Answer:
Step-by-step explanation:
Answer:
Liquid R has a mass of of 1 kg at a temperature of 30°c kept in a refrigerator to freeze . Given the specific heat capacity is 300 J kg-¹ °c-1 and the freezing point is 4°c . Calculate the heat release by liquid R.
Step-by-step explanation:
Liquid R has a mass of of 1 kg at a temperature of 30°c kept in a refrigerator to freeze . Given the specific heat capacity is 300 J kg-¹ °c-1 and the freezing point is 4°c . Calculate the heat release by liquid R.
∆volume = 4/3 *pi * (R^3 -r^3 )
= 4/3 *pi*((6)^3 -(3)^3)
= 252*pi
= 791.68 m^3
Let's say that in the beginning he weighted x and at the end he weighted x-y, y being the number of kg he wanted to loose.
first month he lost
y/3
then he lost:
(y-y/3)/3
this is
(2/3y)/3=2/9y
explanation: ((y-y/3) is what he still needed to loose: y minus what he lost already
and then he lost
(y-2/9y-1/3y)/3+3 (the +3 is his additional 3 pounts)
(y-2/9y-1/3y)/3-3=(7/9y-3/9y)/3+3=4/27y+3
it's not just y/3 because each month he lost one third of what the needed to loose at the current time, not in totatl
and the weight at the end of the 3 months was still x-y+3, 3 pounds over his goal weight!
so: x -y/3-2/9y-4/27y-3=x-y+3
we can subtract x from both sides:
-y/3-2/9y-4/27y-3=-y+3
add everything up:
-19/27y=-y+6
which means
-19/27y=-y+6
y-6=19/27y
8/27y=6
4/27y=3
y=20.25
so... that's how much he wanted to loose, but he lost 3 less than that, so 23.25
ps. i hope I didn't make a mistake in counting, let me know if i did. In any case you know HOW to solve it now, try to do the calculations yourself to see if they're correct!
Write i in trigonometric form. Since |i| = 1 and arg(i) = π/2, we have
i = exp(i π/2) = cos(π/2) + i sin(π/2)
By DeMoivre's theorem,
i² = exp(i π/2)² = exp(i π) = cos(π) + i sin(π)
and it follows that i² = -1 since cos(π) = -1 and sin(π) = 0.