<h3><u>Answer</u>;</h3>
≈ 5 Kgm²/sec
<h3><u>Explanation</u>;</h3>
Angular momentum is given by the formula
L = Iω, where I is the moment of inertia and ω is the angular speed.
I = mr², where m is the mass and r is the radius
= 0.65 × 0.7²
= 0.3185
Angular speed, ω = v/r
= (2 × 3.142 × r × 2.5) r
= 15.71
Therefore;
Angular momentum = Iω
= 0.3185 × 15.71
= 5.003635
<u>≈ 5 Kgm²/sec</u>
In my view, correct answer should look like this: Although wave power does not produce pollution, some people may not want to invest in it because it is <span>prone to storm damage and limited to particular areas of the ocean.</span>
<span><span>anonymous </span> 4 years ago</span>Any time you are mixing distance and acceleration a good equation to use is <span>ΔY=<span>V<span>iy</span></span>t+1/2a<span>t2</span></span> I would split this into two segments - the rise and the fall. For the fall, Vi = 0 since the player is at the peak of his arc and delta-Y is from 1.95 to 0.890.
For the upward part of the motion the initial velocity is unknown and the final velocity is zero, but motion is symetrical - it takes the same amount of time to go up as it does to go down. Physiscists often use the trick "I'm going to solve a different problem, that I know will give me the same answer as the one I was actually asked.) So for the first half you could also use Vi = 0 and a downward delta-Y to solve for the time.
Add the two times together for the total.
The alternative is to calculate the initial and final velocity so that you have more information to work with.
The answer is B. Friction is going to the RIGHT because friction works against where you are trying to go
Given Information:
Current = I = 20 A
Diameter = d = 0.205 cm = 0.00205 m
Length of wire = L = 1 m
Required Information:
Energy produced = P = ?
Answer:
P = 2.03 J/s
Explanation:
We know that power required in a wire is
P = I²R
and R = ρL/A
Where ρ is the resistivity of the copper wire 1.68x10⁻⁸ Ω.m
L is the length of the wire and A is the area of the cross-section and is given by
A = πr²
A = π(d/2)²
A = π(0.00205/2)²
A = 3.3x10⁻⁶ m²
R = ρL/A
R = 1.68x10⁻⁸*(1)/3.3x10⁻⁶
R = 5.09x10⁻³ Ω
P = I²R
P = (20)²*5.09x10⁻³
P = 2.03 Watts or P = 2.03 J/s
Therefore, 2.03 J/s of energy is produced in 1.00 m of 12-gauge copper wire carrying a current of 20 A
