(a) The moment of inertia of the wheel is 78.2 kgm².
(b) The mass (in kg) of the wheel is 1,436.2 kg.
(c) The angular speed (in rad/s) of the wheel at the end of this time period is 3.376 rad/s.
<h3>
Moment of inertia of the wheel</h3>
Apply principle of conservation of angular momentum;
Fr = Iα
where;
- F is applied force
- r is radius of the cylinder
- α is angular acceleration
- I is moment of inertia
I = Fr/α
I = (200 x 0.33) / (0.844)
I = 78.2 kgm²
<h3>Mass of the wheel</h3>
I = ¹/₂MR²
where;
- M is mass of the solid cylinder
- R is radius of the solid cylinder
- I is moment of inertia of the solid cylinder
2I = MR²
M = 2I/R²
M = (2 x 78.2) / (0.33²)
M = 1,436.2 kg
<h3>Angular speed of the wheel after 4 seconds</h3>
ω = αt
ω = 0.844 x 4
ω = 3.376 rad/s
Thus, the moment of inertia of the wheel is 78.2 kgm².
The mass (in kg) of the wheel is 1,436.2 kg.
The angular speed (in rad/s) of the wheel at the end of this time period is 3.376 rad/s.
Learn more about moment of inertia here: brainly.com/question/14839816
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We know, Potential Energy = Force * Height
Here, F = 40 N
h = 5 m
Substitute their values,
U = 40 * 5
U = 200 J
In short, Your Answer would be Option A
Hope this helps!
Answer:
<em>radius of the loop = 7.9 mm</em>
<em>number of turns N ≅ 399 turns</em>
Explanation:
length of wire L= 2 m
field strength B = 3 mT = 0.003 T
current I = 12 A
recall that field strength B = μnI
where n is the turn per unit length
vacuum permeability μ = 
= 1.256 x 10^-6 T-m/A
imputing values, we have
0.003 = 1.256 x 10^−6 x n x 12
0.003 = 1.507 x 10^-5 x n
n = 199.07 turns per unit length
for a length of 2 m,
number of loop N = 2 x 199.07 = 398.14 ≅ <em>399 turns</em>
since there are approximately 399 turns formed by the 2 m length of wire, it means that each loop is formed by 2/399 = 0.005 m of the wire.
this length is also equal to the circumference of each loop
the circumference of each loop = 
0.005 = 2 x 3.142 x r
r = 0.005/6.284 = 
= 0.0079 m =<em> 7.9 mm</em>
