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4 years ago

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true or false incident reports, such as situation reports and status reports enhance situational awareness and ensure that perso

nnel can access needed information.

1 answer:

Degger [83]4 years ago

4 0

True becauseidbajqkfkmxzmbakwlgof

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The electron beam in a TV picture tube carries 1015 electrons per second. As a design engineer, determine the voltage needed to

leonid [27]

Answer:

The voltage needed to accelerate the electron beam is 2.46 x 10^16 Volts

Explanation:

The rate of electron flow is given as:

q = 1015 electrons per second

The total current is given by:

Total Current = (Rate of electron flow)(Charge on one electron)

Total Current = I = (1015 electrons/s)(1.6 x 10^-19 C/electron)

I = 1.624 x 10^-16 A

Now, we know that electric power is given as:

Electric Power = Current x Voltage

P = IV

V = P/I

V = 4 W/1.624 X 10^-16 A

<u>V = 2.46 x 10^16 Volts</u>

6 0

3 years ago

Teaching how to characterize and implement high speed power devices for tomorrow's engineers

Simora [160]

Answer: Teaching how to characterize and implement high speed power devices for tomorrow's engineers

Explanation:

6 0

3 years ago

For a bolted assembly with eight bolts, the stiffness of each bolt is kb = 1.0 MN/mm and the stiffness of the members is km = 2.

rjkz [21]

Answer:

a) 0.978

b) 0.9191

c) 1.056

d) 0.849

Explanation:

Given data :

Stiffness of each bolt = 1.0 MN/mm

Stiffness of the members = 2.6 MN/mm per bolt

Bolts are preloaded to 75% of proof strength

The bolts are M6 × 1 class 5.8 with rolled threads

Pmax =60 kN, Pmin = 20kN

<u>a) Determine the yielding factor of safety</u>

$n_{p} = \frac{S_{p}A_{t} }{CP_{max}+ F_{i} }$ ------ ( 1 )

Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N

Input the given values into the equation above

equation 1 becomes ( np ) = $\frac{380*20.1}{0.277*7500*5728.5}$ = 0.978

note : values above are derived values whose solution are not basically part of the required solution hence they are not included

<u>b) Determine the overload factor of safety</u>

$n_{L} = \frac{S_{p}A_{t}-F_{i} }{C(P_{max} )}$ ------- ( 2 )

Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N

input values into equation 2 above

hence : $n_{L}$ = 0.9191 $n_{L} = 0.9191$

<u>C) Determine the factor of safety based on joint separation</u>

$n_{0} = \frac{F_{i} }{P_{max}(1 - C ) }$

Fi = 5728.5 N, Pmax = 7500 N, C = 0.277,

input values into equation above

Hence $n_{0}$ = 1.056

<u>D) Determine the fatigue factor of safety using the Goodman criterion.</u>

nf = 0.849

attached below is the detailed solution .

4 0

3 years ago

(a) Aluminum foil used for storing food weighs about 0.3 grams per square inch. How many atoms of aluminum are contained in one

atroni [7]

Answer:

note:

solution is attached due to error in mathematical equation. please find the attachment

3 0

3 years ago

A balanced three-phase inductive load is supplied in steady state by a balanced three-phase voltage source with a phase voltage

Kobotan [32]

Answer:

Following are the solution to the given question:

Explanation:

Line voltage:

$V_L=\sqrt{3}V_{ph}=\sqrt{3}(120) \ v$

Power supplied to the load:

$P_{L}=\sqrt{3}V_{L}I_{L} \cos \phi$

$10\times 10^3=\sqrt{3}(120 \sqrt{3}) I_{L}\ (0.85)\\\\I_{L}= 32.68\ A$

Check wye-connection, for the phase current:

$I_{ph}=I_L= 32.68\ A$

Therefore,

Phasor currents: $32.68 \angle 0^{\circ} \ A \ ,\ 32.68 \angle 120^{\circ} \ A\ ,\ and\ 32.68 -\angle 120^{\circ} \ A$

Magnitude of the per-phase load impedance:

$Z_{ph}=\frac{V_{ph}}{I_{ph}}=\frac{120}{32.68}=3.672 \ \Omega$

Phase angle:

$\phi = \cos^{-1} \ (0.85) =31.79^{\circ}$

Please find the phasor diagram in the attached file.

8 0

3 years ago