-dn.net/?f=%5Ctheta%3D%5Comega_i.t%2B%5Cfrac%7B1%7D%7B2%7D%20%5Calpha.t%5E2)
![\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2](https://tex.z-dn.net/?f=%5Ctheta%3D15%5Cpi%5Ctimes%2055.68225-0.5%5Ctimes%200.8463%5Ctimes%2055.68225%5E2)
revolutions are made before stopping.
Answer: 704
Explanation:Vi = 0 m/s
vf = 65 m/s
a = 3 m/s2
d = ??
vf2 = vi2 + 2*a*d
(65 m/s)2 = (0 m/s)2 + 2*(3 m/s2)*d
4225 m2/s2 = (0 m/s)2 + (6 m/s2)*d
(4225 m 2/m2)/(6 m/s2) = d
d = 704 m
Answer:
14 newtons is the answer
Explanation:
because if they act in the same direction they are added together