Spinning a marshmallow over a fire is effective maybe if you hang it over the fire and heat it up equally on each side
Answer:
the yield of product is YP=46.835 % and the concentration of solids is
Cs = 27.33%
Explanation:
Assuming that all the solids and fats remains in the milk after the evaporation, then the mass of product mP will be
Mass of fat in 100 kg of milk = 100 kg* 0.037 = mP* 0.079
mP = 100 kg* 0.037/0.079 = 46.835 kg
then the yield YP of the product is
YP= mP / 100 kg = 46.835 kg / 100 kg = 46.835 %
YP= 46.835 %
the concentration of solids Cs is
Mass of solids in 100 kg of milk = 100 kg* 0.128 = 46.835 kg * Cs
Cs = 100 kg* 0.128 / 46.835 kg = 0.2733 = 27.33%
Cs = 27.33%
Use newtons second law F=ma, plug in the given values which gives us the answer of 22 kg for the mass
The buoyant force must be greater than water.
Answer:
F = GMmx/[√(a² + x²)]³
Explanation:
The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is
dF = GmdM/L²
Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.
So, the horizontal components add from two symmetrically opposite mass elements dM,
Thus, the horizontal component of the force is
dF' = dFcosФ where Ф is the angle between L and the x axis
dF' = GmdMcosФ/L²
L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.
L = √(a² + x²)
cosФ = x/L
dF' = GmdMcosФ/L²
dF' = GmdMx/L³
dF' = GmdMx/[√(a² + x²)]³
Integrating both sides we have
∫dF' = ∫GmdMx/[√(a² + x²)]³
∫dF' = Gm∫dMx/[√(a² + x²)]³ ∫dM = M
F = GmMx/[√(a² + x²)]³
F = GMmx/[√(a² + x²)]³
So, the force due to the sphere of mass m is
F = GMmx/[√(a² + x²)]³
