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muminat
3 years ago
5

Where are the ar manufacturers not fitting the engine in the high end sport cars

Engineering
1 answer:
fomenos3 years ago
8 0

Answer:

it depends on the but i would recommend check in the front next to the turbo intake.

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A Carnot cooler operates with COP = 11, whose ambient temperature is 300K. Determine the temperature at which the refrigerator a
SashulF [63]

Answer:

275 Kelvin

Explanation:

Coefficient of Performance=11

T_H=\text {Absolute Temperature of high temperature reservoir=300 K}

T_L=\text {Absolute Temperature of low temperature reservoir}

\text {Coefficient of performance for carnot cooler}\\=\frac {T_L}{T_H-T_L}\\\Rightarrow 11=\frac{T_L}{300-T_L}\\\Rightarrow 11(300-T_L)=T_L\\\Rightarrow 3300-11T_L=T_L\\\Rightarrow 3300=T_L+11T_L\\\Rightarrow 3300=12T_L\\\Rightarrow T_L=\frac {3300}{12}\\\Rightarrow T_L=275\ K\\\Therefore \text{Temperature at which the refrigerator absorbs heat=275 Kelvin}

8 0
3 years ago
The maximum stress that a bar will withstand before failing is called • Rapture Strength • Yield Strength • Tensile Strength • B
konstantin123 [22]

Answer: Rupture strength

Explanation: Rupture strength is the strength of a material that is bearable till the point before the breakage by the tensile strength applied on it. This term is mentioned when there is a sort of deformation in the material due to tension.So, rupture will occur before whenever there are chances of failing and the material is still able to bear stresses before failing.  

7 0
3 years ago
The human eye, as well as the light-sensitive chemicals on color photographic film, respond differently to light sources with di
jeka57 [31]

Answer:

a) at T = 5800 k  

  band emission = 0.2261

at T = 2900 k

  band emission = 0.0442

b) daylight (d) = 0.50 μm

    Incandescent ( i ) =  1 μm

Explanation:

To Calculate the band emission fractions we will apply the Wien's displacement Law

The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as

F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )

<em>Values are gotten from the table named: blackbody radiati</em>on functions

<u>a) Calculate the band emission fractions for the visible region</u>

at T = 5800 k  

  band emission = 0.2261

at T = 2900 k

  band emission = 0.0442

attached below is a detailed solution to the problem

<u>b)calculate wavelength corresponding to the maximum spectral intensity</u>

For daylight ( d ) = 2898 μm *k / 5800 k  = 0.50 μm

For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm

3 0
3 years ago
The acceleration of a particle is given by a = 2t − 10, where a is in meters per second squared and t is in seconds. Determine t
tensa zangetsu [6.8K]

Answer

given,

a = 2 t - 10

velocity function

we know,

\dfrac{dv}{dt}=a

\dfrac{dv}{dt}=(2t-10)

integrating both side

\int dv =\int (2t -10) dt

 v = t² - 10 t + C

at t = 0   v = 3

so, 3 = 0 - 0 + C

     C = 3

Velocity function is equal to v = t² - 10 t + 3

Again we know,

\dfrac{dx}{dt}=v

\dfrac{dx}{dt}=(t^2-10t + 3)

integrating both side

\int dx =\int (t^2-10t + 3)dt

x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t + C

now, at t= 0 s = -4

-4 = \dfrac{0^3}{3}- 10\dfrac{0^2}{2} + 0 + C

C = -4

So,

x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t-4

Position function is equal to x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t-4

8 0
3 years ago
What types of problems might an electrical engineer need to solve?
aleksandr82 [10.1K]

Answer:

sectores industriales, comerciales o públicos, o para el uso doméstico.

Explanation:

8 0
3 years ago
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