Solution:
Mass of liquid water and water vapor in the insulated tank initially = 1.4 kg
Temperature = 200 °C
And 25% of the volume by liquid water is steam.
State 1



(taking the value of
and
at 200°C )

Now quality of vapor


Internal energy at state 1 can be found out by


= 856.54 kJ/kg
After heating with the resistor for 20 minutes, at state 2, the tank contains saturated water vapor 
Tank is rigid, so volume of tank is constant.



Now interpolate the value to get temperature at state 2 with specific volume value to get final temperature

= 369.11° C
Internal energy at state 2

Now power rating of the resistor


= 1.51 kW
A formula unit is the empirical formula for an ionic compound. It is the lowest possible ratio of the cations and anions of a given ionic compound.
Answer:
a) v₃ = 19.54 km, b) 70.2º north-west
Explanation:
This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition
vector 1 moves 26 km northeast
let's use trigonometry to find its components
cos 45 = x₁ / V₁
sin 45 = y₁ / V₁
x₁ = v₁ cos 45
y₁ = v₁ sin 45
x₁ = 26 cos 45
y₁ = 26 sin 45
x₁ = 18.38 km
y₁ = 18.38 km
Vector 2 moves 45 km north
y₂ = 45 km
Unknown 3 vector
x3 =?
y3 =?
Vector Resulting 70 km north of the starting point
R_y = 70 km
we make the sum on each axis
X axis
Rₓ = x₁ + x₃
x₃ = Rₓ -x₁
x₃ = 0 - 18.38
x₃ = -18.38 km
Y Axis
R_y = y₁ + y₂ + y₃
y₃ = R_y - y₁ -y₂
y₃ = 70 -18.38 - 45
y₃ = 6.62 km
the vector of the third leg of the journey is
v₃ = (-18.38 i ^ +6.62 j^ ) km
let's use the Pythagorean theorem to find the length
v₃ = √ (18.38² + 6.62²)
v₃ = 19.54 km
to find the angle let's use trigonometry
tan θ = y₃ / x₃
θ = tan⁻¹ (y₃ / x₃)
θ = tan⁻¹ (6.62 / (- 18.38))
θ = -19.8º
with respect to the x axis, if we measure this angle from the positive side of the x axis it is
θ’= 180 -19.8
θ’= 160.19º
I mean the address is
θ’’ = 90-19.8
θ = 70.2º
70.2º north-west
Answer:
0.33 Amp
8 Watts
Explanation:
The current through the 12 Ω resistor will be given by Ohm's law:
4 V / 12 Ω = 1/3 Amp ≈ 0.33 Amp
The power dissipated on the 2 Ω resistor can be calculated via the formula:
which gives:
4^2 / 2 = 8 Watts