The correct answer is the third, It reflects the green light waves and absorbs most of the rest.
Explanation:
(a) Draw a free body diagram of the cylinder at the top of the loop. At the minimum speed, the normal force is 0, so the only force is weight pulling down.
Sum of forces in the centripetal direction:
∑F = ma
mg = mv²/RL
v = √(g RL)
(b) Energy is conserved.
EE = KE + RE + PE
½ kd² = ½ mv² + ½ Iω² + mgh
kd² = mv² + Iω² + 2mgh
kd² = mv² + (m RC²) ω² + 2mg (2 RL)
kd² = mv² + m RC²ω² + 4mg RL
kd² = mv² + mv² + 4mg RL
kd² = 2mv² + 4mg RL
kd² = 2m (v² + 2g RL)
d² = 2m (v² + 2g RL) / k
d = √[2m (v² + 2g RL) / k]
Answer:
For areas marked X, Y, Z, X is attractive only, Y has a very small range, and Z is attractive and repulsive
Explanation:
Solution
Given that:
From the question stated, Anna drew a diagram to compare forces that are strong and weak.
Now,
We are to find which labels are grouped in areas marked as X, Y, Z respectively.
Thus,
For X, Y, Z it is marked as:
X: Always attractive or attractive only
Y: Very small range
Z: Repulsive and attractive
<span>65W * 8h * 3600s/h = 1.9e6 J = 447 Cal </span>
