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Is the sliding friction related to the speed? Thank you for your help!
Yes, it's generally thought to be inversely proportional to the square of the velocity
Let the ten digit number be x and the single digit number be y, X-Y = 3 4 / 7 (10x + y) = 10Y + X
X-Y=3 X=3+Y
Substituting x = 3 + y into 4 / 7 (10x + y) = 10Y + X leads to
4/7(30+10Y+Y)=10Y+3+Y
It is reduced to 33y = 99 y = 3
X=3+Y=3+3=6
X=6 Y=3
The number is 63
The distance between a and B is 40km, the speed of motorcycle is 45km an hour, the speed of truck is 35km an hour, the two vehicles start from a place at the same time, the speed of motorcycle is 45km an hour
Arrive at B several hours earlier than the truck
40 △ 35-40 △ 45 = 16 / 63 ≈ 0.25 (H)
Two integers a, B, a × B = (a + b) a, a + B = a × B + 1, find [(- 2) × (- 5)] + (- 4)
Solution
a×b=(a+b)a
∴(-2)×(-5)
=[(-2)+(-5)]×(-2)
=(-7)×(-2)
=14
∴[(-2)×(-5)]+(-4)=14+(-4)
a+b=a×b+1
∴14+(-4)=14×(-4)+1=-55
The original formula = - 55
Calculation & nbsp; 9.8 + 99.8 + 999.8 + 9999.8 + 99999.8=______ ．
9.8 + 99.8 + 999.8 + 9999.8 + 99999.8 = (9.8 + 0.2) + (99.8 + 0.2) + (999.8 + 0.2) + (9999.8 + 0.2) + (99999.8 + 0.2) - 0.2 × 5 = 10 + 100 + 1000 + 10000 + 100000-0.2 × 5 = 111110-1 = 111109; so the answer is: 111109
The total weight of two tons of sand is 1757 kg. The first pile uses 60%, and the second pile uses 574 kg. The remaining sand is as much as two tons
x.(1-60%)=(1757-x)-574
X = 845 first pile
1757-x = 912 reactor 2
What is the meaning of the square of the algebraic formula A-B?
(2) Find the two odd numbers of 2x-1. (3) what is the original price of a commodity if the price after 10% discount is a yuan? (4) the distance between a and B is s kilometers. If someone plans to arrive in an hour, if he wants to arrive 2 hours in advance, he has to go every hour?
(1) (2) 2x - 3, 2x + 1. (3) a divided by 90%
(4) S divided by (A-2) I don't know if it's right or wrong,
The solution of A4 (b2-c2) + B4 (c2-a2) + C4 (A2-B2)
The factorization of
The original formula = a4b2-a4c2 + b4c2-b4a2 + c4a2-c4b2 = a2b2 (A2-B2) - C2 (b4-a4) + C4 (A2-B2) = a2b2 (a + b) (a-b) - C2 (B2 + C2) (B + a) (B-A) + C4 (a + b) (a-b) = (a + B) (a-b) (a2b2-b2c2 + C4) = (a + b) (a-b) (a-b) (b2-c2) (a2-c2) = (a + b) (a-b) (B + C) (a + C) = (a + b) (A-C) = (a + b) (B-C) = (a + C) = (a + b) (A-C) = (a-b-b-c) = (a + B-C) = (A-C) = (a-b-c) = (a-b-c) = (A-C) = (a-C
The store bought a batch of fruit, which sold 30% on the first day and 120 kg on the second day. It sold 20% more than the first day. How many kg are these fruits?
Because the first day sold 30%, the next day sold 120 kg. Sold 20% more than the first day
So the next day they sold 30% + 30% × 20% = 36%
So 6% of the fruit is 120 kg
120 △ 6% = 2000 kg
This batch of fruit is 2000 kg
The three apprentices cooperate to process a batch of parts in five days. Apprentice a completes 12 of apprentice B's and apprentice B completes 12 of master's. If apprentice a works alone for two days, how many days will apprentice B cooperate with master to complete the rest of the work?
The efficiency of apprentice a: the efficiency of apprentice B: the efficiency of master is 1:2:4. The total number of shares: 1 + 2 + 4 = 7. The efficiency of apprentice a: 15 × 17 = 135; the efficiency of apprentice B and master: 15 × 2 + 47 = 15 × 67 = 635; 1-135 × 2 = 1-235 = 33353335 △ 635 = 5.5 (days) a: it will take 5.5 days to complete ．
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