Answer:
One way to look at this is to consider the forces acting on any point in a string.
For a very small portion of string F = M a must still hold. As M approaches zero the small portion of string would have to approach infinite acceleration if the net force on that portion of string were not zero.
One generally considers the net force acting on the center of mass of an object not the individual forces acting on each infinitesimal mass composing
the object.
Answer:
because
Explanation:
Mass and velocity are both directly proportional to the momentum. If you increase either mass or velocity, the momentum of the object increases proportionally. If you double the mass or velocity you double the momentum.
Answer:
0.89534 kgm²
Explanation:
= Final angular velocity = 12.9 rad/s
= Initial angular velocity = 5.5 rad/s
= Initial moment of inertia = 2.1 kgm²
= Final moment of inertia
As there is no external torque the angular momentum is conserved
![I_i\omega_i=I_f\omega_f\\\Rightarrow I_f=\dfrac{I_i\omega_i}{\omega_f}\\\Rightarrow I_f=\dfrac{2.1\times 5.5}{12.9}\\\Rightarrow I_f=0.89534\ kgm^2](https://tex.z-dn.net/?f=I_i%5Comega_i%3DI_f%5Comega_f%5C%5C%5CRightarrow%20I_f%3D%5Cdfrac%7BI_i%5Comega_i%7D%7B%5Comega_f%7D%5C%5C%5CRightarrow%20I_f%3D%5Cdfrac%7B2.1%5Ctimes%205.5%7D%7B12.9%7D%5C%5C%5CRightarrow%20I_f%3D0.89534%5C%20kgm%5E2)
The final moment of inertia is 0.89534 kgm²
We are given
m = mass of the object
r = distance from the center of the planet
r <span>≥ rplanet
We are asked for the
ve = escape velocity
The escape velocity
ve = </span>√ (2Gm/r)
<span>
</span>
Answer:
a. ![\alpha =2122.22\: rev/s^{2}](https://tex.z-dn.net/?f=%5Calpha%20%3D2122.22%5C%3A%20rev%2Fs%5E%7B2%7D)
b. ![\Delta \theta =9,550.02\: rev](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%20%3D9%2C550.02%5C%3A%20rev)
Explanation:
The computation is shown below:
data provided in the question
The initial angular velocity
= 0 rev/s,
f = Final angular velocity =
= 382000 rpm i.e =
= 6,366.67
And time = t = 3.0s
Based on the above information
a. For angular acceleration of drill
![\small \omega =\omega _{o}+\alpha t \\\\\ 6,366.67 = 0 + \alpha (3.0) \\\\ \alpha =2122.22\: rev/s^{2}](https://tex.z-dn.net/?f=%5Csmall%20%5Comega%20%3D%5Comega%20_%7Bo%7D%2B%5Calpha%20t%20%5C%5C%5C%5C%5C%206%2C366.67%20%3D%200%20%2B%20%5Calpha%20%283.0%29%20%5C%5C%5C%5C%20%5Calpha%20%3D2122.22%5C%3A%20rev%2Fs%5E%7B2%7D)
b. For the number of revolutions
![\small \omega ^{2}-\omega _{o}^{2}=2\alpha \Delta \theta \\\\(6,366.67) ^{2}-(0)^{2}=2(2122.22) \Delta \theta \\\\ \Delta \theta =9,550.02\: rev](https://tex.z-dn.net/?f=%5Csmall%20%5Comega%20%5E%7B2%7D-%5Comega%20_%7Bo%7D%5E%7B2%7D%3D2%5Calpha%20%5CDelta%20%5Ctheta%20%5C%5C%5C%5C%286%2C366.67%29%20%5E%7B2%7D-%280%29%5E%7B2%7D%3D2%282122.22%29%20%5CDelta%20%5Ctheta%20%5C%5C%5C%5C%20%5CDelta%20%5Ctheta%20%3D9%2C550.02%5C%3A%20rev)
We simply applied the above formulas for determining each parts