At its current short-run level of production, a firm's average variable costs equal $20 per unit, and its average fixed costs equal $30 per unit. Its total costs at this production level equal $2,500. A. What is the firm's current output level? B. What are its total variable costs at this output level? C. What are its total fixed costs?

Answer:

b) 70°C

Explanation:

Given that super heat temperature at 101.325  KPa=  170°C.

We know that saturation temperature at 101.325  KPa is 100°C.

So

Degree of super heat =Super heat temperature - saturation temperature (  at constant pressure)

Now by putting the values

Degree of super heat=170-100

Degree of super heat=70°C.

So our option b is right.  

Answer:

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2

Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)

T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)

T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)

T1 = 1/3 * SGw * sin(30) * L * h^3

To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)

1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

In an equilateral triangle h = b * cos(30)

1/4 * b^3 * cos(30) * L * SG  > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

SG > 1/2 * SGw

For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

This is avergae specific gravity, including holes.

Answer:

(a) 1/L∫Vdt; integral t [0,1]

(b) 1/L∫Vdt; integral t [ 1, infinity]

Explanation:

An Inductor current I, flowing through an inductor depends on the voltage, V, across the inductor and the inductance, L, of the inductor. The switch 1, 2 timing varies the voltage V with time t

The expression for inductor current is given as:

I= 1/L∫Vdt,

where I is equal to the current flowing through the inductor, L is equal to the inductance of the inductor, and V is equal to the voltage across the inductor.

The formula can also be written as:

I= I0 + 1/L∫Vdt, where I is inductor current at time t, and io is inductor current at t = 0. Time can be varied by controlling the switch

Answer Explanation : The general principles for design for assembly (DFA) are,

• MINIMIZE NUMBER OF COMPONENT
• USE STANDARD COMMERCIALLY AVAILABLE COMPONENTS
• USE COMMON PARTS ACROSS PRODUCT LINES
• DESIGN FOR EASE OF PART FABRICATION
• DESIGN PARTS WITH TOLERANCE THAT ARE WITHIN PROCESS CAPABILITY
• MINIMIZE USE OF FLEXIBLE COMPONENT
• DESIGN FOR EASE OF ASSEMBLY
• USE MODULAR DESIGN
• REDUCE ADJUSTMENT REQUIRED

STEPS TO MINIMIZE THE NUMBER OF PARTS

• USE OF INCORPORATE HINGS
• USE OF INTEGRAL SPRINGS
• USE OF SNAP FITS
• USE OF GUIDES BEARINGS
• USE OF COVERS

Answer:

D) AND gate.

Explanation:

Given that:

A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print

These conditions are:

1. The printer's electronic circuits must be energized.

2. Paper must be loaded and ready to advance.

3. The printer must be "on line" with the microprocessor.

Now; if these conditions are met  the logic gate produces a HIGH output indicating readiness to print.

The objective here is to determine the basic logic gate used in this circuit.

Now;

For NOR gate;

NOR gate gives HIGH only when all the inputs are low. but the question states it that "a HIGH is generated and applied to a 3-input logic gate". This already falsify NOR gate to be the right answer.

For NOT gate.

NOT gate operates with only one input and one output device but here; we are dealing with 3-input logic gate.

Similarly, OR gate gives output as a high if any one of the input signals is high but we need "a HIGH that is generated and applied to a 3-input logic gate".

Finally, AND gate output is HIGH only when all the input signal is HIGH and vice versa, i.e AND gate output is LOW only when all the input signal is LOW. So AND gate satisfies the given criteria that; all the three conditions must be true for the final signal to be HIGH.

Answer:

Minimum standard diameter for the PVC pipe = 14.26 inches

Minimum standard diameter for the steel pipe = 14.70 inches

Explanation:

Head loss = 6/1000...................................................(1)

Head loss = hf/l

............................(2)

Q = 2500 gal/min

a) Minimum standard diameter for PVC

C for PVC = 130

Equating (1) and (2) and putting C = 130

b) Minimum standard diameter for steel

C for steel = 120

Equating (1) and (2) and putting C = 120