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In a cuboid basin with a length of CM, a width of CM, and a depth of CM, the water depth is 20 cm. If you put a cube stone with an edge length of 10 cm in the basin, it will be very difficult How many centimeters will basin reed water go up? It is 50 cm long 40 cm wide 30 cm deep
Complete the title to help you work it out
Rising height = (bottom area of cuboid × height of water depth + volume of cube) × bottom area of cuboid - height of water depth
（50×40×20 + 10×10×10）÷ (50×40）—20
= 41000÷2000 —20
=0.5 (CM)
There is 270 ml of water in a cuboid container. Put a cube iron block with 3 cm edge length into the water completely. At this time, the water depth is 6.6 cm. What is the original water depth
3 × 3 × 3 = 27 CC
270 ml = 270 CC
270 + 27 = 297 CC
297 △ 6.6 = 45 square centimeter
270 △ 45 = 6 cm
Q: chickens and rabbits are in the same cage. The number of chickens and rabbits is the same. The total number of legs of the two animals is 84. How many chickens and rabbits are there?
Because the numbers are the same, a chicken + a rabbit's feet = 6,
84 / 6 = 14
So, 14 chickens, 14 rabbits
Find the derivative of the function f (x) = sin (x),
The basic scheme of derivation is: 1) to find the increment of function Δ y = f (x0 + Δ x) - f (x0); 2) to find the average rate of change; 3) to take the limit to get the derivative
Δ x tends to 0 LIM (f (x + Δ x) - f (x)) / Δ X
After LIM (sin (x + Δ x) - SiNx) / Δ X and difference product:
=lim2(sin(Δx/2)cos（x+Δx/2)/Δx
=Limsin (Δ X / 2) / (Δ X / 2) * limcos (x + Δ X / 2) is applied here to limsin X / x = 1, X tends to 0
=1*cosx
=cosx
In △ ABC, a, B and C are its three sides. Try to compare the size of a 2 + B 2 + C 2 and 2 (AB + BC + AC)
A2 + B2 + C2 + C2-2 (AB + BC + BC + AC) = (a-b) 2 + c2-2c (a + b) (a-b-b-2c) (a-b-b-2c (a + b) (a-b-b-2c) (a + b) (a-b-b-2c) < C (a + b) (a-b) (a-b-b-2c) < C (a-b) 2 + C2 (a + b) (a-b) (a-b) (a-b-b-b-2c) (a-b-b-b-2c) (a-b-b-b) (a-b-b-b-b-2c) (a-b-b-b-b-b-b-b-2c) (a-b-b-b-b-b-b-b-b-b-b-b-2c) (a-b-b-b) (a-b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-2c(2) a 2 + B 2 + C 2 < 2 （ab+bc+ac）．
According to the speed effect of special relativity, if the spacecraft approaches the speed of light, will it do harm to its members?
According to Lorentz transformation, when the speed reaches 99.9% of the speed of light, the mass will increase by tens of times, and the volume will shrink by tens of times, and it seems that the speed will not be able to recover after slowing down. However, after such drastic changes in the astronaut's body, the person may have already died. However, some countries are still keen on "sub light speed spaceship", which does not consider the consequences at all?
I want to know whether the mass and volume of an object will recover if it moves at a speed close to the speed of light and then slows down. If it really can't recover, then the "subluminal spaceship" is not feasible at all. The reduction of the spaceship's volume will cause damage to the instrument, Astronauts may die if they gain weight or decrease in size (even "bloody") -- why did the "brickers" who proposed the "subluminal spaceship" so ignore the actual possible harm!
This is a misunderstanding of the theory of relativity. In fact, the increase of mass and the decrease of length are only the measurement appearances in another inertial reference frame. For the astronauts themselves, everything is as usual. For example, we can often see sublight particles, so for these particles, we are in the sublight spaceship - the earth. Do you see any flesh and blood flying?
Find the value of X: 7 × (x + 6) - 3x = 4 × (2x + 5); 153x + 5 = 2.5
7×（x+6）-3x=4×（2x+5）， 7x+42-3x=8x+20， 42+4x=8x+20， &nbs...
The perimeter of a rectangle is 26cm. The length of the rectangle is reduced by 1cm, and the width is increased by 2cm, forming a square. What is the area of the rectangle?
Let the original rectangle be x in length and Y in width
SO 2 (x + y) = 26
x-1=y+2
The solution is x = 8, y = 5
8*5=40(cm^2)
The area of this rectangle is 40 cm ^ 2
The radius of the wheel is 1.2m, the speed is 12m / s, and the angular speed of the wheel is 1.2m______ Rad / s, its period is______ s．
From the formula of linear velocity and angular velocity v = ω R, we get: ω = VR = 12m / s1.2m = 10s-1 = 10rad / S; from ω = 2 π T, we get t = 2 π ω = 2 π rad10rad / S = 0.2 π s; so the answer is: 10, 0.2 π
How to solve the equation 1.8 / 1.2-8x-1.3-3x / 2 = 5x-0.4/0.3
1.8/1.2-8x-1.3-3x/2=5x-0.4/0.3
3/2-8x-1.3-3x/2=5x-4/3
1.5-1.3+4/3=5x+8x+3x/2
0.2+4/3=13x+3x/2
1/5+4/3=26x/2+3x/2
3/15+20/15=29x/2
23/15=29x/2
x=23/15 X 2/29
x=46/435
RELATED INFORMATIONS
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