In the United States, fire codes are developed primarily by two model code organizations, the International Code Council (ICC) and the National Fire Protection Association (NFPA).
Answer:Vb=-6i-(-0.1ωab+8)j m/s
Explanation:
Va=V0+Va0
Va=V0+(ra0 x ωao)
ω=Angular velocity of link A0
Using r0a=0.1m;
Va=V0+(0.1i x ω0a K)
Va=0
ixk=j
Va=0+0.1ω0aj
Calculating te velocity of using te equation below
Vb=Va+Vba
Vb=Va+ωab x rba
ωab=40rad/s
rab=-0.21i+0.15j
Va=0.1ω0aj
Vb=Va+ωabxrba
Vb=0.1ω0aj+40k x -(0.21i+0.15j)
Vb=0.1ω0aj-8j-6i
Vb=-6i-(-0.1ωab+8)j m/s
Answer:
a = 40
b = 29
Explanation:
Give a place holder for the numbers that we don't know.
Lets call the two numbers a and b.
From the given info, we can write an expression and solve it:
"one number is 11 more than another number"
a = 11 + b
from this, we know that a > b.
''three times the larger number exceeds four times the smaller number by 4"
3a = 4b + 4
Now we have 2 equations, we can use them to solve using whatever method you want.
a = 11 + b
3a = 4b + 4
I will be using matrices RREF to solve for this.
a - b = 11
3a - 4b = 4


a = 40
b = 29
Answer:
0.304 L of Freon is needed
Explanation:
Q = mCT
Q is quantity of energy that must be removed = 47 BTU = 47×1055.06 = 49587.82 J
C is specific heat of Freon = 74 J/mol.K = 74 J/mol.K × 1 mol/120 g = 0.617 J/g.K
T is temperature in the area of Mars = 189 K
m = Q/CT = 49587.82/(0.617×189) = 452.23 g = 452.24/1000 = 0.45223 kg
Density of Freon = specific gravity of Freon × density of water = 1.49 × 1000 kg/m^3 = 1490 kg/m^3
Volume of Freon = mass/density = 0.45223/1490 = 0.000304 m^3 = 0.000304×1000 = 0.304 L
Answer:
Power required to overcome aerodynamic drag is 50.971 KW
Explanation:
For explanation see the picture attached