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Dmitrij [34]
2 years ago
9

Help please i will give brainlist

Engineering
2 answers:
vlada-n [284]2 years ago
5 0

Answer:

Open Circuit

Explanation:

Because the circuit isnt closed, the current wont flow

alekssr [168]2 years ago
3 0
The last one
And nice to meet you I am fatima and I think we are from the same place

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True or False; If I was trying to find the Voltage of my computer, and I was given the Watts and Amps it uses, I would use Watt'
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true

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19. A circuit contains four 100 S2 resistors connected in series. If you test the circuit with a digital VOM,
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6 0
3 years ago
Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320
lesya [120]

Answer:

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

Explanation:

The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:

n = \frac{S_{uts}}{\tau_{max}}

Where:

n - Safety factor, dimensionless.

S_{uts} - Ultimate shear strength, measured in pascals.

\tau_{max} - Maximum allowable shear stress, measured in pascals.

The maximum allowable shear stress is consequently cleared and computed: (n = 4.2, S_{uts} = 320\times 10^{6}\,Pa)

\tau_{max} = \frac{S_{uts}}{n}

\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}

\tau_{max} = 76.190\times 10^{6}\,Pa

Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:

\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}

Where:

\tau_{max} - Maximum allowable shear stress, measured in pascals.

V - Shear force, measured in kilonewtons.

A - Cross section area, measured in square meters.

As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:

V = \frac{P}{5}

V = \frac{450\,kN}{5}

V = 90\,kN

The minimum allowable cross section area is cleared in the shearing stress equation:

A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}

If V = 90\,kN and \tau_{max} = 76.190\times 10^{3}\,kPa, the minimum allowable cross section area is:

A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}

A = 1.640\times 10^{-3}\,m^{2}

The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:

A = \frac{\pi}{4}\cdot D^{2}

The diameter is now cleared and computed:

D = \sqrt{\frac{4}{\pi}\cdot A}

D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})

D = 0.0457\,m

D = 45.7\,mm

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

5 0
3 years ago
How does the two-stroke Otto cycle differ from the four-stroke Otto cycle?
Digiron [165]

Answer:

Two stroke cycle                                               Four stroke cycle

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2.Complete the cycle in 2 stroke                 2.Complete the cycle in 4 stroke    

3.It have ports                                                3.It have vales

                                                                         

4.Greater requirement of cooling              4.Lesser requirement of cooling  

5.Less thermal efficiency                            5.High thermal efficiency

6.Less volumetric efficiency                       6.High volumetric efficiency    

7.Size of flywheel is less.                           7.Size of flywheel is more.

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3 years ago
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