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
Answer:
The minimum wall thickness Tmin required for the spherical tank is 65.90mm
Explanation:
Solution
Recall that,
Tmin = The minimum wall thickness =PD/2бp
where D = diameter of 8.0 m
Internal pressure = 1.62 MPa
Then
The yield strength = 295MPa/3.0 = 98.33
thus,
PD/2бp = 1.62 * 8000/ 2 *98.33
= 12960/196.66 = 65.90
Therefore the wall thickness Tmin required for the spherical tank is 65.90mm
Well it seems like this problem gives you what you need. You said the car was going 4m/s and then accelerated to 60m/s... so 4m/s would be your answer for the initial velocity
<span>The isotope of an atom containing 40 protons and 51 neutrons suddenly has 2 neutrons added to it
That is X-93 so it will be
</span><span>Zirconium-93
</span>hope it helps
Answer:
1.7 L
Explanation:
PV = nRT
If P, n, and R are constant:
V₁ / T₁ = V₂ / T₂
(2.0 L) / (293.15 K) = V / (255.15 K)
V = 1.7 L
