Ask question

Ask question

All categories

2 years ago

10

What is the voltage across the load, RL, in

1 answer:

JulsSmile [24]2 years ago

5 0

Answer:

(c) 5.71 V

Explanation:

The circuit can be redrawn to a Thevenin equivalent that is 6V through a 5-ohm resistor into a 100-ohm load. Then the voltage at the load is ...

(6 V)(100/(100 +5) ≈ 5.71 V

You might be interested in

A colorful design for a product meant to appeal to a preschool audience reflects which of the following points in industrial des

lord [1]

Appeal to emotion

the product doesn’t need to be colorful to function but it would be more appealing to the intended audience if it was

6 0

3 years ago

Ann’s Retail, a women’s clothing store, hires female attendants to assist clients in the store’s dressing rooms. Larry, a male,

mojhsa [17]

Answer:

A bona fide occupational qualification defense

Explanation:

Since the store is for women clothing, the retail may prefer to employ only female to assist the customers. Under a bona fide occupational qualification defense, an employer is allowed to discriminate if a characteristic is a necessity for the performance of the job and for the business. Therefore, the store has a bona fide occupational qualification defense.

3 0

3 years ago

public interface Frac { /** @return the denominator of this fraction */ int getDenom(); /** @return the numerator of this fracti

True [87]

Answer:

The Full details of the answer is attached.

7 0

3 years ago

134a refrigerant enters an adiabatic compressor at 140kPa and -10C, the refrigerant is compressed at 0.5kW up to 700kPa and 60C.

vichka [17]

Answer:

(a) 65.04%

(b) 16.91%

Solution:

As per the question:

At inlet:

Pressure of the compressor, P = 140 kPa

Temperature, T = $- 10^{\circ}C$ = 263 K

Isentropic work, W = 700 kPa

At outlet:

Pressure, P' = 700 kPa

Temperature, T' = $60^{\circ}C$ = 333 K

Now, from the steam table;

At the inlet , at a P = 700 kPa, T = $60^{\circ}C$ :

h = 243.40 kJ/kg, s = 0.9606 kJ/kg.K

At outlet, at P = 140 kPa, T = $- 10^{\circ}C$ :

h' = 296.69 kJ/kg, s' = 1.0182 kJ/kg.K

Also in isentropic process, s = $s'_{s}$ and $h'_{s} = 278.06 kJ/kg.K$ at 700kPa

(a) Isentropic efficiency of the compressor, $\eta_{s} = \frac{Work\ done\ in\ isentropic\ process}{Actual\ work\ done}$

$\eta_{s} = \frac{h'_{s} - h}{h' - h} = frac{278.06 - 243.40}{296.69 - 243.40} = 0.6504 = 65.04%$

(b) The temperature of the environment, $T_{e} = 27^{\circ}C$ = 273 + 27 = 300 K

Availability at state 1, $\Psi = h - T_{e}s = 243.40 - 300\times 0.9606 = - 44.78 kJ/kg$

Similarly for state 2, $\Psi' = h' - T_{e}s' = 296.69 - 300\times 1.0182 = - 8.77 kJ/kg$

Now, the efficiency of the compressor as per the second law;

$\eta' = \frac{\Psi' - \Psi}{h' - h} = \frac{- 8.77 - (- 44.78)}{296.69 - 243.40} = 0.6757$ = 67.57%

4 0

4 years ago

Savabuck University has installed standard pressure-operated flush valves on their water closets. When flushing, these valves de

Dvinal [7]

Answer:

Cost = $2527.2 per month.

Explanation:

Given that

Discharge ,Q = 130 L/min

So

$Q=0.13\ m^3/min$

Cost = $0.45 per cubic meter

1 month = 30 days

1 days = 24 hr = 24 x 60 min

1 month = 30 x 24 x 60 min

1 month = 43,200 min

Lets x $m^3\ water\ waste\ in\ a\ month$

x = 0.13 x 43,200

$x=5616\ m^3$

So the total cost = 5616 x 045 $

Cost = $2527.2 per month.

7 0

4 years ago