Answer:
Para x=0:
Para x=30 cm:
Explanation
Podemos utilizar la ley de Fourier par determinar el flujo de calor:
(1)
Por lo tanto debemos encontrar la derivada de T(x) con respecto a x primero.
Usando la ley de potencia para la derivda, tenemos:
Remplezando esta derivada en (1):
Para x=0:
Para x=30 cm:
Espero que te haya ayudado!
Answer:
a) 1.16 m/s
b) 1/216000
c) (√15)/6480000
Explanation:
The parameters given are;
Length of boat prototype, lp = 30 m
Speed of boat prototype = 9 m/s
Length of boat model, lm= 0.5 m
a) lm/lp = 0.5/30 = 1/60 = ∝
(vm/vp) = ∝^(1/2) = √∝ = (1/60)^(1/2)
vm = 9 × (1/60)^(1/2) = 1.16 m/s
b) The ratio of the model to prototype drag, Fm/Fp, is given as follows;
Fm/Fp = (vm/vp)²×(lm/lp)² = ∝³
Fm/Fp = (1/60)³ = 1/216000
c) The ratio of the model to prototype power pm/p = (Fm/Fp) × (vm/vp) = ∝³×√∝
The ratio of the model to prototype power pm/p = √(1/60) × (1/60)³
pm/p = √(1/60) × (1/60)³ = (√15)/6480000
Answer:
Weight of cement = 10968 lb
Weight of sand = 18105.9 lb
Weight of gravel = 28203.55 lb
Weight of water = 5484 lb
Explanation:
Given:
Entrained air = 7.5%
Length, L = 40 ft
Width,w = 12 ft
thickness,b= 6 inch, convert to ft = 6/12 = 0.5 ft
Specific gravity of sand = 2.60
Specific gravity of gravel = 2.70
The volume will be:
40 * 12 * 0.5 = 240 ft³
We need to find the dry volume of concrete.
Dry volume = wet volume * 1.54 (concrete)
Dry volume will be = 240 * 1.54 = 360ft³
Due to the 7% entarained air content, the required volume will be:
V = 360 * (1 - 0.07)
V = 334.8 ft³
At a ratio of 1:2:3 for cement, sand, and gravel respectively, we have:
Total of ratio = 1+2+3 = 6
Their respective volume will be =
Volume of cement =
Volume of sand =
Volume of gravel =
To find the pounds needed the driveway, we have:
Weight = volume *specific gravity * density of water
Specific gravity of cement = 3.15
Weight of cement =
55.8 * 3.15 * 62.4 = 10968 pounds
Weight of sand =
111.6 * 2.60 * 62.4 = 18105.9 lb
Weight of gravel =
167.4 * 2.7 * 62.4 = 28203.55 lb
Given water to cement ratio of 0.50
Weight of water = 0.5 of weight of cement
= 1/2 * 10968 = 5484 lb