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3 years ago

13

Opinion: Choose a side and defend it in four to five sentences. Given the choice between a vehicle that runs on gasoline or one

that runs on electricity such as the Aptera 2e, which would you choose and why? Think about fuel consumption, design of the vehicles, and other factors that may influence your decision.

2 answers:

vaieri [72.5K]3 years ago

6 0

Answer:

my opinion would be electric

Explanation:

because when it comes down to the bare minimum the best choice for the world in the long run would be electric because it puts a big dent back for global warming and the burning of gasses so the more people who drive electric cars are the people who are trying to save the world a little at a time by preventing the burning of gasses. hope this helps :)

IRISSAK [1]3 years ago

6 0

I would say electric

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Thermal energy storage systems commonly involve a packed bed of solid spheres, through which a hot gas flows if the system is be

hammer [34]

Answer:

A) i) 984.32 sec

ii) 272.497° C

B) It has an advantage

C) attached below

Explanation:

Given data :

P = 2700 Kg/m^3

c = 950 J/kg*k

k = 240 W/m*K

Temp at which gas enters the storage unit = 300° C

Ti ( initial temp of sphere ) = 25°C

convection heat transfer coefficient ( h ) = 75 W/m^2*k

<u>A) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere</u>

First step determine the Biot Number

characteristic length( Lc ) = ro / 3 = 0.0375 / 3 = 0.0125

Biot number ( Bi ) = hLc / k = (75)*(0.0125) / 40 = 3.906*10^-3

Given that the value of the Biot number is less than 0.01 we will apply the lumped capacitance method

attached below is a detailed solution of the given problem

<u>B) The physical properties are copper</u>

Pcu = 8900kg/m^3)

Cp.cu = 380 J/kg.k

It has an advantage over Aluminum

C<u>) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere</u>

Given that:

P = 2200 Kg/m^3

c = 840 J/kg*k

k = 1.4 W/m*K

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3 years ago

Pressure sensor rated at 0-500 psi and has an output of 0-10 Volts DC (10 V correspond to

scoray [572]

Answer:

Explanation:

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5 0

3 years ago

(a) Consider a message signal containing frequency components at 100, 200, and 400 Hz. This signal is applied to a SSB modulator

MakcuM [25]

Answer:

Explanation:

The frequency components in the message signal are

f1 = 100Hz, f2 = 200Hz and f3 = 400Hz

When amplitude modulated with a carrier signal of frequency fc = 100kHz

Generates the following frequency components

Lower side band

$100k - 100 = 99.9kHz\\\\100k - 200 = 99.8kHz\\\\100k - 400 = 99.6kHz\\\\$

Carrier frequency 100kHz

Upper side band

$100k + 100 = 100.1kHz\\\\100k + 200 = 100.2kHz\\\\100k + 400 = 100.4kHz$

After passing through the SSB filter that filters the lower side band, the transmitted frequency component will be

$100k, 100.1k, 100.2k\ \texttt {and}\ 100.4kHz$

At the receive these are mixed (superheterodyned) with local ocillator frequency whichh is 100.02KHz, the output frequencies will be

$100.02 - 100.1k = 0.08k = 80Hz\\\\100.02 - 100.2k = 0.18k = 180Hz\\\\100.02 - 100.4 = 0.38k = 380Hz$

After passing through the SSB filter that filters the higher side band, the transmitted frequency component will be

$100k, 99.9k, 99.8k\ \ and \ \99.6kHz$

At the receive these are mixed (superheterodyned) with local oscillator frequency which is 100.02KHz, and then fed to the detector whose output frequencies will be

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3 0

3 years ago

Do the coil resistances have any effect on the plots?

PolarNik [594]

Because of the skin depth effect, the current at high frequency tends to flow at very low depth from radius. Then at high frequency the effective cross section of the wire is narrower than at DC.

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8 0

3 years ago

An adiabatic gas turbine expands air at 1300 kPa and 500◦C to 100 kPa and 127◦C. Air enters the turbine through a 0.2-m2 opening

Viktor [21]

Given:

Pressure, $P_{1}$ = 1300 kPa

Temperature, $T_{1}$ = $500^{\circ}$

$P_{2}$ = 100 kPa

$T_{2} = 127^{\circ}$

velocity, v = 40 m/s

A = 1 $m^{2}$

Solution:

For air propertiess at

$P_{1}$ = 1300 kPa

$T_{1}$ = $500^{\circ}$

$h_{1}$ = 793kJ/K

$v_{1}$ = $0.172\frac{m^{3}}{kg}$

and also at

$P_{2}$ = 100 kPa

$T_{2} = 127^{\circ}$

$h_{2}$ = 401 KJ/K

$v_{2}$ = $1.15\frac{m^{3}}{kg}$

a) Mass flow rate is given by:

$m' = \frac{Av}{v_{1}}$

Now,

$m = \frac{0.2\times 40}{0.172}$ = 46.51 kg/s

b) for the power produced by turbine, $P = m'(h_{1} - h_{2})$

$P = 46.51\times(793 - 401)$ = 18.231 MW

5 0

4 years ago