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3 years ago

14

Technician A says that an A-pillar may be designed to transfer collision energy

1 answer:

svetoff [14.1K]3 years ago

7 0

C both A and b cause they are technician both technicians so they both measure out the floor pan reinforcement be designed to transfer collision energy so I say both A and B

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A system is linear if it has

Nadusha1986 [10]

A system is a linear system if all equations inside the system can be simplified into the form,

$y=mx+n$

aka linear form of a linear equation.

So to sum up, a system is linear if all equations are linear equations.

Hope this helps :)

6 0

4 years ago

Fluids at rest possess no flow energy. a)- True b)- False

gtnhenbr [62]

Answer:

True.

Explanation:

According the engineering flow they don not possess flow energy when they are in rest.

When they are in motion they show a translation energy.

The features if fluids may be different according the variables of pressure and temperature.

8 0

3 years ago

Three spheres are subjected to a hydraulic stress. The pressure on spheres 1 and 2 is the same, and they are made of the same ma

r-ruslan [8.4K]

Answer:

"150000 N/m²" is the right approach.

Explanation:

According to the question, the pressure on the two spheres 1 and 2 is same.

Sphere 1 and 2:

Then,

⇒ $P_1=P_2$

⇒ $\frac{\Delta V_1}{V_1}=\frac{\Delta V_2}{V_2}$

and the bulk modulus be,

⇒ $B_1=B_2$

Sphere 3:

⇒ $\frac{\Delta V_3}{V_3} =\frac{\frac{\Delta V_1}{V_1} }{\frac{\Delta V_2}{V_2} } =1$

then,

⇒ $P_3=B\times \frac{\Delta V_3}{V_3}$

⇒ $=B\times 1$

⇒ $=150000\times 1$

⇒ $=150000 \ N/m^2$

5 0

3 years ago

Air at 1600 K, 30 bar enters a turbine operating at steady state and expands adiabatically to the exit, where the pressure is 2.

djyliett [7]

Solution :

The isentropic efficiency of the turbine is given as :

$\eta = \frac{\text{actual work done}}{\text{isentropic work done}}$

$=\frac{m(h_1-h_2)}{m(h_1-h_{2s})}$

$=\frac{h_1-h_2}{h_1-h_{2s}}$

The entropy relation for the isentropic process is given by :

$0=s^\circ_2-s^\circ_1-R \ln \left(\frac{P_2}{P_1}\right)$

$\ln \left(\frac{P_2}{P_1}\right)=\frac{s^\circ_2-s^\circ_1}{R}$

$\frac{P_2}{P_1}=exp\left(\frac{s^\circ_2-s^\circ_1}{R}\right)$

$\left(\frac{P_2}{P_1}\right)_{s=constant}=\frac{P_{r2}}{P_{r1}}$

Now obtaining the properties from the ideal gas properties of air table :

At $T_1 = 1600 \ K,$

$P_{r1}=791.2$

$h_1=1757.57 \ kJ/kg$

Calculating the relative pressure at state 2s :

$\frac{P_{r2}}{P_{r1}}=\frac{P_2}{P_1}$

$\frac{P_{r2}}{791.2}=\frac{2.4}{30}$

$P_{r2}=63.296$

Obtaining the properties from Ideal gas properties of air table :

At $P_{r2}=63.296$ , $T_{2s}\approx 860 \ K$

Considering the isentropic relation to calculate the actual temperature at the turbine exit, we get:

$\eta=\frac{h_1-h_2}{h_1-h_{2s}}$

$\eta=\frac{c_p(T_1-T_2)}{c_p(T_1-T_{2s})}$

$\eta=\frac{T_1-T_2}{T_1-T_{2s}}$

$0.9=\frac{1600-T_2}{1600-860}$

$T_2= 938 \ K$

So, at $T_2= 938 \ K$ , $h_2=975.66 \ kJ/kg$

Now calculating the work developed per kg of air is :

$w=h_1-h_2$

= 1757.57 - 975.66

= 781 kJ/kg

Therefore, the temperature at the exit is 938 K and work developed is 781 kJ/kg.

4 0

3 years ago

Which traits are common in all four career pathways of the Information Technology field? Check all that apply.

Irina-Kira [14]

2,4,5, I think are the answers

7 0

3 years ago

Read 2 more answers