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The little monkey ate four more peaches than the rest. After another one, he ate three times the rest. How many peaches did the little monkey have?
(4+1+1)÷(3-1)*(3+1)=12
The little monkey ate four more peaches than the rest. After another one, he ate three times as many peaches as the rest. How many peaches are there in all
Let x peaches be eaten at the beginning, then the total number = x + x-4 = 2X-4, leaving x-4,
If you eat one more, you eat x + 1, and the remaining x-4-5 = X-5
x+1=3(x-5)
x=8
Total number of peaches = 2X-4 = 12
There are 300 peaches. The big monkey takes 13 and the little monkey takes the rest 14. How many peaches did the little monkey take?
300 × [(1-13) × 14], = 300 × [23 × 14], = 300 × 16, = 50; a: the little monkey took 50 peaches
The vehicle starts to move with an acceleration of 1 m / S 2 at rest_____ m. What is the average velocity in the second second____ The displacement is_
Displacement s of the first 5 S
=1/2at²
=1/2×1×25
=12.5m
The displacement of the second second second
=Displacement in the first two seconds - displacement in the first second
=1/2at1²-1/2at2²
=1/2×1×4-1/2×1×1
=1.5m
Two complete square formula problems in the second grade of junior high school
How to use the complete square formula to solve 63 and 98?
First = (60 + 3) ^ 2
=60^2+60x2x3+3^2
=3600+360+9
=3969
Two of = (100-2) ^ 2
=100^2-2x100x2+2^2
=10004-400
=9604
As shown in the figure, the board with length L = 1m and mass m = 0.25kg is placed on a smooth horizontal plane, and the small block with mass m = 2kg (can be regarded as a particle) is located on the left side of the board
The dynamic friction coefficient between the board and the object block is μ = 0.1. The initial velocity V0 of the board to the left is suddenly given to be 2m / s, and a constant horizontal right pulling force F = 10N is applied to the small object block. After a period of time, the object block and the board are relatively static, and g = 10m / S2 is taken to calculate:
(1) The final position of the block on the board;
(2) The work done by the pulling force F in the above process
1. The friction between the small block and the board is FF = um, g = 0.1 * 2 * 10 = 2n,
So the acceleration of the block a object = (f-ff) / m object = 8 / 2 = 4m / S & # 178;,
The displacement of the block S1 = v0t + 1 / 2a, T & # 178; = 2T & # 178;,
The velocity of the block V1 = V0 + A, t = 4T;
The acceleration of the board is a wood = FF / M wood = 2 / 0.25 = 8m / S & #;,
The displacement of the board S2 = v0t + 1 / 2A wood T & # 178; = - 2T + 4T & # 178;,
The speed of wood is V2 = V0 + A, t = - 2 + 8t;
When the block and the board are relatively stationary, V1 = v2,
That is 4T = - 2 + 8t, t = 0.5s,
Then S1 = 2T & # 178; = 0.5,
S2=-2t+4t²=0,
The displacement of the small block relative to the board is △ s = S1-S2 = 0.5,
That is, the midpoint of the block on the board;
2. The work of pulling force F is w = FS1 = 10 * 0.5 = 5 J
How many kilograms is one kilogram? How many kilograms is one kilogram?
1 kg = 1 kg
1kg = 0.5kg
An object in the air is weighed 8N with a spring scale, and half of it in the water is weighed 6N with a spring scale to calculate the volume. Release the spring scale to calculate the buoyancy of the object in the water
Half buoyancy = 8-6 = 2n, water volume = 2 / 1 * 9.8 = 0.2 cubic decimeter, object volume = 2 * 0.2 = 0.4 cubic decimeter, density = 8 / 9.8 / 0.4 = 2.04 kg / cubic decimeter > water density, submergence buoyancy = gravity = 8N
The area of a triangle is 40cm2, the height is 5cm, and its bottom is () cm
A. 4B. 8C. 10D. 16
40 × 2 / 5, = 80 / 5, = 16 (CM), answer: its bottom is 16 cm, so choose: D
The object a with mass of 2.0kg standing on the horizontal ground is pulled by the spring along the horizontal direction, and the elongation of the spring is kept at 4.0cm. After the object slides for 2.0m, the velocity becomes 4.0m/s, and the dynamic friction factor between the object a and the ground is 0.2 (g = 10m / S2)
Please give me the formula and the specific steps,
The useful forces for an object are drag F and pull f (pull) (1) ∵ V (T) ^ 2-V (0) ^ 2 = 2As. Because the original object is still, V (0) = 0 ∵ V (T) ^ 2 = 2as4m / s * 4m / S = 2 * a * 2mA = 4m / S ^ 2 (2). According to Newton's second law, f = am = 4m / S ^ 2 * 2kg = 8nf = μ n = 0.2 * 2kg * 10N / kg = 4NF (pull) = F + F
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