Answer:
x = A sin ω t describes the displacement of the particle
v = A ω cos ω t
a = -A ω^2 sin ω t
a (max) = -A ω^2 is the max acceleration (- can be ignored here)
ω = (K/ m)^1/2 for SHM
F = - K x^2 restoring force of spring
K = 4.34 / .0745^2 = 782 N / m
ω = (782 / .297)^1/2 = 51.3 / sec
a (max) = .0745 * 782 / .297 = 196 m / s^2
Answer:
Time, I believe. Pretty sure it's time lol
Answer:
correct option is b. 31.3 m/s
Explanation:
given data
artificial gravity a1 = 1 g
artificial gravity a2 = 2 g
diameter = 100 m
radius r= 50 m
speed v1 = 22.1 m/s
solution
As acceleration is ∝ v²
so we can say
.....................1
put here value
solve it
v2 = 
× 22.1
v2 = 31.25 m/s
so correct option is b. 31.3 m/s
Assuming Earth's gravity, the formula for the flight of the particle is:
<span>s(t) = -16t^2 + vt + s = -16t^2 + 144t + 160. </span>
<span>This has a maximum when t = -b/(2a) = -144/[2(-16)] = -144/(-32) = 9/2. </span>
<span>Therefore, the maximum height is s(9/2) = -16(9/2)^2 + 144(9/2) + 160 = 484 feet. </span>
