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tankabanditka [31]
2 years ago
13

Two basic types of mechanical fuel injector systems?​

Engineering
2 answers:
dem82 [27]2 years ago
7 0

Answer:

excretory system oxgen

REY [17]2 years ago
4 0

Answer:

Single-point or throttle body injection. Port or multipoint fuel injection.

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Calculate and plot the radial and circumferential stress distribution in the left ventricle at the end of systole (p 5 80 mmHg;
alexandr402 [8]

Answer:

62990.08 N/M

Explanation:

Circumferential stress, or hoop stress, a normal stress in the tangential (azimuth) direction. axial stress, a normal stress parallel to the axis of cylindrical symmetry. radial stress, a stress in directions coplanar with but perpendicular to the symmetry axis.

See attachment for the step by step solution of the given problem..

8 0
3 years ago
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

4 0
3 years ago
Name some technical skills that are suitable for school leavers .​
Natalka [10]

Answer:

Welding, carpentry, masonry, construction worker, barber

Explanation:

7 0
3 years ago
To compute the energy used by a motor, multiply the power that it draws by the time of operation. Con- sider a motor that draws
ehidna [41]

Answer:

E=52000Hp.h

E=38724920Wh

E=1.028x10^11 ftlb

Explanation:

To solve this problem you must multiply the engine power by the time factor expressed in h / year, to find this value you must perform the conventional unit conversion procedure.

Finally, when you have the result Hp h / year you convert it to Ftlb and Wh

E=(12.5hp)(\frac{16h}{day} )(\frac{5 days}{week} )(\frac{52week}{year} )\\

E=52000Hp.h

E=52000Hp.h(\frac{744.71Wh}{Hp.h} )\\

E=38724920Wh

E=52000Hph(\frac{1977378.4  ft lb}{1Hph}

E=1.028x10^11 ftlb

3 0
3 years ago
Argue the importance to society of incorporating green building into an engineer's designs, with at least two examples.
stellarik [79]

Answer:

Green Design is your answer

Explanation:

4 0
3 years ago
Read 2 more answers
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