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2 years ago

8

A sports car has a drag coefficient of 0.29 and a frontal area of 20 ft2, and is travelling at a speed of 120 mi/hour. How much

power is required to overcome aerodynamic drag if 휌= 0.002378 slugs/ft3

1 answer:

Andrej [43]2 years ago

4 0

Answer:

Power required to overcome aerodynamic drag is 50.971 KW

Explanation:

For explanation see the picture attached

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An op-amp is connected in an inverting configuration with R1 = 1kW and R2 = 10kW, and a load resistor connected at the output, R

Svetllana [295]

Answer:

View Image

Explanation:

You didn't provide me a picture of the opamp.

I'm gonna assume that this is an ideal opamp, therefore the input impedance can be assumed to be ∞ . This basically implies that...

no current will go in the inverting(-) and noninverting(+) side of the opamp
V₊ = V₋ , so whatever voltage is at the noninverting side will also be the voltage at the inverting side

Since no current is going into the + and - side of the opamp, then

i₁ = i₂

Since V₊ is connected to ground (0V) then V₋ must also be 0V.

V₊ = V₋ = 0

Use whatever method you want to solve for v_out and v_in then divide them. There's so many different ways of solving this circuit.

You didn't give me what the input voltage was so I can't give you the entire answer. I'll just give you the equations needed to plug in your values to get your answers.

8 0

3 years ago

Earth whose in situ weight is 105lb/cf and whose compacted weight is 122 lb/cf is placed in a fill at the rate of 260 cy/hr, mea

Stells [14]

Answer:

Number of rollers required to complete the compaction are 2

Explanation:

The solution is given in the attachments.

6 0

3 years ago

What is the primary water source for a water cooled recovery unit's condensing coll?

nataly862011 [7]

A) chilled water from evaporator

7 0

3 years ago

500 flights land each day at San Jose’s airport. Assume that each flight has a 5% chance of being late, independently of whether

BARSIC [14]

Answer:

a.0.0199

b.0.1765

c.0.0785

d.0.1268

e.Yes

Explanation:

It is given that X follows a Binomial distribution with (n= 500, p = 0.05)

The probabilities are computed using the EXCEL .

a) The required probability here is:

P(X less of equal to 15)

= binom.dist(15,500,0.05,TRUE)

=0.0199

Therefore the probability is 0.0199 .

b) The required probability here is:

P(X greater or equal to 30) = 1 - P(X less or equal to 29)

=1 - binom.dist(29,500,0.05,TRUE)

=0.1765

Therefore the probability is 0.1765

c) P(X = 26 )

= binom.dist(26,500,0.05,FALSE)

=0.0785

Therefore the probability is 0.0785

d) The required probability here is computed as:

P(10 less or equal to X less or equal to 20 ) = P(X less or equal to 19) - P(X less or equal to 10)

= binom.dist(19,500,0.05,TRUE) - binom.dist(10,500,0.05,TRUE)

=0.1268

Therefore the probability 0.1268

e) Yes . Therefore the probability because that is the assumption used to apply binomial distribution .

6 0

3 years ago

Water vapor at 6 MPa, 600 degrees C enters a turbine operating at steady state and expands to 10kPa. The mass flow rate is 2 kg/

kirill115 [55]

Answer:

Explanation:

Obtain the following properties at 6MPa and 600°C from the table "Superheated water".

$h_1=3658.8KL/Kg\\s_1=7.1693kJ/kg.k$

Obtain the following properties at 10kPa from the table "saturated water"

$h_{f2}=191.81KJ/Kg.K\\h_{fg2}=2392.1KJ/Kg\\s_{f2}=0.6492KJ/Kg.K\\s_{fg2}=7.4996KJ/Kg.K$

Calculate the enthalpy at exit of the turbine using the energy balance equation.

$\frac{dE}{dt}=Q-W+m(h_1-h_2)$

Since, the process is isentropic process $Q=0$

$0=0-W+m(h_1-h_2)\\h_2=h_1-\frac{W}{m}\\\\h_2=3658.8-\frac{2626}{2}\\\\=2345.8kJ/kg$

Use the isentropic relations:

$s_1=s_{2s}\\s_1=s_{f2}+x_{2s}s_{fg2}\\7.1693=6492+x_{2s}(7.4996)\\x_{2s}=87$

Calculate the enthalpy at isentropic state 2s.

$h_{2s}=h_{f2}+x_{2s}.h_{fg2}\\=191.81+0.87(2392.1)\\=2272.937kJ/kg$

a.)

Calculate the isentropic turbine efficiency.

$\eta_{turbine}=\frac{h_1-h_2}{h_1-h_{2s}}\\\\=\frac{3658.8-2345.8}{3658.8-2272.937}=0.947=94.7%$

b.)

Find the quality of the water at state 2

since $h_f$ at 10KPa < $h_2$ < $h_g$ at 10KPa

Therefore, state 2 is in two-phase region.

$h_2=h_{f2}+x_2(h_{fg2})\\2345.8=191.81+x_2(2392.1)\\x_2=0.9$

Calculate the entropy at state 2.

$s_2=s_{f2}+x_2.s_{fg2}\\=0.6492+0.9(7.4996)\\=7.398kJ/Kg.K$

Calculate the rate of entropy production.

$S=\frac{Q}{T}+m(s_2-s_1)$

since, Q = 0

$S=m(s_2-s_1)\\=2\frac{kg}{s}(7.398-7.1693)kJ/kg\\=0.4574kW/k$

6 0

3 years ago