The moving point a starts from the origin and moves along the negative direction of the number axis, while the moving point C on the number axis always moves along the number axis at a speed of 10 unit length / s Moving point a starts from the origin and moves in the negative direction along the number axis, while moving point B also starts from the origin and moves in the positive direction along the number axis. After 4 seconds, the two points are 20 unit lengths apart. It is known that the velocity ratio of moving point a and B is 2:3 (1). Calculate the velocity of two moving points (2). If two points a and B start from the positions marked in (1) at the same time, they move in the negative direction along the number axis according to the original velocity, A few seconds later, the origin is just in the middle of the two moving points (3) on the number axis, and the moving point C has been moving along the number axis at a speed of 10 unit length / second. When a and B start from the position marked in (1) and move in the negative direction of the number axis, the moving point will catch up with a and then return to B, and then reverse a when B meets. In this way, when B catches up with a, C stops moving immediately, How many unit lengths does point C move

1) Let the velocity of a be x, then the velocity of B be 1.5x
(1.5x+x)×4=20
The solution is x = 2 unit length per second
The velocity of B is 1.5 × 2 = 3 unit length per second
2) Let the distance between the two moving points be exactly in the middle after x seconds
(12-3x)-0=0-(-8-2x)
x=0.8s
3) Let B catch up with a in x seconds
3X-2X=20
X = 20 seconds
Then the unit length of C is 10 × 20 = 200
It is known that there are three points a, B and C on the number axis, which represent the rational numbers - 24, - 10 and 10 respectively. The moving point P starts from a and moves to the end point C at the speed of 1 unit per second. Let the moving time be T seconds. (1) the distance from P to point a and point C is expressed by an algebraic expression containing T: PA=______ ，PC=______ When the velocity from the starting point a to the end point B is the same as that from the starting point a to the end point B, can it immediately return to the starting point a and the end point B in the same speed unit? If yes, request the number represented by point p; if not, please explain the reason
(1) ∵ the moving point P starts from a and moves to the end point C at the speed of 1 unit per second. Let the moving time be T seconds, the distance from P to point a is pa = t, and the distance from P to point C is PC = (24 + 10) - t = 34-t; so the answer is t, 34-t; (2) when p is on the right side of Q and Q has not caught up with P, 3T + 2 = 14
A starts from the origin and moves to the negative direction of the number axis. At the same time, moving point B also moves from the origin to the positive direction of the number axis. Their speed is 1:4, after a period of time
Their speed is 1:4. After a period of time, A.B. returns at the same time. Q: when is the origin in the middle of A.B/
One point eight
It is known that there are three points a, B and C on the number axis, representing numbers - 24, - 10 and 10 respectively. Two electronic ants a and B are facing each other from two points a and C at the same time. The speed of a is 4 units / second, and that of B is 6 units / second. (1) which point on the number axis do a and B meet? (2) How many seconds later is the sum of the distances from a to a, B and C 40 units? If a turns back at this time, can a and B still meet on the number axis? If you can, find the meeting point; if not, please explain the reason
(1) Let a and B meet after x seconds, then 4x + 6x = 34 4, 4 × 3.4 = 13.6, - 24 + 13.6 = - 10.4, so a and B meet at - 10.4 on the number axis; (2) let the sum of the distances from a to a, B and C be 40 units after y seconds, the distance from B to a and C is 14 + 20 = 34 ＜ 40, the distance from a to B and C is 14 + 34 = 48 ＞ 40, and the distance from C to a and B is 34 + 20 = 54 ＞ 40, so a should be between AB and BC (2) between BCS: 4Y + (4y-14) + (34-4y) = 40, the solution is y = 5. ① when a moves from a to the right for 2 seconds, let y seconds later meet B. at this time, a and B represent the same point on the number axis, and the numbers are the same. The number of a is: - 24 + 4 × 2-4y; the number of B is: 10-6 × 2-6y, according to the meaning of the question: - 24 + 4 × 2-4y = 10-6 × 2-6y, the solution is: y = 7, The number represented by the meeting point is: - 24 + 4 × 2-4y = - 44 (or: 10-6 × 2-6y = - 44). ② when a moves from a to the right for 5 seconds, let y seconds later meet B. the number represented by a is: - 24 + 4 × 5-4y; the number represented by B is: 10-6 × 5-6Y. According to the meaning of the question, we can get: - 24 + 4 × 5-4y = 10-6 × 5-6Y, and the solution is: y = - 8 (not suitable for the meaning of the question), that is, when a moves from a to the right for 2 seconds, a can return with B on the number axis The number of the meeting point is - 44
It is known that there are three points a, B and C on the number axis, representing numbers - 24, - 10 and 10 respectively. Two electronic ants a and B are facing each other from two points a and C at the same time. The speed of a is 4 units / second, and that of B is 6 units / second. (1) which point on the number axis do a and B meet? (2) How many seconds later is the sum of the distances from a to a, B and C 40 units? If a turns back at this time, can a and B still meet on the number axis? If you can, find the meeting point; if not, please explain the reason
(1) Let a and B meet after x seconds, then 4x + 6x = 34, the solution is x = 3.4, 4 × 3.4 = 13.6, - 24 + 13.6 = - 10.4, so a and B meet at - 10.4 on the number axis; (2) let the sum of the distances from a to a, B, C after y seconds be 40 units, the distance between B and a, C is 14 + 20 = 34 ＜ 40, and the distance between a and B, C is 14 + 20 = 34 ＜ 40
Physical energy and electric energy conversion
There is a water pump with a mechanical efficiency of 70%, which needs to inject 46 tons of water into a 30 m high water tank every hour. How much is the working current of the motor (v = 380V cos & Oslash; = 0.8) matched with the water pump calculated?
P=((mgh)/η)/t={(46000*10*30)/0.7}/3600≈5476.2W
Also, P = UI * cos & Oslash;
Therefore, I = P / (U * cos & Oslash;) = 5476.2 / (380 * 0.8) ≈ 18.0a
Useful mechanical work per hour = 30g × 46 × 10 ^ 3 = 13524000j (g = 9.8)
Total power consumption of water pump P = useful mechanical work △ mechanical efficiency △ 1 hour (3600s) = 5366.67w
Working current = P / (VCOs & Oslash;) = 17.65a
If a is greater than zero, is a greater than 2A or less than or equal to 2A?
If a
A is less than 2A
Less than 2A: why?
How can I get 18 by adding, subtracting, multiplying and dividing five nines and brackets?
9×（9÷9+9÷9）=18
（9÷9+9÷9）×9=18
(9-9)×9＋9＋9=18
tana=2,sin^2a-2cos^2a+1=?
(sin ^ 2a-2cos ^ 2A + 1) divide the whole by 1
Make 1 sin ^ 2A + cos ^ 2A
Then the upper and lower expressions divide cos ^ 2A together
You can figure out the answer
The answer is 24
Use the universal exchange method!
A physics problem: why can power conversion work when charging a mobile phone
Why can power conversion work when charging a mobile phone?
Chemical energy, battery will learn later
electric energy
electric energy
Chemical energy! When used, the chemical energy of the battery is converted into electric energy again!
Because electricity turns into chemical energy
chemical energy
Chemical energy and internal energy

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