A 150 cubic centimeter iron block is completely immersed in a cylindrical container containing water. The diameter of its bottom is 10 cm. How much has the water surface risen People's education sixth grade matching exercise book page 12

A 150 cubic centimeter iron block is completely immersed in a cylindrical container containing water. The diameter of its bottom is 10 cm. How much has the water surface risen People's education sixth grade matching exercise book page 12


Water level rise = 150 ÷ (5 * 5 * 3.14) ≈ 1.91cm



The diameter of the bottom surface of a cylindrical glass container is 8 cm. Take a piece of iron from the water of the container, and the water drops by 3 cm. What's the volume of this piece of iron
kuai


V = π R ^ 2 * H = 3.14 * 4 ^ 2 * 3 = 150.72 CC



The edge of a cube container is 2 decimeters long. The container contains water. When a stone sinks into the water, the water in the container rises by 2 cm. What is the volume of the stone
Cubic centimeter?
Less cubic centimeter?


2 decimeters = 20 cm
The volume of stone is 20 × 20 × 2 = 800 cubic centimeters



How many microamperes is 0.6 a? How many microamperes is 70 ma? How many microamperes is 750 ma?


0.6A=600mA
70 Ma = 70000 micro a = 0.07 a
750 μ a = 0.75 Ma = 0.00075 a



Which one of balance or spring dynamometer can't be used in space


Balance
Because the balance is an equal arm lever, it uses the lever balance condition, and the pressure of the object on the tray is equal to the pressure of the weight on the tray,
Objects in space are completely weightless and will not exert corresponding pressure on the tray
The spring dynamometer can measure the pulling force, but it can't measure the force related to gravity



If the sum of length and width of a cuboid is 14cm, the sum of width and height is 11cm, and the sum of length and height is 13cm, its surface area is? Cm square,
What about the volume?


Length + width = 14cm,
Width + height = 11cm,
Length + height = 13cm,
Length = 8cm width = 6cm height = 5cm
So the surface area is s = 2 (8 * 6 + 8 * 5 + 6 * 5) = 2 (48 + 40 + 30) = 2 * 118 = 236 square centimeters
The volume is 8 * 6 * 5 = 240 cubic centimeters



As shown in the figure, one end of a spring is fixed on the ceiling, and the other end of B spring is fixed on the horizontal ground. When a weight g is hung on the other end of a and a weight g is pressed on the other end of B, the length of the two springs is L. now the two springs are connected in parallel and a weight g is tied under them. At this time, the length of the spring should be ()
A. L+G2kB. L+GkC. L-G2kD. L-Gk


(1) Let the stiffness coefficient of the spring be K. according to the expression of the spring length, if the length of the spring is L = l0 + x = l0 + FK, then the length of the first spring is L = l0 + GK, then the original length is l0 = l-gk; if the length of the second spring is L = l0 GK, then the original length is l0 = L + GK; (2) if the length of the first spring and the second spring are set to be s, then the elongation of the first spring is X1 = s - (l-gk), and the tensile force is kx1 = k [S - (l-gk)]; the elongation of the second spring is X 2 = s - (L + GK), producing tension kx2 = k [S - (L + GK)]; the sum of the tension of a and B springs should be equal to g, that is, K [S - (l-gk)] + K [S - (L + GK)] = g. the length of spring s = L + g2k is obtained by solving the equation



Give you four natural numbers 3,5,4,7. Through four operations, you can exchange the position of numbers and add operation symbols at will, but each number can only be used once. Write a comprehensive formula, and the result is equal to 24


Four times five plus seven minus three



Under the action of pulling force F, an object with mass m moves uniformly along the horizontal plane to the right?


The object with mass m moves uniformly along the horizontal direction to the right under the action of tension F, so the force is balanced
Friction f = tension f
f=uMg
So UMG = F
u=F/Mg
The dynamic friction factor between M and horizontal plane is f / mg



Let Sn be the sum of the first n terms of the arithmetic sequence an, if S3 / S7 = 1 / 3, then S6 / S7=


S3/S7=1/3
3﹙3a1+3q﹚=7a1+21q
a1=6q
S6/S7=﹙6a1+15q﹚/﹙7a1+21q﹚=51/63=17/21