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Natali5045456 [20]

3 years ago

9

A lightweight electric car is powered by ten 12-V batteries. At a speed of 100 km/h, the average frictional force is 1500 N. Wha

t must be the power of the electric motor if the car is to travel at a speed of 100 km/h?

1 answer:

Aleksandr [31]3 years ago

6 0

Answer:41.67kW

Explanation: a) power = force * velocity

Convert km/hr to m/s=1000m/(60×60s)

=1000/3600

P=F×vel

Power =1500×100× 1000/3600

=1500×100×5/18

=41,666.67W

=41.67kW

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The answer is "0.0728"

Explanation:

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if $P>P^{*} \to$ flow is not chocked

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match number:

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$\ velocity \ of \ sound \ (C)=\sqrt{KRT}\\\\$

$=\sqrt{1.4\times1716\times464.6}\\\\=1057 ft^3\\\\$

R= gas constant=1716

$m=PAV\\\\$

$=0.001808\times0.05\times(Ma.C)\\\\=0.001808\times0.05\times0.7625\times 1057\\\\=0.0728\frac{slug}{s}$

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A simple non-ideal Rankine cycle with water as the working fluid operates between the pressure limits of 15 MPa in the boiler an

konstantin123 [22]

Answer:

The right solution is "28.45%".

Explanation:

The given values are:

$P_4=50\ kPa$

$h_4=0.7(2304.7)+340.5$

$=1953.83 \ KJ/Kg$

and,

$P_3=15 \ mPa$

$h_3=hg$

$=2610.8 \ KJ/Kg$

$s_3=sg$

$=5.3108 \ KJ/Kgh$

At 45,

⇒ $x_{45} = \frac{5.3108-1.0912}{6.5019}$

$=0.66$

At $P_4=50 \ Kpa$ ,

$h_f=340.54$

or,

$V_f=0.001030 \ m^3/Kg$

then,

⇒ $h_2=340.54+0.001030(15\times 10^{3}-50)$

$=355.94 \ kJ/kg$

hence,

The isentropic efficiency of turbine will be:

⇒ $n_T=\frac{h_3-h_4}{h_3-h_{45}}$

$=\frac{2610.8-1953.83}{2610.8-1836.26}$

$=84.818$ (%)

The thermal efficiency of cycle will be:

⇒ $n_C=\frac{W_T-W_P}{2_{in}}$

$=\frac{(2610.8-1953-83)-(355.93-340.54)}{2610.8-355.93}$

$=28.45$ (%)

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3 years ago