ENGINEERING COLLEGE

How to find magnitude of net electric charge when current flows through a capacitor with initial charge?

Answers

Answer 1

Answer:

Answer:

Q=q $e^(-t/RC)$

also

q=it

Explanation:

How to find magnitude of net electric charge when current flows through a capacitor with initial charge?

A capacitor is a device used in storing electric charges. The unit of capacitance is measured in farad

To find the magnitude of net electric charge in a capacitor, we use the following relation

Q=q $e^(-t/RC)$

Q=the magnitude of net electric charge in coulombs

t=the time for an electron tpo pass through the capacitor

R=the resistance of the capacitor , measured in ohms

C=the capacitance of the capacitor measures in Farad

q=initial quantity of charge

Having all the parameters above will make it possible to determine the net electric charge

Recall also that

q=it

quantity (coulombs)=current*time(s)

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If the total energy change of an system during a process is 15.5 kJ, its change in kinetic energy is -3.5 kJ, and its potential energy is unchanged, calculate its change in specificinternal energy if its mass is 5.4 kg. Report your answer in kJ/kg to one decimal place.

Answers

Answer:

The change in specific internal energy is 3.5 kj.

Explanation:

Step1

Given:

Total change in energy is 15.5 kj.

Change in kinetic energy is –3.5 kj.

Change in potential energy is 0 kj.

Mass is 5.4 kg.

Step2

Calculation:

Change in internal energy is calculated as follows:

$\bigtriangleup E=\bigtriangleup KE+\bigtriangleup PE+\bigtriangleup U$ $15.5=-3.5+0+\bigtriangleup U$

$\bigtriangleup U=19$ kj.

Step3

Specific internal energy is calculated as follows:

$\bigtriangleup u=(\bigtriangleup U)/(m)$

$\bigtriangleup u=(19)/(5.4)$

$\bigtriangleup u=3.5$ kj/kg.

Thus, the change in specific internal energy is 3.5 kj/kg.

The structure of a house is such that it loses heat at a rate of 5400 kJ/h per degree Cdifference between the indoors and outdoors. A heat pump that requires a power input of 6 kW isused to maintain this house at 21 C. Determine the lowest outdoor temperature for which the heatpump can meet the heating requirements of this house

Answers

Answer: Tl = - 13.3°C

the lowest outdoor temperature is - 13.3°C

Explanation:

Given that;

Temperature of Th = 21°C = 21 + 273 = 294 K

the rate at which heat lost is Qh = 5400 kJ/h°C

the power input to heat pump Wnet = 6 kw

The COP of a reversible heat pump depends on the temperature limits in the cycle only, and is determined by;

COPhp = Th/(Th - Tl)

COPhp = Qh/Wnet

Qh/Wnet = Th/(Th -Tl)

the amount of heat loss is expressed as

Qh = 5400/3600(294 - Tl)

the temperature of sink

( 5400/3600(294 - Tl)) / 6 = 294 / ( 294 - Tl)

now solving the equation

Tl = 259.7 - 273

Tl = - 13.3°C

so the lowest outdoor temperature is - 13.3°C

If you answer the whole question and show your work/coding I will rate 5 stars/brainliest!!! Walnut Orchard has two farms that grow wheat and corn. Because of different soil conditions, there are differences in the yields and costs of growing crops on the two farms. The yields and costs are shown in the following table. Each farm has 100 acres available for cultivation. 11,000 bushels of wheat and 7,000 bushels of corn must be grown. Please have an LP model to minimize the total cost while meeting the demand and solve it with Lindo or Excel. You need to have all parts of a model: notation, objective function, constraints, and sign restrictions.

Answers

Answer: 18,100

Explanation: two farms that grow wheat and corn.

In this type of projection, the angles between the three axes are different:- A) Isometric B) Axonometric C) Trimetric D) Dimetnic

Answers

Answer:

The correct answer is C) Trimetric

Explanation:

The most suitable answer is a trimetric projection because, in this type of projection, we see that the projection of the three angles between the axes are not equal. Therefore, to generate a trimetric projection of an object, it is necessary to have three separate scales.

In a home, air infiltrates from the outside through cracks around doors and windows. Consider a residence where the total length of cracks is 62 m and the total internal volume is 210 m3 . Due to the wind, 9.4 x 10-5 kg/s of air enters per meter of crack and exits up a chimney. Assume air temperature is the same inside and out and air density is constant at 1.186 kg/m3 . If windows and doors are not opened or closed, estimate the time required for one complete air change in the building.

Answers

Answer:

Time period = 41654.08 s

Explanation:

Given data:

Internal volume is 210 m^3

Rate of air infiltration $9.4 * 10^(-5) kg/s$

length of cracks 62 m

air density = 1.186 kg/m^3

Total rate of air infiltration $= 9.4* 10^(-5) * 62 = 582.8* 10{-5} kg/s$

total volume of air infiltration $= \frac{582.8* 10{-5}}{1.156} = 5.04* 10^(-3) m^3/s$

Time period $= (210)/(5.04* 10^(-3)) = 41654.08 s$

What is the minimum hose diameter of an ideal vacuum cleaner that could lift a 14 kg dog off the floor?

Answers

The diameter would be 267km speed suction i think

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