Yes, a small child can indeed play with a heavier child on a seesaw. This is because reducing the distance from the fulcrum for the heavier child increases the effort required, allowing the smaller child to play without much strain.
<span>The primary factor behind these currents is the </span>variability in the density of ocean waters.
Answer:
v = 54.2 m/s
Explanation:
We can utilize conservation of energy to solve this issue.
Initial condition Higher
Em₀ = U = m g h
Final condition. Lower
= K = ½ m v²
Em₀ = Em_{f}
m g h = ½ m v²
v² = 2gh
v = √ (2gh)
Now let's perform the calculation
v = √ (2 * 9.8 * 150)
v = 54.2 m/s
Let L be the length of the inclined plane. The work done by gravity on the block is calculated as the product of force and distance traveled, which amounts to mg sinθ x L, where m stands for the mass of the block and θ denotes the angle of inclination. This translates into the potential energy of the compressed spring represented by 1/2 k x² = mgL sin31, with k as the spring constant. Compression x measures 0.37. Solving this gives: 0.5 x 3400 x 0.37² = 33 x 9.8 x sin31 L yields L = 1.4 m, indicating the incline measures 1.4 meters.
a.) 10 Hz b.) 0.1 s c.) 187.4 m/s d.) -412.6 m/s²
