Investigators are most likely to use the case history method when they study <span>a rare behavior or an unusual person.
They do this to obtain some sort of basis that they could use as a pointer to make their decision regarding the similar case (after figuring out the difference in situation between each period)</span>
If you want to change the thermos into an open energy system, you have to remove the lid. Once the lid is removed, the energy is no longer contained inside the thermos bottle. From the bottle, the energy dissipates to the environment.
El beneficio que obtuvo es de 2 L
Explanation:
12) q = mCΔT
125,600 J = (500 g) (4.184 J/g/K) (T − 22°C)
T = 82.0°C
13) Solving for ΔT:
ΔT = q / (mC)
a) ΔT = 1 kJ / (0.4 kg × 0.45 kJ/kg/K) = 5.56°C
b) ΔT = 2 kJ / (0.4 kg × 0.45 kJ/kg/K) = 11.1°C
c) ΔT = 2 kJ / (0.8 kg × 0.45 kJ/kg/K) = 5.56°C
d) ΔT = 1 kJ / (0.4 kg × 0.90 kJ/kg/K) = 2.78°C
e) ΔT = 2 kJ / (0.4 kg × 0.90 kJ/kg/K) = 5.56°C
f) ΔT = 2 kJ / (0.8 kg × 0.90 kJ/kg/K) = 2.78°C
14) q = mCΔT
q = (2000 mL × 1 g/mL) (4.184 J/g/K) (80°C − 20°C)
q = 502,000 J
20) q = mCΔT
q = (2000 g) (4.184 J/g/K) (100°C − 15°C) + (400 g) (0.9 J/g/K) (100°C − 15°C)
q = 742,000 J
24) q = mCΔT
q = (0.10 g) (0.14 J/g/K) (8.5°C − 15°C)
q = -0.091 J
<span>net work = change in kinetic energy
for Block B, we just have the force from block A acting on it
F(ab)d= .5(1)vf² - .5(1)(2²)
F(ab)d= .5vf² - 2
Block A, we have the force from the hand going in one direction and the force of block B on A going the opposite direction
10-F(ba)d = .5(4)vf² - .5(4)(2²)
10-F(ba)d = 2vf² - 8
F(ba)d = 18 - 2vf²
now we have two equations:
F(ba)d = 18 - 2vf²
F(ab)d= .5vf² - 2
since the magnitude of F(ba) and F(ab) is the same, substitute and find vf (I already took into account the direction when solving for F(ab)
10-.5vf² + 2 = 2vf² - 8
12 - .5vf² = 2vf² - 8
20 = 2.5vf²
vf² = 8
they both will have the same velocity
KE of block A= .5(4)(2.828²) = 16 J
KE of block B=.5(1)(2.828²) = 4 J</span>
