Answer:
q = 0.1086 micro Coulombs
Explanation:
By Coulombs law, we have;
Where;
F = The electric force = 120 mgm
q₁ and q₂ = Charge
r = The separating distance = 30 cm = 0.3 m
k = 8.9876×10⁹ kg·m³/(s²·C²)
Where, q₁ and q₂, we have;
Whereby the force is the force of 120 milligram mass, we have;
0.00012 × 9.81 = 000011772 N
Substituting the values, we have;
q = 0.1086 micro Coulombs.
Answer:
The rate of work output = -396.17 kJ/s
Explanation:
Here we have the given parameters
Initial temperature, T₁ = 355°C = 628.15 K
Initial pressure, P₁ = 350 kPa
h₁ = 763.088 kJ/kg
s₁ = 4.287 kJ/(kg·K)
Assuming an isentropic system, from tables, we look for the saturation temperature of saturated air at 4.287 kJ/(kg·K) which is approximately
h₂ = 79.572 kJ/kg
The saturation temperature at the given
T₂ = 79°C
The rate of work output =
×
×(T₂ - T₁)
Where;
= The specific heat of air at constant pressure = 0.7177 kJ/(kg·K)
= The mass flow rate = 2.0 kg/s
Substituting the values, we have;
= 2.0 × 0.7177 × (79 - 355) = -396.17 kJ/s
= -396.17 kJ/s