Answer:
Temperature of water leaving the radiator = 160°F
Explanation:
Heat released = (ṁcΔT)
Heat released = 20000 btu/hr = 5861.42 W
ṁ = mass flowrate = density × volumetric flow rate
Volumetric flowrate = 2 gallons/min = 0.000126 m³/s; density of water = 1000 kg/m³
ṁ = 1000 × 0.000126 = 0.126 kg/s
c = specific heat capacity for water = 4200 J/kg.K
H = ṁcΔT = 5861.42
ΔT = 5861.42/(0.126 × 4200) = 11.08 K = 11.08°C
And in change in temperature terms,
10°C= 18°F
11.08°C = 11.08 × 18/10 = 20°F
ΔT = T₁ - T₂
20 = 180 - T₂
T₂ = 160°F
Answer:
2.2 s
Explanation:
Hi!
Let's consider the origin of the coordinate system at the ground, and consider that the clam starts with zero velocity, the equation of motion of the clam is given by

We are looking for a time t for which x(t) = 0

Solving for t:

Rounding at the first decimal:
t = 2.2 s
Kinetic energy than parked
Answer:
0.035 N
Explanation:
Parameters given:
Charge q1 = -3.31x10^(-7) C
Charge q2 = -5.7x10^(-7) C.
Distance between them, R = 22 cm = 0.22 m
Electrostatic force between to particles is given as:
F = (k* q1 * q2) / R²
F = (9 * 10^9 * -3.31 * 10^(-7) * -5.7 * 10^(-7)) / 0.22²
F = 0.035 N
The answer is 3
I hope I helped