Answer:
293 kg
Explanation:
Let's say the tension in each cable is Tb, Tc, and Td.
First, find the length of cable AD:
r = √(2² + 2² + 1²)
r = 3
Using similar triangles:
Tdx = 2/3 Td
Tdy = 2/3 Td
Tdz = 1/3 Td
Sum of the forces in the x direction:
∑F = ma
Tb − 2/3 Td = 0
Td = 3/2 Tb
Sum of the forces in the y direction:
∑F = ma
2/3 Td − Tc = 0
Td = 3/2 Tc
Sum of the forces in the z direction:
∑F = ma
1/3 Td − mg = 0
Td = 3mg
From the first two equations, we know Td is greater than Tb or Tc. So we need to set Td to 8.6 kN, or 8600 N.
8600 N = 3mg
m = 8600 N / (3 × 9.8 m/s²)
m ≈ 292.5 kg
Rounded to three significant figures, the maximum mass of the crate is 293 kg.
Answer:
Cyclical
Explanation:
I looked at the next question on edgenuity and it said it in the question.
Answer:
The volume flow rate necessary to keep the temperature of the ethanol in the pipe below its flashpoint should be greater than 1.574m^3/s
Explanation:
Q = MCp(T2 - T1)
Q (quantity of heat) = Power (P) × time (t)
Density (D) = Mass (M)/Volume (V)
M = DV
Therefore, Pt = DVCp(T2 - T1)
V/t (volume flow rate) = P/DCp(T2 - T1)
P = 20kW = 20×1000W = 20,000W, D(rho) = 789kg/m^3, Cp = 2.44J/kgK, T2 = 16.6°C = 16.6+273K = 289.6K, T1 = 10°C = 10+273K = 283K
Volume flow rate = 20,000/789×2.44(289.6-283) = 20,000/789×2.44×6.6 = 1.574m^3/s (this is the volume flow rate at the flashpoint temperature)
The volume flow rate necessary to keep the ethanol below its flashpoint temperature should be greater than 1.574m^3/s