Helpppp!!! Used to secure lights to lighting bars and can be used when working alone to

Oleg is using a multimeter to test the circuit branch you just installed. After turning off the current to the circuit at the se

irakobra [83]

Answer:

Option D, Ground Wire

Explanation:

Oleg is a mustimeter or multi tester or a basic voltage tester which is used to test the ground. The main purpose of the tester is to ensure that the outlet is connected to the ground of the circuit, thereby ensuring its proper functioning.

Hence, option D is correct

7 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

Select the level of education that is best demonstrated in each example.

Nastasia [14]

Answer:

masters

associate

bachelors

Explanation:

8 0

3 years ago

Read 2 more answers

Air exits a compressor operating at steady-state, steady-flow conditions at 150 oC, 825 kPa, with a velocity of 10 m/s through a

ioda

Answer:

a) Qe = 0.01963 m^3 / s , mass flow rate m^ = 0.1334 kg/s

b) Inlet cross sectional area = Ai = 0.11217 m^2 , Qi = 0.11217 m^3 / s

Explanation:

Given:-

- The compressor exit conditions are given as follows:

Pressure ( Pe ) = 825 KPa

Temperature ( Te ) = 150°C

Velocity ( Ve ) = 10 m/s

Diameter ( de ) = 5.0 cm

Solution:-

- Define inlet parameters:

Pressure = Pi = 100 KPa

Temperature = Ti = 20.0

Velocity = Vi = 1.0 m/s

Area = Ai

- From definition the volumetric flow rate at outlet ( Qe ) is determined by the following equation:

Qe = Ae*Ve

Where,

Ae: The exit cross sectional area

Ae = π*de^2 / 4

Therefore,

Qe = Ve*π*de^2 / 4

Qe = 10*π*0.05^2 / 4

Qe = 0.01963 m^3 / s

- To determine the mass flow rate ( m^ ) through the compressor we need to determine the density of air at exit using exit conditions.

- We will assume air to be an ideal gas. Thus using the ideal gas state equation we have:

Pe / ρe = R*Te

Where,

Te: The absolute temperature at exit

ρe: The density of air at exit

R: the specific gas constant for air = 0.287 KJ /kg.K

ρe = Pe / (R*Te)

ρe = 825 / (0.287*( 273 + 150 ) )

ρe = 6.79566 kg/m^3

- The mass flow rate ( m^ ) is given:

m^ = ρe*Qe

= ( 6.79566 )*( 0.01963 )

= 0.1334 kg/s

- We will use the "continuity equation " for steady state flow inside the compressor i.e mass flow rate remains constant:

m^ = ρe*Ae*Ve = ρi*Ai*Vi

- Density of air at inlet using inlet conditions. Again, using the ideal gas state equation:

Pi / ρi = R*Ti

Where,

Ti: The absolute temperature at inlet

ρi: The density of air at inlet

R: the specific gas constant for air = 0.287 KJ /kg.K

ρi = Pi / (R*Ti)

ρi = 100 / (0.287*( 273 + 20 ) )

ρi = 1.18918 kg/m^3

Using continuity expression:

Ai = m^ / ρi*Vi

Ai = 0.1334 / 1.18918*1

Ai = 0.11217 m^2

- From definition the volumetric flow rate at inlet ( Qi ) is determined by the following equation:

Qi = Ai*Vi

Where,

Ai: The inlet cross sectional area

Qi = 0.11217*1

Qi = 0.11217 m^3 / s

- The equations that will help us with required plots are:

Inlet cross section area ( Ai )

Ai = m^ / ρi*Vi

Ai = 0.1334 / 1.18918*Vi

Ai ( V ) = 0.11217 / Vi .... Eq 1

Inlet flow rate ( Qi ):

Qi = 0.11217 m^3 / s ... constant Eq 2

6 0

4 years ago

Army people are good people right

Phantasy [73]

Answer: Yes army people are good people but it also depends on how you fraze that some have been in trouble before but it doesnt mean there bad people we all make mistakes

5 0

3 years ago