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3 years ago

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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]3 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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At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho

victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

$\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2} }{2})^{2}+\tau _{xy}^{2} }$

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

$32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2} }$

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

$\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}$

Where

σp1 = 20 ksi

σp2 = -30 ksi

$\sigma _{p1} +\sigma _{p2}=-10 ksi$ (equation 1)

$\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi$ equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

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3 years ago

A student engineer is given a summer job to find the drag force on a new unmaned aerial vehicle that travels at a cruising speed

yan [13]

Answer:

b. 1232.08 km/hr

c. 1.02 kn

Explanation:

a) For dynamic similar conditions, the non-dimensional terms R/ρ V2 L2 and ρVL/ μ should be same for both prototype and its model. For these non-dimensional terms , R is drag force, V is velocity in m/s, μ is dynamic viscosity, ρ is density and L is length parameter.

See attachment for the remaining.

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3 years ago

An inventor claims to have invented a heat engine that operates between the temperatures of 627°C and 27°C with a thermal effici

Oksana_A [137]

Answer Explanation:

the efficiency of the the engine is given by=1- $\frac{T_2}{T_1}$

where T₂= lower temperature

T₁= Higher temperature

we have given efficiency =70%

lower temperature T₂=27°C=273+27=300K

higher temperature T₁=627°C=273+627=900K

efficiency=1- $\frac{T_2}{T_1}$

=1- $\frac{300}{900}$

=1-0.3333

=0.6666

=66%

66% is less than 70% so so inventor claim is wrong

3 0

3 years ago

A constant-head permeability test gives the following information: - Water flows horizontally through the soil sample. - The hei

Marysya12 [62]

Answer:

Complete answer to the question is explained in the attached files.please have a look on it.

Explanation:

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3 years ago

An insulated rigid tank is initially evacuated. A valve is opened, and atmospheric air at 95 kPa and 17 ºC enters the tank until

Darya [45]

Answer:

$T2= 406k$

Explanation:

The temperature can be defined as the measurement of the intensity of the heat present in the object. Fahrenheit, kelvin and centigrade are the common scale used for measuring Temperature.

Given:

T1=170C

To convert to Kelvin

= 17+273 =290K

T1 = 290K

Pressure (P)= 95KPa

Specific heat ratio = CP/CV= K

WhereK=1.005/0.718

K = 1.4

The final temperature can be calculated using the formula below.

T2 = CP/CV × T1

=. K × T1

T2 = 1.4 × 290

$T2= 406k$

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2 years ago