Answer:
The tangential speed of the tack is 8.19 m/s.
Explanation:
The wheel rotates 3.37 times a second that means wheel complete 3.37 revolutions in a second. Therefore, the angular speed ω of the wheel is given as follows:

Use the relation of angular speed with tangential speed to find the tangential speed of the tack.
The tangential speed v of the tack is given by following expression
v = ω r
Here, r is the distance to the tack from axis of rotation.
Substitute 21.174 rad/s for ω, and 0.387 m for r in the above equation to solve for v.
v = 21.174 × 0.387
v = 8.19m/s
Thus, The tangential speed of the tack is 8.19 m/s.
The Moment of Inertia of the Disc is represented by
. (Correct answer: A)
Let suppose that the Disk is a Rigid Body whose mass is uniformly distributed. The Moment of Inertia of the element is equal to the Moment of Inertia of the entire Disk minus the Moment of Inertia of the Hole, that is to say:
(1)
Where:
- Moment of inertia of the Disk.
- Moment of inertia of the Hole.
Then, this formula is expanded as follows:
(1b)
Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole (
):


And the resulting equation is:



The moment of inertia of the Disc is represented by
. (Correct answer: A)
Please see this question related to Moments of Inertia: brainly.com/question/15246709
Answer:
<em>a. 4.21 moles</em>
<em>b. 478.6 m/s</em>
<em>c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>
Explanation:
Volume of container = 100.0 L
Temperature = 293 K
pressure = 1 atm = 1.01325 bar
number of moles n = ?
using the gas equation PV = nRT
n = PV/RT
R = 0.08206 L-atm-

Therefore,
n = (1.01325 x 100)/(0.08206 x 293)
n = 101.325/24.04 = <em>4.21 moles</em>
The equation for root mean square velocity is
Vrms = 
R = 8.314 J/mol-K
where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol
Vrms =
= <em>478.6 m/s</em>
<em>For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship</em>
= 
where
Voxy = root mean square velocity of oxygen = 478.6 m/s
Vnit = root mean square velocity of nitrogen = ?
Moxy = Molar mass of oxygen = 31.9 g/mol
Mnit = Molar mass of nitrogen = 14.00 g/mol
= 
= 0.66
Vnit = 0.66 x 478.6 = <em>315.876 m/s</em>
<em>the root mean square velocity of the oxygen gas is </em>
<em>478.6/315.876 = 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>