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3 years ago

What happens if your son makes a spark in a outlet and then the room goes dark

Engineering

1 answer:

Kitty [74]3 years ago

3 0

Answer:

He probably tripped the wiring, when metal hits the electricity it creates a reaction that burns the wiring.

Explanation:

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Calculate the Lee for the same wing if we increase the span to 0.245 m. By increasing the span we also increase the glider weigh

alexira [117]

Answer:

0.21

Explanation:

This would have been a fairly easy one, except for that the first part of the question is missing, and as such, I'd assume a value.

We need to use chord, so, I'm assuming the length of the chord to be 0.045 m

The Area is given by the formula

Area = span * chord

Area = 0.245 * 0.045

Area = 0.011 m²

This area gotten, is what we then divide the glider weight by to get our answer.

Lee = area / weight

Lee = 0.011 / 0.0523

Lee = 0.21

Therefore, using the values of the chord I'd assumed, the Lee of the same wing is 0.21

8 0

3 years ago

Read 2 more answers

Provide an argument justifying the following claim: The average (as defined here) of two Java ints i and j is representable as a

ahrayia [7]

Answer:

public static int average(int j, int k) {

return (int)(( (long)(i) + (long)(j) ) /2 );

}

Explanation:

The above code returns the average of two integer variables

Line 1 of the code declares a method along with 2 variables

Method declared: average of integer data type

Variables: j and k of type integer, respectively

Line 2 calculates the average of the two variables and returns the value of the average.

The first of two integers to average is j

The second of two integers to average is k

The last parameter ensures average using (j+k)/2

3 0

3 years ago

Three ways to advertise for AVID

lilavasa [31]

Answer:

newspaper, radio, televison

Explanation:

had avid in 7th :)

8 0

3 years ago

Read 2 more answers

For a bolted assembly with six bolts, the stiffness of each bolt is kb=Mlbf/in and the stiffness of the members is km=12Mlbf/in.

makkiz [27]

Answer:

nP ≈ 4.9

nL = 1.50

Explanation:

GIVEN DATA

external load applied (p) = 85 kips

bolt stiffness ( Kb ) = 3(10^6) Ibf / in

Member stiffness (Km) = 12(10^6) Ibf / in

Diameter of bolts ( d ) = 1/2 in - 13 UNC grade 8

Number of bolts = 6

assumptions

for unified screw threads UNC and UNF

tensile stress area ( A ) = 0.1419 in^2

SAE specifications for steel bolts for grade 8

we have

Minimum proff strength ( Sp) = 120 kpsi

Minimum tensile strength (St) = 150 Kpsi

Load Bolt (p) = external load / number of bolts = 85 / 6 = 14.17 kips

Given the following values

Fi = 75%* Sp*At = (0.75*120*0.1419 ) = 12.771 kip

Preload stress

αi = 0.75Sp = 0.75 * 120 = 90 kpsi

stiffness constant

C = $\frac{Kb}{Kb + Km}$ = $\frac{3}{3+2}$ = 0.2

A) yielding factor of safety

nP = $\frac{sPAt}{Cp + Fi}$ = $\frac{120* 0.1419}{0.2*14.17 + 12.771}$

nP = 77.028 / 15.605 = 4.94 ≈ 4.9

B) Determine the overload factor safety

$nL = \frac{SpAt - Fi}{CP}$ = ( 120 * 0.1419) - 12.771 / 0.2 * 14.17

= 17.028 - 12.771 / 2.834

= 1.50

3 0

3 years ago

Who works alongside and assists the engineers?

nika2105 [10]

Answer:

Assistants works alongside and assists the engineers.

5 0

3 years ago