Is 6 square copper wire and 1.5 square copper wire OK?

Is 6 square copper wire and 1.5 square copper wire OK?

Absolutely!

A cuboid 40 cm long has a square cross section. If its length is increased by 5 cm, its surface area will be increased by 80 square cm
Finding the surface area and volume of the original cuboid

80 △ 5 = 16 cm
16 △ 4 = 4cm Side length of the base of a cuboid
The surface area of the original cuboid: 4 × 4 × 2 + 40 × 4 × 4 = 672 square centimeters
The volume of cuboid: 4 × 4 × 40 = 640 cubic centimeter

A cuboid with a square cross section is 40 cm long. If the length is increased by 5 cm, the surface area will be increased by 80 square percent
A cuboid with a square cross section is 40 cm long. If the length is increased by 5 cm, the surface area will be increased by 80 square percent

Examples of answers:
What is the circumference of the cross section of a cuboid
80 △ 5 = 16 (CM)
What is the width and height of a cuboid
16 △ 4 = 4 (CM)
What is the volume of a cuboid
40 × 4 × 4 = 640 (cm3)
I hope my answer is helpful to your study,
If you don't understand this question, you can ask,

400 fourth grade mental arithmetic questions
For example:
90*10=
Three lines on a page and fifty questions on a page. Please don't use the multiply sign as I did

640÷80=
15×5=
23×3=
12×2×5=
480÷80=
16×5=
27×3=
90÷15=
48÷4=
640÷16=
39÷3=
24×20=
32×3=
48÷16=
12×8=
27×3=
56÷14=
24÷8=
14×2=
83－45=
560÷80=
96÷24=
40÷20=
40×30=
37＋26=
76－39=
605＋59=
30×23=
12×8=
27＋32=
48＋27=
4500×20=
73＋15 =
120×600 =
200×360=
6800×400=
280＋270=
4×2500=
6000÷40=
5×1280=
310－70=
400×14=
470＋180=
1000÷25=
160×600=
20×420=
290×300=
8100÷300=
7600÷200=
7600÷400=
680＋270=
980÷14=
4200÷30=
6×1300=
1300×50=
200×48=
930－660=
530＋280=
9200÷400=
840÷21=
180×500=
8000÷500 =
1900÷20=
200×160=
8700÷300=
300×330=
3×1400=
7000÷14=
600÷12=
9600÷80=
140×300=
8800÷40=
9600÷800=
750－290=
5×490=
760×20=
7500÷500=
370×200=
650÷13=
8600－4200=
240×4=
640÷80=
15×10=
12×11=
160×30=
220×40=
104×5=
4500÷50=
120×2=
90÷30=
270÷30=
270×30=
84÷21=
76÷9=
66÷7=
100－54=
123＋15=
360÷4=
55÷5=
32×6= 7000÷70＝
200÷40＝
180÷30=
240÷40=
35×2=
140×7=
13×6=
280×3=
350×2=
50×11=
250×6=
7200+900=
410-201=
125×8=
48×20=
6600÷600=
390+140=
11×80=
24×50=
3600÷400= 4000÷50=
530－70=
420－90=
9600÷30=
7×700=
203+98=
1800÷300=
240+570=
4800÷400=
370+580=
580－490=
910－370=
25×8=
270－190=
36×2=
75÷25=
330÷11=
6×800=
5400÷9=
420÷60= 9×800=
3×330=
300×7=
9×500=
390÷13=
6300÷700=
5600÷700=
4800÷12=
3500÷7=
370＋560=
520＋490＝
450－90＝
80＋330＝
70×700＝
7000÷70＝
4000÷80＝
2400÷200＝
420－90＝
170＋320＝
1000－51＝ 520－260＝
910－190＝
35×200＝
22×200＝
8800÷400＝
9300÷300＝
6×300＝
1800÷200＝

The weight of the car and the passengers of an elevator is 1.2t, from one floor to seven floors in 10 seconds. If each floor is 3 meters high, what is the power of the elevator motor at least? （g=10N/kg）

M = 1.2t = 1.2 × 1000kg = 1200kg, g = mg = 1200kg × 10N / kg = 12000n. There are six floors from the first floor to the seventh floor, so the elevator running height is: H = 6 × 3M = 18m, w = GH = 12000n × 18m = 2.16 × 105jp = wt = 2.16 × 105j10s = 2.16 × 104w

If the number of items in the arithmetic sequence an is odd, A1 = 1, the sum of odd items in an is 175, and the sum of even items is 150, the tolerance of the arithmetic sequence can be obtained

The average values of odd and even terms are the same, and the odd term is one more than the even term,
The number ratio of the two items = 175:150 = 7:6, that is, there are 6 even items
In addition, except for the first item, the order of odd items has more "tolerance" values than that of even items,
Tolerance = (175-1-150) / 6 = 4

When driving on the expressway, the car first advances for 10s at the speed of 54km / h, and then uniformly accelerates for 15s at the acceleration of 0.6m/s
Find (1) the distance the car passes in 25 seconds (2) the instantaneous speed of the car at the end of the 20s (3) the average speed of the car in the first 20s

1、V0=54km/h=15m/s V2=V0+at2=15+0.6*15=24m/s S=S1+S2=V0t1+(V0+V2)t2 /2 =15*10+(15+24)15/2 =442.5m2、V3=V0+at3=15+0.6*10=21m/s3、S=S1+S3=V0t1+(V0+V3)t3...

Xiaogang and Xiaoyong start from two places by bike at the same time. Xiaogang meets Xiaoyong at 5 / 8 of the whole race. Xiaoyong continues to advance at the speed of 10km / h, and runs the rest of the race in 2.5 hours to find Xiaogang's speed
Use the method of primary school, don't compare, because you haven't learned yet, please write your analysis

Xiaogang met Xiaoyong when he finished five eighths of the whole race, so Xiaoyong ran 1-5 / 8 = 3 / 8 of the whole race when he met. The speed ratio of the two was (5 / 8): (3 / 8) = 5:3, so Xiaogang's speed was: 10 * 5 / 3 = 50 / 3km / h. Xiaogang met Xiaoyong when he finished 5 / 8 of the whole race, so Xiaoyong ran 5 / 8 of the whole race after meeting: 10 * 2

Given that the equation | x | = ax + 1 has a negative root but no positive root, the value range of a is ()
A. A ≥ 1b. A ＜ 1C. - 1 ＜ a ＜ 1D. A ＞ - 1 and a ≠ 0

∵ the equation | x | = ax + 1 has a negative root but no positive root, ∵ x ＜ 0, the equation is changed to: - x = ax + 1, X (a + 1) = - 1, x = − 1A + 1 ＜ 0, ∵ a + 1 ＞ 0, ∵ a ＞ - 1, and a ≠ 0. If x ＞ 0, | x | = x, x = ax + 1, x = 11 − a ＞ 0, then 1-A ＞ 0, the solution is a ＜ 1. ∵ a ＜ 1 is not established without positive root, ∵ a ＞ 1. Therefore, select a

On the number axis, points a and B represent the rational number a and B respectively. The origin o is just the midpoint of AB, and the value of 1 / 2 of 2009a × 2010b is calculated

a+b=0 a=-b a/b=-1 2009a/2010b=-2009/2010