Answer:
Explanation:
350 N force stretches the spring by 30 cm
spring constant K = 350 / 0.30 = (350 / 0.3) N / m
To calculate work done by a spring force we proceed as follows
spring force when the spring is stretched by x = Kx
This force is variable so work done by it can be calculated by integration
Work done by it in stretching from x₁ to x₂
W = ∫ F dx
= ∫ Kx dx with limit from x₁ to x ₂
= 1/2 K ( x₂² - x₁² )
Putting the given values of x₁ = 0.50 m , x₂ = 0.8 m
Work done
= 1/2 x (350 / 0.3)x ( 0.80² - 0.50² )
= 227.50 J
Pnet = Po + dgh
<span>Density of saltwater = 1030 kg/m^3. </span>
<span>Disregard the thickness. Assuming it's a circular window, then the area is pi(r^2). </span>
<span>d = 20 cm = 0.2 m </span>
<span>r = d/2 = 0.1 m </span>
<span>A = pi(r^2) </span>
<span>A = 3.14159265(.1^2) </span>
<span>A = 0.0314159265 m^2 </span>
<span>p = F/A </span>
<span>p = (1.1 x 10^6) / (0.0314159265) </span>
<span>p = 35,014,087.5 Pa </span>
<span>1 atm = 101,325 Pa </span>
<span>P = Po + dgh </span>
<span>h = (P - Po) / dg </span>
<span>h = (35,014,087.5 - 101,325) / (1030 x 9.81) </span>
<span>h = 3 455.23812 m </span>
<span>h = 3.5 km</span>
The part that moves are called anti-nodes. The stationary pars are nodes
I remember c/d. That's not a problem. But if you want 'c', you'll have to give me 'd'.
