Answer:
Cp = 4756 [J/kg*°C]
Explanation:
In order to calculate the specific heat of water, we must use the equation of energy for heat or heat transfer equation.
Q = m*Cp*(T_f - T_i)/t
where:
Q = heat transfer = 2.6 [kW] = 2600[W]
m = mass of the water = 0.8 [kg]
Cp = specific heat of water [J/kg*°C]
T_f = final temperature of the water = 100 [°C]
T_i = initial temperature of the water = 18 [°C]
t = time = 120 [s]
Now clearing the Cp, we have:
Cp = Q*t/(m*(T_f - T_i))
Now replacing
Cp = (2600*120)/(0.8*(100-18))
Cp = 4756 [J/kg*°C]
Answer:
16∠45° Ω
Explanation:
Applying,
Z = V/I................... Equation 1
Where Z = Impedance, V = Voltage output, I = current input.
Given: V = 120cos(10t+75°), = 120∠75°, I = 7.5cos(10t+30) = 7.5∠30°
Substitute these values into equation 1
Z = 120cos(10t+75°)/7.5cos(10t+30)
Z = 120∠75°/ 7.5∠30°
Z = 16∠(75°-30)
Z = 16∠45° Ω
Hence the impedance of the linear network is 16∠45° Ω
Answer:
the magnitude of Vpg = 493.711 km/h
Explanation:
given data
speed Vpg = 560 km/h
speed Vwg = 80 km/h
solution
we get here magnitude of the plane velocity w.r.t. ground is
we know that the Vpg = Vpw + Vwg .....................1
writing the component of the velocity that is
Vpw = (0 km/h î - 560 km/h j )
Vwg = (80 cos 45 km/h î + 80 sin 45 km/h j)
adding these
Vpg = (0+80 cos 45 km/h ) î + ( -560 + 80 sin 45 km/h j)i
Vpg = (42.025 ) î (-491.92 km/h)j
now we take magnitude
the magnitude of Vpg = 
the magnitude of Vpg = 493.711 km/h
