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DedPeter [7]
3 years ago
8

9 b. A sign (including the post and base) weighs 40 pounds and is

Engineering
1 answer:
tigry1 [53]3 years ago
8 0

Answer:

The weight should be added to the base of the sign to keep it from tipping is 65.6 lb

Explanation:

Given data:

A sigh weighs 40 pounds

Suported by an 18 in x 18 in square

Force of the wind 13.2 lb

Questions: Will the sign tip over, if yes, how much evelnly distributed weight should be added to the base of the sign to keep it from tipping, W = ?

The sign and the post have a length of 6 ft. You need to calculate the distance from the edge to the middle point:

18/2 = 9 in = 0.75 ft

Force acting in the base (40 lb):

F=\frac{40*0.75}{6} =5lb

The weight should be added to the base:

(40+W)*\frac{0.75}{6} =13.2\\W=65.6lb

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If i build thing a and thing a builds thing b did i build thing b
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Answer:

<h3>Yes</h3>

Explanation:

If you build thing "a" and thing "a" builds thing "b" you <u>indirectly</u> build thing "b".

3 0
3 years ago
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B1) 20 pts. The thickness of each of the two sheets to be resistance spot welded is 3.5 mm. It is desired to form a weld nugget
kap26 [50]

Answer:

minimum current level required =  8975.95 amperes

Explanation:

Given data:

diameter = 5.5 mm

length = 5.0 mm

T = 0.3

unit melting energy = 9.5 j/mm^3

electrical resistance = 140 micro ohms

thickness of each of the two sheets = 3.5mm

Determine the minimum current level required

first we calculate the volume of the weld nugget

v = \frac{\pi }{4} * D^2 * l = \frac{\pi }{4} * 5.5^2 * 5 = 118.73 mm^3

next calculate the required melting energy

= volume of weld nugget * unit melting energy

= 118.73 * 9.5 = 1127.94 joules

next find the actual required electric energy

= required melting energy / efficiency

= 1127 .94 / ( 1/3 )  = 3383.84 J

TO DETERMINE THE CURRENT LEVEL REQUIRED  use the relation below

electrical energy =  I^2 * R * T

3383.84 / R*T = I^2

3383.84 / (( 140 * 10^-6 ) * 0.3 ) = I^2

therefore  8975.95 = I ( current )

4 0
3 years ago
15. Whether technology is good or bad depends on how it is used.
Reptile [31]

Answer:

true

Explanation:

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2 years ago
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3 years ago
Ammonia enters the expansion valve of a refrigeration system at a pressure of 1.4 MPa and a temperature of 32degreeC and exits a
AveGali [126]

Answer:

the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %

Explanation:

given data

pressure p1 = 1.4 MPa = 14 bar

temperature t1 = 32°C

exit pressure = 0.08 MPa = 0.8 bar

to find out

the quality of the refrigerant exiting the expansion valve

solution

we know here refrigerant undergoes at throtting process so

h1 = h2

so by table A 14 at p1 = 14 bar

t1 ≤ Tsat

so we use equation here that is

h1 = hf(t1) = 332.17 kJ/kg

this value we get from table A13

so as h1 = h2

h1 = h(f2)  + x(2) * h(fg2)

so

exit quality  = \frac{h1 - h(f2)}{h(fg2)}

exit quality  = \frac{332.17- 9.04}{1382.73)}

so exit quality = 0.2337 = 23.37 %

the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %

5 0
3 years ago
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