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MAXImum [283]
3 years ago
13

The drag coefficient of a vehicle increases when its windows are rolled down or its sunroof is opened. A sports car has a fronta

l area of 18 ft2 and a drag coefficient of 0.32 when the windows and sunroof are closed. The drag coefficient increases to 0.41 when the sunroof is open. Determine the additional power consumption of the car when the sunroof is opened at (a) 35 mi/h and (b) 70 mi/h. Take the density of air to be 0.075 lbm/ft3 .

Engineering
2 answers:
Ne4ueva [31]3 years ago
5 0

Answer:

the additional power consumption is a) 0.4669 hp, b) 3.7353 hp

Explanation:

the solution is in the attached Word document

Download docx
klio [65]3 years ago
3 0

Answer:

Explanation:

The additional power consumption of the car when v :

35 mi/h = 0.464hp

70 mi/h = 3.71hp.

For the step by step explanation of how we arrived at these answers, please go through the attached files.

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You are in charge of ordering the concrete for a basement wall concrete pour. The wall forms are all set up and ready. The wall
choli [55]

Answer:

189.15cy

Explanation:

To understand this problem we need to understand as well the form.

It is clear that there is four wall, two short and two long.

The two long are \rightarrow 120ft5in+2(10ft)

The two long are \rightarrow 122ft1in=122.08ft

The two shors are \rightarrow 86ft4.5in = 86.375ft

The height and the thickness are 14ft and 0.83ft respectively.

So we only calculate the Quantity of concrete,

Q_c = [(2*122.08)+(2*86-375)]*14*0.833\\Q_c=4864.02ft^3

That in cubic yards is equal to 180.15 (1cy=27ft^3)

Hence, we need order 5% plus that represent with the quantity

Q_{ordered}=1.05*180.15=189.15cy

8 0
3 years ago
If welding is being done in the vertical position, the torch should have a travel angle of?
siniylev [52]

Answer:

Between 35°– 45°

Explanation:

In the vertical position, Point the flame in the direction of travel. Keep the flame tip at the correct height above the base metal. An angle of 35°–45° should be maintained between the torch tip and the base metal. This angle may be varied up or down to heat or cool the weld pool if it is too narrow or too wide

4 0
2 years ago
The privilege of driving comes with
julsineya [31]
The privilege of driving comes with responsibility
5 0
3 years ago
Tahir travel twice as far as ahmed, but onley one third as fast. Ahmed starts travel on tuesday at noon at point x to point z 30
shepuryov [24]

Answer:

6:00 pm the next day

Explanation:

Given that

Tahir traveled twice as far as Ahmed. We say,

Ahmed traveled a distance, D

Tahir would travel a distan, 2D

Tahir traveled 1/3 as fast as Ahmed, so we say

Ahmed traveled at a speed, S

Tahir would travel at a speed, S/3

If Ahmed starts travel on tuesday at noon at point x to point z 300km, by 9:00pm,

Time taken by Ahmed to travel is

9:00 pm - 12:00 pm = 9 hours

Ahmed, traveled 300 km in 9 hours, meaning he traveled at 33.3 km in an hour.

Speed, S that Ahmed traveled with is 33.3 km/h

Remember, we stated that Tahir travels at a speed of S/3, that is, The speed of Tahir is

33.3/3 = 11.1 km/h.

300 km would then be traveled in 300 km/11.1 km/h = 27 hours.

Tahir started traveling, 3 hours after Ahmed, that is 12:00 pm + 3:00 hrs = 3:00 pm, and if he's to spend 27 hours on the journey he would reach destination z at 6:00 pm the next day

7 0
3 years ago
The Stefan-Boltzmann law can be employed to estimate the rate of radiation of energy H from a surface, as in
Mazyrski [523]

Explanation:

A.

H = Aeσ^4

Using the stefan Boltzmann law

When we differentiate

dH/dT = 4AeσT³

dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³

= 8.4085

Exact error = 8.4085x20

= 168.17

H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴

= 1366.376watts

B.

Verifying values

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴

= 1542.468

H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴

= 1205.8104

Error = 1542.468-1205.8104/2

= 168.329

ΔT = 40

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴

= 1735.05

H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴

= 1735.05-1059.83/2

= 675.22/2

= 337.61

5 0
3 years ago
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