On problem 2.5
i also need a short (or detailed) explanation. you

Explanation:

Problem 2.5, mixing dough.

Please note dimensional units and standard formulas for energies.

Input:

Electrical energy = VIT = 110 V * 1.5 A * 5 min. * 60 s/min = 49500 joules = 49.5 kJ

Output:

Warming of dough = mCΔT = 1 kg  * 4.2 kJ/(kg*K) * 5 deg.K = 21 kJ

Dissipation of energy = remainder of input energy = (49.5-21) kJ = 28.5 kJ

Fraction of energy converted to energy of dough = 21 kJ / 49.5 kJ

= forty-two percent (to the nearest percent)

The last item, dissipation of energy could be attributed to:

1.Loss in efficiency of mixer in the form of mechanical friction, heating up due to resistance of motor circuit, etc.

2.Loss in mechanical mixing of dough due to friction between mixer paddle and dough in the form of heat, part of which stays with dough (thus heating up of dough), and the remainder heats up the dough container, loss of heat to environment, air, heat of vaporization of water content of dough, etc.

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14