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:
Explanation:
Ignoring friction, the acceleration will double
F = ma
2F = m(2a)
Its prominent ring system which is composed of primarily ice particles with smaller amounts of rocky detbris. Hope this helped!
Answer:
a) S = 1.69 10⁹ W/m², b) P = 5.63 Pa
, c) F = 20.6 10⁻¹² N
Explanation:
a) The intensity defined as the energy per unit area
S = U / A
Area of a circle is
W = 6.2 mw = 6.2 10-3 W
R = 1080 nm = 1080 10⁻⁹ m = 1.080 10⁻⁶ m
A = π R2
A = π (1,080 10⁻⁶)²
A = 3.66 10 -12 m²
S = 6.2 10-3 / 3.66 10-12
S = 1.69 10⁹ W / m²
b) The radiation pressure
P = 1 / c (dU / dt) / A
S = (dU / dt) / A
P = S / c
P = 1.69 10 9 / 3. 108
P = 5.63 Pa
c) the definition of pressure is force over area
P = F / A
F = P A
F = 5.63 3.66 10⁻¹²
F = 20.6 10⁻¹² N
d) for this we use Newton's second law
F = ma
a = F / m
The answer should be 741 N.
W=m*g
9.5*78 is 741