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Pavlova-9 [17]
2 years ago
11

The most important rating for batteries is the what

Engineering
1 answer:
kifflom [539]2 years ago
7 0

Answer:

I'm completely sure that the answer is: The most important rating for batteries is the ampere-hour rating. Ampere-hour is the battery discharge rating. It's used as a measure of charge in your device. It indicates how long your device will work without charging.

Explanation:

Hope this helped!

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What major advancement in machine tools occurred in the 1970s and what benefits did it provide? describe in your own words.
mixer [17]

Answer:

I'm just a seventh grader

4 0
3 years ago
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A center-point bending test was performed on a 2 in. x d in. wood lumber according to ASTM D198 procedure with a span of 4 ft an
Zigmanuir [339]

Answer:

3.03 INCHES

Explanation:

According to ASTM D198 ;

Modulus of rupture = ( M / I ) * y  ----- ( 1 )

M ( bending moment ) = R * length of span / 2

                                     = (120 * 10^3 ) * 48 / 2 = 288 * 10^4 Ib-in

I ( moment of inertia ) = bd^3 / 12

                                    = ( 2 )*( d )^3  / 12 =  2d^3 / 12

b = 2 in ,  d = ?

length of span = 4 * 12 = 48 inches

R = P  / 2 =  240 * 10^3 / 2 =   120 * 10^3 Ib

y ( centroid distance ) = d / 2  inches

back to equation ( 1 )

( M / I ) * y

940.3 ksi = ( 288 * 10^4 / 2d^3 / 12 ) * d / 2

                = ( 288 * 10^4 * 12 ) / 2d^3 )  * d / 2

940300  = 34560000* d / 4d^3

4d^3 ( 940300 ) = 34560000 d  ( divide both sides with d )

4d^2 = 34560000 / 940300

d^2 = 9.188   ∴ Value of d ≈ 3.03 in

8 0
3 years ago
Which of the following would not be considered hot work? A chipping B soldering C
tankabanditka [31]
I believe the answer is D: brazing
Hope this helps you have a good night
5 0
2 years ago
A residential heat pump has a coefficient of performance of 1.49 How much heating effect, in kJ/h, will result when 4 kW is supp
AfilCa [17]

Answer:

21.456 kJ/h

Explanation:

See the figure attached. In this case  

W_{cycle} = 4 kW

Q_{out} = \text{heating effect}

Coefficient of performance in heat pump is defined by

COP = \frac{Q_{out}}{W_{cycle}}

Q_{out} =COP*W_{cycle}

Q_{out} =1.49*4 \, W

Q_{out} = 5.96 \, W

Now it is necessary to change units, remember that Watt (W) is defined as J/s

Q_{out} = 5.96 \frac{J}{s} \frac{3600s}{1 h} \frac{1 kJ}{1000 J}

Q_{out} = 21.456 \frac{kJ}{h}

3 0
3 years ago
SPELL HELLO PLZ I NEED HELP BIG TIME
QveST [7]

Answer:

hello

Explanation:

8 0
3 years ago
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