The magnetic field of a bar magnet is strongest at either pole of the magnet. It is equally strong at the north pole when compared with the south pole. The force is weaker in the middle of the magnet and halfway between the pole and the center.
I’ve answered this problem before and there were 2 parts in
this problem.
The solution would be like this for this specific problem:
<span>A.
</span><span>Vf = Vi +
Vex*ln(Mi / Mf) </span><span>
<span>0.002 * 3e8m/s = 0 + 2000m/s * ln(Mi / Mf) </span>
<span>300 = ln(Mi / Mf) </span>
<span>1.9e130 = Mi / Mf </span></span>
<span>B.
</span><span>4000m/s =
2000m/s * ln(Mi / Mf) </span><span>
<span>2 = ln(Mi / Mf) </span>
<span>7.389 = Mi / Mf </span>
<span>Mf = Mi / 7.389 = 0.135*Mi<span> </span></span></span>
Answer:
The answer is Muscular endurance
Explanation:
It is this because your seeing how long your muscles can with stand. I answered by guessing lol.
So the formula for density is mass divided by volume, so it would be
Answer:
The angular acceleration α = 14.7 rad/s²
Explanation:
The torque on the rod τ = Iα where I = moment of inertia of rod = mL²/12 where m =mass of rod and L = length of rod = 4.00 m. α = angular acceleration of rod
Also, τ = Wr where W = weight of rod = mg and r = center of mass of rod = L/2.
So Iα = Wr
Substituting the value of the variables, we have
mL²α/12 = mgL/2
Simplifying by dividing through by mL, we have
mL²α/12mL = mgL/2mL
Lα/12 = g/2
multiplying both sides by 12, we have
Lα/12 × 12 = g/2 × 12
αL = 6g
α = 6g/L
α = 6 × 9.8 m/s² ÷ 4.00 m
α = 58.8 m/s² ÷ 4.00 m
α = 14.7 rad/s²
So, the angular acceleration α = 14.7 rad/s²