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harina [27]
1 year ago
12

The thrust angle is checked by referencing

Engineering
1 answer:
anygoal [31]1 year ago
5 0

In Engineering, the thrust angle is checked by referencing: C. vehicle centerline.

<h3>What is a thrust angle?</h3>

A thrust angle can be defined as an imaginary line which is drawn perpendicularly from the centerline of the rear axle of a vehicle, down the centerline.

This ultimately implies that, the thrust angle is a reference to the centerline (wheelbase) of a vehicle, and it confirms that the two wheels on both sides are properly angled within specification.

Read more on thrust angle here: brainly.com/question/13000914

#SPJ1

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Brrunno [24]

Well, I do know that polarity affects the voltage.

6 0
2 years ago
The minimum recommended standards for the operating system, processor, primary memory (RAM), and storage capacity for certain so
nlexa [21]

Answer:System requirements

Explanation:

its right.

8 0
2 years ago
Read 2 more answers
Hot carbon dioxide exhaust gas at 1 atm is being cooled by flat plates. The gas at 220 °C flows in parallel over the upper and l
sergeinik [125]

The local convection heat transfer coefficient at 1 m from the leading edge is  0.44 \frac{W}{m^{2} \times K} ,  the average convection heat transfer coefficient over the entire plate is  0.293 \frac{W}{m^{2} \times K}and the total heat flux transfer to the plate is 61.6 KJ.

Explanation:

It is case of heat and mass transfer in which due to temperature difference between gas  and surface. Further temperature  boundary layer will developed on flat plate in longitudinal direction.  

Hot carbon dioxide exhaust gas

physical properties

r= 1.05 \frac{kg}{m^{3}}

c_p = 1.02 \frac{kJ}{Kg \times K}

m= 231 \times 10^{7}  \frac{N \times s }{m^2}

υ = 21.8 \times 10^{6}  \frac{m^2}{s}

k = 32.5 \times 10^{3} \frac{W}{m \times K}

\alpha = 30.1 \times 10^{6} \frac{m^{2}}{s}

Pr = 0.725

Apart from these other data arr given below,

v= 3 \frac{m}{s}  \\ p= 1 atm \\ L_c = 1.5m \\T_g= 220 C \\ T_s = 80 C

To find the local convection heat transfer coefficient at 1 m from the leading edge, we use correlation used for laminar flow over flat plate,

Nu = \frac{ h \times L }{k}  = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })

where h= Average heat transfer coefficient

           L= Length of a plate

           k= Thermal Conductivity of carbon dioxide

           Re = Reynold's Number

           Pr  = Prandtle Number

(a) Convection heat transfer coefficient at 1 m from the leading edge

    is referred as local convection heat transfer coefficient.

   

   To find convection heat transfer coefficient at 1 m from leading edge,

  Nu = \frac{ h_local \times L }{k}  = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })

  Here, first we have to find Re and Pr,

   Re = \frac{r \times v \times L}{m}

   Re = \frac{1.0594 \times 3 \times 1}{231 \times 10^{7}}

   Re = 20.63 \times  10^{-10}

   Pr number is take from physical property data and Pr is 0.725.

   Putting value of Re and Pr in main equation,

   we get

   Nu = \frac{ h_local \times 1 }{32.5 \times 10^{3}}  = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })

    h_local   = 32.5 \times 10^{3} \times  0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })

    h_local   =  0.44 \frac{W}{m^{2} \times K}

(b)  To find average convection heat transfer coefficient,

      it can be find out as case (a), only difference is that instead of L=1 m,        L=1.5 m would come,  

   Therefore,

    Nu = \frac{ h \times 1.5 }{32.5 \times 10^{3}}  = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })

    Finally,

      h  = \frac{0.44}{1.5}

      h  = 0.293 \frac{W}{m^{2} \times K}

(C) Total heat flux transfer to the plate is found out by,

     Q = h \times (T_g - T_s)

     Q = 0.293 \times (220-80) \\ Q= 0.293 \times 140  \\ Q= 61.6 KJ

     

     

   

   

     

   

     

   

   

 

   

   

   

   

8 0
2 years ago
A sample of sand weighs 490 g in stock and 475 in Oven Dry (OD) condition, respectively. If absorption capability of the sand is
Ivahew [28]

The weight of the specimen in SSD condition is 373.3 cc

<u>Explanation</u>:

a) Apparent specific gravity = \frac{A}{A-C}

Where,

A = mass of oven dried test sample in air = 1034 g

B = saturated surface test sample in air = 1048.9 g

C = apparent mass of saturated test sample in water = 975.6 g

apparent specific gravity = \frac{A}{A-C}

                                         = \frac{1034}{1034-675 \cdot 6}

Apparent specific gravity = 2.88

b) Bulk specific gravity G_{B}^{O D}=\frac{A}{B-C}

G_{B}^{O D}=\frac{1034}{1048.9-675 \cdot 6}

       =  2.76

c) Bulk specific gravity (SSD):

G_{B}^{S S D}=\frac{B}{B-C}

=\frac{1048 \cdot 9}{1048 \cdot 9-675 \cdot 6}

G_{B}^{S S D} = 2.80

d) Absorption% :

=\frac{B-A}{A} \times 100 \%

=\frac{1048 \cdot 9-1034}{1034} \times 100

Absorption = 1.44 %

e) Bulk Volume :

v_{b}=\frac{\text { weight of dispaced water }}{P \omega t}

=\frac{1048 \cdot 9-675 \cdot 6}{1}

= 373.3 cc

5 0
2 years ago
There is an electric field near the Earth's surface whose magnitude is about 145 V/m . How much energy is stored per cubic meter
weqwewe [10]

Answer:

u_e = 9.3 * 10^-8 J / m^3  ( 2 sig. fig)

Explanation:

Given:

- Electric Field strength near earth's surface E = 145 V / m

- permittivity of free space (electric constant) e_o =  8.854 *10^-12 s^4 A^2 / m^3 kg

Find:

- How much energy is stored per cubic meter in this field?

Solution:

- The solution requires the energy density stored between earth's surface and the source of electric field strength. The formula for charge density is given by:

                                        u_e = 0.5*e_o * E^2

- Plug in the values given:

                                        u_e = 0.5*8.854 *10^-12 *145^2

                                        u_e = 9.30777 * 10^-8  J/m^3

5 0
3 years ago
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