All categories

2 years ago

Can I get an answer to this question please

Engineering

1 answer:

crimeas [40]2 years ago

7 0

Answer:

(i) 12 V in series with 18 Ω.

(ii) 0.4 A; 1.92 W

(iii) 1,152 J

(iv) 18Ω — maximum power transfer theorem

Explanation:

As seen by the load, the equivalent source impedance is ...

10 Ω + (24 Ω || 12 Ω) = (10 +(24·12)/(24+12)) Ω = 18 Ω

The open-circuit voltage seen by the load is ...

(36 V)(12/(24 +12)) = 12 V

The Thevenin's equivalent source seen by the load is 12 V in series with 18 Ω.

The load current is ...

(12 V)/(18 Ω +12 Ω) = 12/30 A = 0.4 A . . . . load current

The load power is ...

P = I^2·R = (0.4 A)^2·(12 Ω) = 1.92 W . . . . load power

10 minutes is 600 seconds. At the rate of 1.92 J/s, the electrical energy delivered is ...

(600 s)(1.92 J/s) = 1,152 J

The load resistance that will draw maximum power is equal to the source resistance: 18 Ω. This is the conclusion of the Maximum Power Transfer theorem.

The power transferred to 18 Ω is ...

((12 V)/(18 Ω +18 Ω))^2·(18 Ω) = 144/72 W = 2 W

You might be interested in

The difference in quantity between the add and full marks on an engine oil dipstick is typically

svetoff [14.1K]

Answer:

1 quart (0.9 liters).

Explanation:

A proper inspection of various systems and components in a vehicle at regular intervals is very important and necessary because it helps to ensure that the vehicle is in a safe and reliable condition.

Generally, these inspection includes tyres, lighting systems, fan belts, shock absorbers, fluid (oil and water) level, etc. If any fault or concern is detected in the course of an inspection, it should be noted for quick repair or servicing by an expert technician.

All automobile engine requires an adequate amount of engine oil as a lubricant so as to mitigate friction and enhance proper functionality of the vehicle. Thus, the proper functionality of an engine is largely dependent on the level of the engine oil; it shouldn't be too low or high.

Basically, the engine oil should be checked at regular intervals (periodically) and should be on the level indicated or chosen by the manufacturer of the vehicle.

A dipstick is designed to be used for checking the engine oil level in a vehicle and it is marked with lines indicating minimum and maximum, low and high or add and full.

The difference in quantity between the add and full marks on an engine oil dipstick is typically 1 quart (0.9 liters).

5 0

2 years ago

I need solution for this question please

AlladinOne [14]

Answer:

a true b false thos is the solution

6 0

3 years ago

Read 2 more answers

A satellite orbits the Earth every 2 hours at an average distance from the Earth's centre of 8000km. (i) What is the average ang

AlexFokin [52]

Answer:

i)ω=3600 rad/s

ii)V=7059.44 m/s

iii)F=1245.8 N

Explanation:

We know that angular speed given as

$\omega =\dfrac{d\theta}{dt}$

We know that for one revolution

θ=2π

Given that time t= 2 hr

ω=θ/t

ω=2π/2 = π rad/hr

ω=3600 rad/s

ii)

Average speed V

$V=\sqrt{\dfrac{GM}{R}}$

Where M is the mass of earth.

R is the distance

G is the constant.

Now by putting the values

$V=\sqrt{\dfrac{GM}{R}}$

$V=\sqrt{\dfrac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{8000\times 10^3}}$

V=7059.44 m/s

iii)

We know that centripetal fore given as

$F=\dfrac{mV^2}{R}$

Here given that m= 200 kg

R= 8000 km

so now by putting the values

$F=\dfrac{mV^2}{R}$

$F=\dfrac{200\times 7059.44^2}{8000\times 10^3}$

F=1245.8 N

3 0

3 years ago

Water flows at a rate of 10 gallons per minute in a new horizontal 0.75?in. diameter galvanized iron pipe. Determine the pressur

ruslelena [56]

Answer:

$\frac{\delta p }{l} = 30.4 lb/ft^3$

Explanation:

Given data:

flow rate = 10 gallon per minute = 0.0223 ft^3/sec

diameter = 0.75 inch

we know discharge is given as

Q = VA

solve for velocity V = \frac{Q}{A}[/tex]

$V = \frac{0.223}{\frac{\pi}{4} \frac{0.75}{12}}$

V = 7.27 ft/sec

we know that Reynold number

$Re = \frac{VD}{\nu}$

$Re = \frac{7.27 \times \frac{0.75}{12}}{1.21\times 10^{-5}}$

$Re = 3.76 \times 10^4$

calculate the $\frac{\epsilon }{D}$ ratio to determine the fanning friction f

$\frac{\epsilon }{D} = \frac{0.0005}{\frac{0.75}{12}} = 0.008$

from moody diagram f value corresonding to Re and $\frac{\epsilon }{D}$ is 0.037

for horizontal pipe

$\delta p = \frac{f l \rho v^2}{2D}$

$\frac{\delta p }{l} = \frac{1 \times 0.037 \times 1.94 \times 7.27}{\frac{0.75}{12}}$

where 1.94 slug/ft^3is density of water

$\frac{\delta p }{l} = 30.4 lb/ft^3$

3 0

3 years ago

In a production turning operation, the foreman has decreed that a single pass must be completed on the cylindrical workpiece in

stellarik [79]

Answer:

V = 125.7m/min

Explanation:

Given:

L = 400 mm ≈ 0.4m

D = 150 mm ≈ 0.15m

T = 5 minutes

F = 0.30mm ≈ 0.0003m

To calculate the cutting speed, let's use the formula :

$T = \frac{pi* D * L}{V*F}$

We are to find the speed, V. Let's make it the subject.

$V = \frac{pi* D * L}{F*T}$

Substituting values we have:

$V = \frac{pi* 0.4 * 0.15}{0.0003*5}$

V = 125.68 m/min ≈ 125.7 m/min

Therefore, V = 125.7m/min

7 0

3 years ago