All categories

3 years ago

9. Calculate the total resistance and current in a parallel cir-

Engineering

1 answer:

Taya2010 [7]3 years ago

5 0

Answer:

d. 2.3 ohms (5.3 amperes)

Explanation:

The calculator's 1/x key makes it convenient to calculate parallel resistance.

Req = 1/(1/4 +1/8 +1/16) = 1/(7/16) = 16/7 ≈ 2.3 ohms

This corresponds to answer choice D.

<em>Additional comment</em>

This problem statement does not tell the applied voltage. The answer choices suggest that it is 12 V. If so, the current is 12/(16/7) = 21/4 = 5.25 amperes.

You might be interested in

Between chain, belt and gear drive, which causes more friction?

loris [4]

Answer:

All of them cause friction

Explanation:

google

6 0

3 years ago

Convert 850 nm wavelength into frequency, eV, wavenumber, joules and ergs.

Sholpan [36]

Answer:

Frequency = $3.5294\times 10^{14}s^{-1}$

Wavenumber = $1.1765\times 10^6m^{-1}$

Energy = $2.3365\times 10^{-19}J$

Energy = 1.4579 eV

Energy = $2.3365\times 10^{-12}erg$

Explanation:

As we are given the wavelength = 850 nm

conversion used : $(1nm=10^{-9}m)$

So, wavelength is $850\times 10^{-9}m$

The relation between frequency and wavelength is shown below as:

$Frequency=\frac{c}{Wavelength}$

Where, c is the speed of light having value = $3\times 10^8m/s$

So, Frequency is:

$Frequency=\frac{3\times 10^8m/s}{850\times 10^{-9}m}$

$Frequency=3.5294\times 10^{14}s^{-1}$

Wavenumber is the reciprocal of wavelength.

So,

$Wavenumber=\frac{1}{Wavelength}=\frac{1}{850\times 10^{-9}m}$

$Wavenumber=1.1765\times 10^6m^{-1}$

Also,

$Energy=h\times frequency$

where, h is Plank's constant having value as $6.62\times 10^{-34}J.s$

So,

$Energy=(6.62\times 10^{-34}J.s)\times (3.5294\times 10^{14}s^{-1})$

$Energy=2.3365\times 10^{-19}J$

Also,

$1J=6.24\times 10^{18}eV$

So,

$Energy=(2.3365\times 10^{-19})\times (6.24\times 10^{18}eV)$

$Energy=1.4579eV$

Also,

$1J=10^7erg$

So,

$Energy=(2.3365\times 10^{-19})\times 10^7erg$

$Energy=2.3365\times 10^{-12}erg$

5 0

3 years ago

Do you think for security reasons everything that happens on the internet should be analyzed by the public security services ?

givi [52]

Answer: yes

Explanation: People post bad things that i think should get taken down i was on an app the other day and people were posting bad things

8 0

3 years ago

The average age of engineering students at graduation is a little over 23 years. This means that the working career of most engi

RideAnS [48]

Answer:

$916

Explanation:

To solve this, we use the formula

FV = P/i * [(1+i)^n - 1], where

FV = future value of the all the money invested, $5 million

n = time span, = 500 months

P = payment per month

I = interest rate, 9% by 12 months, = 0.0075

Considering that we have been given all in our question, then we substitute directly and solve. So we have,

5000000 = P/0.0075 * [(1+0.0075)^500 -1]

5000000 * 0.0075 = P * [1.0075^500 - 1]

37500 = P * [41.93 - 1]

37500 = P * 40.93

P = 37500/40.93

P = $916.20

Therefore, the engineer needs to save $916 in a month which is the accrued

6 0

3 years ago

Assuming 800 W / m 2 solar irradiance and a 35 % efficient solar panel, how much roof area should be covered to supply 10 A at 1

Korvikt [17]

Answer:

a) 4.286 m² b) $ 55.44/mo

Explanation:

If we assume that the sun is behaving as an isotropic radiator, the power density that is arriving to the house, is constant and equal to the quoted solar irradiance.

If the energy conversion capability of the solar panels were 100%, the roof area needed to supply the power required, would be simply the quotient between the power required and the solar irradiance, as follows:

A = P / SI = 10 A* 120 V / 800 W/m² = 1200 W / 800 W/m²= 1,5 m²

As the solar panels are only 35% efficient in converting the solar energy to useful electrical energy, we will need more roof area, according to this expression:

Ae = At / 0.35 = 1,5 / 0.35 = 4.286 m²

b) If we can get 1200 W during 7 hs/day, the energy supplied by the solar panels will be the product of the power times the time, as follows:

E= 1200 W* 7 hs = 8.4 Kwh

If the cost per Kwh, is $0.22, assuming 7 hs. of use in average during a month (assumed to be of 30 days), we can have savings as follows:

Cost = 0.22($/Kwh)* 8.4 (Kwh/day)*30 (days/mo) = $ 55.44

8 0

4 years ago