With its apparent magnitude
Most likely the answer would be John joined a group that studies astrological signs.
Recall the formula,
∆<em>θ</em> = <em>ω</em>₀ <em>t</em> + 1/2 <em>α</em> <em>t</em> ²
where ∆<em>θ</em> = angular displacement, <em>ω</em>₀ = initial angular speed (which is zero because the disk starts at rest), <em>α</em> = angular acceleration, and <em>t</em> = time. Solve for the acceleration with the given information:
50 rad = 1/2 <em>α</em> (5 s)²
<em>α</em> = (100 rad) / (25 s²)
<em>α</em> = 4 rad/s²
Now find the angular speed <em>ω </em>after 3 s using the formula,
<em>ω</em> = <em>ω</em>₀ + <em>α</em> <em>t</em>
<em>ω</em> = (4 rad/s²) (3 s)
<em>ω</em> = 12 rad/s
4-C. ... 5-D. ... 6-A. ... 7-D. ... 8-C. ... 9-B. ... 10-D. It's really a joy and a delight that you've learned so much by posting all of these questions.
1 Nautical metre=1852 metres
10×1852
=18520 metres
