I need parts 2 and 3 answered please. Thanks! 1) During a rescue operation, a st
ID: 1296697 • Letter: I
Question
I need parts 2 and 3 answered please. Thanks!
1) During a rescue operation, a stranded submarine (m = 20.0x 10^6 kg) is being supported by a 12.0 cm diameter cable 120 m below the surface of the water as shown. If the volume of the submarine is 11, 310 m^3, determine the tension in the cable. Assume the density of the water is 1000 kg/m^3. Draw a FBD of the submarine! 2) If the steel cable is 140.00000 m long when loaded as shown (note the number of significant figures here), determine the length of this same piece of steel cable when it is not loaded. Record your answer to five significant figures just this once. E steel = 19.2x 10^10 N/m^2. Please do not write on this sheet. 3) Now that you know the load on the cable, and how much it stretched under that load, calculate the spring constant (k) of the cable. Use Hooke's law to accomplish this.Explanation / Answer
1)
T = Weight - Buoynt force
= weight of submerin - weight of displaced water
= m*g - rho_water*V*g
= 20*10^6*9.8 - 1000*11310*9.8
= 85162000 N or 8.5162*10^7 N
2)
Let Lo is the actual length.
L is the length of the string when it is loaded.
Y = (F/A)/(delta L/Lo)
delta L = (F*Lo/A*Y)
L - Lo = (F*Lo/A*Y)
L = Lo*(1 + F/A*Y))
140 = Lo(1 + 8.5162*10^7/(( pi*0.12^2/4)*19.2*10^10)))
= Lo*(1 + 0.03924)
Lo = 140/(1+0.03924)
= 134.71383 m
3) F = K*x
K = F/x
= 8.5162*10^7/(140-134.71383)
= 1.61*10^7 N/m