Answer:
a mass of water required is mw= 1273.26 gr = 1.27376 Kg
Explanation:
Assuming that the steam also gives out latent heat, the heat provided should be same for cooling the hot water than cooling the steam and condense it completely:
Q = mw * cw * ΔTw = ms * cs * ΔTw + ms * L
where m = mass , c= specific heat , ΔT=temperature change, L = latent heat of condensation
therefore
mw = ( ms * cs * ΔTw + ms * L )/ (cw * ΔTw )
replacing values
mw = [182g * 2.078 J/g°C*(118°C-100°C) + 118 g * 2260 J/g ] /[4.187 J/g°C * (90.7°C-39.4°C)] = 1273.26 gr = 1.27376 Kg
B) Oxygen combines with nitrogen in the air to form NOx at about 2500 degrees Fahrenheit.
Answer:
A degree in architecture with 60 credit hours.
Explanation:
The requirements need for a student to qualify for a two year master of architecture degree are;
- 60 credit hours in architecture
- Complete 60 credit hours in related area of profession such as; planning, landscape architecture ,public health and others.
- 45 credit hours in architecture course at the level of 500/600
Answer:
B- extreme fit, close fit, adjustable fit
Explanation:
A human-fit design typically involves the process of manufacturing or producing products (tools) that are easy to use by the end users. Therefore, human-fit designs mainly deals with creating ideas that makes the use of a particular product comfortable and convenient for the end users.
The design for human-fit strategies include; extreme fit, close fit and adjustable fit.
Hence, when the aforementioned strategies are properly integrated into a design process, it helps to ensure the ease of use of products and guarantees comfort for the end users.
Answer:
a.) -147V
b.) -120V
c.) 51V
Explanation:
a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).
b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.
c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.
Honestly, these things take practice to get used to. It's really hard to explain this.
