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%
Answer:
b) Rs.901 is the selling price
Explanation:
selling price = cost price+profit percentage/100 x cost price
cost price = 850
profit percentage = 6%
selling price = 850+6/100 x 850 = 901
soln 2
if 100 % = 850
therefore what about 106% = ?
106 x 805/100 = 901
ans = 901
if you need any clarification or more explanation pls do mention on the comment section.i would like to help more
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Answer:
Vi = 8.28 m/s
Explanation:
This problem is related to the projectile motion.
As we know there are two components of motion associated with this, the horizontal component and vertical component.
The horizontal distance covered by the ball is
Vx*t = x
Vx*t = 5.3
Vx = 5.3/t eq. 1
Also we know that
Vx = Vicos(60)
Vx = Vi*0.5 eq. 2
equate eq. 1 and eq. 2
5.3/t = Vi*0.5
5.3/0.5 = Vi*t
Vi*t = 10.6 eq. 3
The vertical distance is
Vy = y1 + Vyi*t - 0.5gt²
also we know that
Vyi = Visin(60)
Vyi = Vi*0.866
It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance
3 = 1.9 + Vi*0.866*t - 0.5gt²
3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
1.1 = 0.866(Vi*t) - 4.9t²
0.866(Vi*t) = 4.9t² + 1.1
substitute Vi*t = 10.6 in above equation
0.866(10.6) = 4.9t² + 1.1
9.18 = 4.9t² + 1.1
4.9t² = 8.08
t² = 8.08/4.9
t² = 1.648
t = 1.28 sec
Finally, initial speed can be found by substituting the value of t into eq. 3
Vi*t = 10.6
Vi = 10.6/t
Vi = 10.6/1.28
Vi = 8.28 m/s