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A vertical projectile moves in a straight line with constant velocity in both ascending and descending stages, and its acceleration is equal to that of a free falling body The initial velocity of the straight upward motion is 30 meters per second. After 2 seconds, 3 seconds and 4 seconds, what is the displacement of the object? What is the distance? What is the velocity at the end of each second?
Take upward as positive direction, initial velocity is 30m / s, acceleration is - g = - 10m / S ^ 2
After 2S, displacement: S = v0t + 0.5at ^ 2 = 30 * 2-5 * 4 = 40m, velocity: v = V0 + at = 30-10 * 2 = 10m / s
After 3S, displacement: S = v0t + 0.5at ^ 2 = 30 * 3-5 * 9 = 45m, velocity: v = V0 + at = 30-10 * 3 = 0m / s
After 4S, the displacement is s = v0t + 0.5at ^ 2 = 30 * 4-5 * 16 = 40m, the velocity is v = V0 + at = 30-10 * 4 = - 10m / s, that is, the direction is downward, and the size is 10m / s
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How to prove that the difference of displacement is equal in the same time t for the uniform linear motion with acceleration a
Let the initial velocity of a period of time t be V1 and the final velocity V2, and then the initial velocity V2 and the final velocity V3 in the same period of time
V2-at=V1 V3=V2+at
From V2 * 2 to V1 * 2 = 2as1
V3*2-V2*2=2aS2
S2-s1 = at * 2 is a certain value
How to prove that the time ratio of equal displacement of uniform speed change linear motion with zero initial velocity is 1: √ 2-1: √ 3 - √ 2
Let the displacements from the beginning of the motion be x, 2x, 3x
According to x = 1 / 2A * T ^ 2
T1:T2:T3=1:√2:√3
Then the time ratio of moving the same displacement is 1: √ 2-1: √ 3 - √ 2
It is known that a multiplied by 7 / 3 equals B multiplied by 5 / 12 equals C multiplied by 1.25 equals D multiplied by 9 equals A. please arrange ABCD in order from big to small
Bcad, of course
The known formula ABCD efgh = 1996, where ABCD and efgh are 4 digits; ABCD efgh is eight different numbers in 0.1.2.3.4.5.6.7.8.9
What's the difference between the maximum and minimum of the sum of ABCD and efgh?
The maximum value: 14998, the minimum value: 5000, because: efgh + 1996 = ABCD, individual H + 9 can't carry, otherwise 10 bits g = C; similarly, 10 bits G + 9 can't carry, otherwise 100 bits f = B. then G can only take 0, C can only take 9. So, the maximum is efgh = 6501, ABCD = 8497, efgh + ABCD = 6501 + 8497 = 14998; minimum
Given that a, B, C and D are mutually unequal integers, and ABCD = 25, what is the value of a + B + C + D?
a. B, C, D are 5, - 5,1, - 1, respectively
They add up to zero
Given that a.b.c.d is an unequal integer and ABCD = 25, find the value of a + B + C + D
-5 * - 1 * 1 * 5 = 25 add up to 0, of course
If the integers a, B, C and D satisfy ABCD = 25 and a > b > C > D, then | a + B | + | C + D | equals ()
A. OB. 10C. 2D. 12
25 = 5 × 5 × 1 × 1 = 5 × (- 5) × 1 × (- 1), then a = 5, B = 1, C = - 1, d = - 5, | a + B | + | C + D | = | 5 + 1 | + | - 1-5 | = 6 + 6 = 12
In the cube abcd-a.b.c.d., the edge length is proved to be a. plane ab.d. / c.bd
It is suggested that BD / / B.D. and ad. / / BC
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