The iron ball with the volume of 5DM & # 179; weighs 39.2n. What is the buoyancy of the iron ball after immersion in water? (the density of iron = 7.9 * 10 & # 179; kg / M & # 179;)

When immersed, buoyancy f = ρ water GV row = 1.0 × 10 & # 179; kg / M & # 179; * 9.8N / kg * 0.005M & # 179; = 49n
G=39.2N
F floating ＞ g, so floating when floating is still, f floating '= g = 39.2n

There is a 7n iron ball. When it is immersed in water, what is the buoyancy? G = 10N / kg

Is that how volume is calculated

Two solid balls, the mass of the wooden ball is half of the iron ball, the radius of the wooden ball is twice of the iron ball, then the density of the wooden ball is ()
I know the answer is 1 / 16. How to solve similar problems

First of all, we need to use two formulas. One is the mass formula, density = mass divided by volume, ρ = m / v. M = ρ v. v = m / ρ, in which we use the formula of ρ = m / v. the other is the spherical volume formula, v = 4 / 3 π R ^ 3 (Four Thirds times π times the third power of R)
We assume that the mass of the wooden ball M1, the density ρ 1, the iron ball m2, ρ 2, the wooden ball R1, the iron ball R2
According to the data you provided, list a proportion test M1 = 1 / 2M2 R1 = 2r2
The volume of the sphere gives v 1 = 3 / 4 π R 1 ^ 3
V2 = 3 / 4 π R2 ^ 3 (Note: R1 = 2r2, so V2 expressed by V1 is V2 = 4 / 3 π 0.5r1 ^ 3 (Four Thirds times π times the power of 0.5r). M2 uses M1 instead of M2 = 2M1
ρ 1 / ρ 2 = m1v2 / m2v1 = M1 times 4 / 3 times 0.5 times R1 to the third power / 2M1 times 4 / 3 times R1 to the third power
In the end, you will get the result of your calculation
If you don't understand, please reply me,

What is the earth's orbit?
Ecliptic
Bai Dao
Silver Road
red line

The zodiac,

I really can't write. Please help me,
【1】 Xiao Ming went out to travel and bought a Yixing teapot. He heard that Yixing teapot was made of Yixing's unique clay material, and wanted to know the density of this material. So he measured the mass of the lid with a balance to be 44.4g, and then put the lid into an overflow cup filled with water, and measured the mass of the overflow water to be 14.8g
What is the density of this material?
If the mass of the whole empty teapot is 159G, what is the volume of the material used in the teapot?
【2】 Haibao is the mascot of the 2010 World Expo. Among the various Haibao ornamenting the streets of Shanghai, there is a Haibao [uniform and solid] with a mass of 3.0 * 10 cubic kg, a density of 1.5 * 10 cubic kg per cubic meter, and a contact area of 1 square meter with the horizontal ground
Find its volume v
Find the gravity g [g is 10 / kg]?
Find the pressure on the ground p?
【3】 A tourist with a mass of 60kg, lying on the surface of the sea and lake with a density of 1.2 * 10 cubic kg per cubic meter, the average density of the human body can be considered to be equal to the density of ordinary lake water, that is, 1.0 * 10 cubic kg per cubic meter, G is 10 / kg
How much buoyancy do you get?
What is the volume of human body exposed on the water?

【1】 Xiao Ming went out to travel and bought a Yixing teapot. He heard that Yixing teapot was made of Yixing's unique clay material, and wanted to know the density of this material. So he measured the mass of the lid with a balance to be 44.4g, and then put the lid into an overflow cup filled with water, and measured the mass of the overflow water to be 14.8g
Question 1: what is the density of this material
Question 2: if the mass of the whole empty teapot [without lid] is 159G, what is the volume of the material used in the teapot? [detailed steps]
(1) The volume of lid v = m water / P water = 14.8g/1g/cm ^ 3 = 14.8cm ^ 3
P cover = m cover / v = 44.4g/14.8cm ^ 3 = 3G / cm ^ 3
(2) The mass of the teapot M = M1 + M2 = 159G + 44.4g = 203.4g
V pot = m / P cover = 203.4g/3g/cm ^ 3 = 67.8cm ^ 3
A
【2】 Haibao is the mascot of the 2010 World Expo. Among the various Haibao ornamenting the streets of Shanghai, there is a Haibao [uniform and solid] with a mass of 3.0 * 10 cubic kg, a density of 1.5 * 10 cubic kg per cubic meter, and a contact area of 1 square meter with the horizontal ground
Problem 1: find its volume v
Question 2: how to find the gravity g [g is 10 / kg]?
Question 3: find the pressure on the ground p?
（1）V=m/p=3.0*10^3kg/1.5*10^3kg/m^3=2m^3
(2)G=mg=3.0*10^3kg*10N/kg=3.0*10^4N
(3)P=F/S=G/S=3.0*10^4N/1m^2=3.0*10^4N/m^2
A
【3】 A tourist with a mass of 60kg, lying on the surface of the sea and lake with a density of 1.2 * 10 cubic kg per cubic meter, the average density of the human body can be considered to be equal to the density of ordinary lake water, that is, 1.0 * 10 cubic kg per cubic meter, G is 10 / kg
Question 1: how buoyant is the applicant?
Question 2: what is the volume of human body exposed on the water?
(1) F = g = mg = 60kg * 10N / kg = 600N
(2) F = PGV
V = f / PG = 600N / (1.2 * 10 ^ 3kg / m ^ 3 * 10N / kg) = 0.05m ^ 3
This tourist's volume V tour = m / P tour = 60kg / 1.0 * 10 ^ 3kg / m ^ 3 = 0.06m ^ 3
0.05m^3/0.06m^3=5/6
That's five sixths
A

If D, e and F are the middle points of the three sides of △ ABC, and the perimeter of △ ABC is 18cm and the area is 36cm, then the perimeter of △ DEF is cm,
The area is cm².

Perimeter = 18 / 2 = 9 cm
Area = 36 / 4 = 9 square centimeter

Simple calculation of 1 / 7 + 1 / 14 + 1 / 28 + 1 / 56

1/7+1/14+1/28+1/56
=1/7+(1/7-1/14)+(1/14-1/28)+(1/28-1/56)
=2/7-1/56
=15/56

Find the range of the function y = x ^ 2 + 2x + 3 in the following given closed interval: ① [- 4, - 3] ② [- 4,1] ③ [- 2,1] ④ [0,1]

The image of quadratic function is shown in the figure. It can be seen from the image that: (1) the image on [- 4, - 3] is a decreasing interval, and the value range is: [6,11]; (2) the image on [- 4,1] first decreases and then increases. When x = - 4, x = 1, the value of Y is calculated, and compared with the lowest point y = 2, the value range is: [2,11]; (3) the image on [- 2,1] is also a decreasing and then increasing. Similarly, the value range is: [2,6] and (4) in [0,1], 1] The value range is [3,6] as long as the image of quadratic function is made, this problem can be solved easily. I'm sorry that the image is not well drawn

The length of a rectangle is xcm, and the width is 4cm less than the length. If the length and width of the rectangle are increased by 3cm, how much is the area increased?

（x+3)(x-4+3)-x(x-4)=x^2+2x-3-(x^2-4x)=6x-3

3. How can 4, 10, - 6 be equal to - 24

4-10 * 3-6 front 4-10 with brackets

The iron ball with the volume of 5DM & # 179; weighs 39.2n. What is the buoyancy of the iron ball after immersion in water? (the density of iron = 7.9 * 10 & # 179; kg / M & # 179;)