Chapter 1 Summary of knowledge points and classical exercises of mass and density,

Chapter 1 Summary of knowledge points and classical exercises of mass and density,

Quality and density
1. Mass (m): the amount of substance in an object is called mass
2. The international unit of mass is kilogram. Others are: ton, gram, milligram, 1 ton = 103 kilogram = 106 gram = 109 milligram
3. The mass of an object does not change with shape, state, position and temperature
4. Mass measuring tools: commonly used balances in laboratory are pallet balance and physical balance
5. The correct use of the balance: (1) put the balance on the level table, and put the traveling weight on the zero line at the left end of the scale; (2) adjust the balance nut, so that the pointer points to the middle line of the dividing plate, and then the balance is balanced; (3) put the object in the left plate, add and subtract weights with tweezers to the right plate, and adjust the position of the traveling weight on the scale, (4) the mass of the object is equal to the total mass of the weight in the right disk plus the scale value of the upstream weight
6. When using the balance, attention should be paid to: (1) the maximum weight should not be exceeded; (2) tweezers should be used to add and subtract weights, and the action should be light; (3) do not put wet objects and chemicals directly on the tray
7. Density: the mass per unit volume of a substance is called the density of the substance. Use ρ to represent the density, m to represent the mass, V to represent the volume, and the formula for calculating density is ρ = m / V; the unit of density is kg / m3, (and: g / cm3), 1 g / cm3 = 1000 kg / m3; the unit of mass m is kg; the unit of volume V is m3
8. Density is a property of matter. Different kinds of matter have different densities
9. The density of water is 1.0 × 103 kg / m3
10. Application of density knowledge
(1) Identify substance: measure mass m with balance and volume V with measuring cylinder, then calculate substance density according to formula: ρ = m / v. then check density table
(2) Find the mass: M = ρ v
(3) Find the volume: v = m / ρ
Strength

A physics problem about mass and density
American astronauts have successfully landed on the moon, and brought back a bag of lunar soil samples from the moon. The mass of the sample on the earth is 0.2kg, and the measured volume is 0.4dm & sup3;, then the density of lunar soil is kg / M & sup3;, and the mass of these samples on the moon is kg. Cut it half, then the mass of the remaining part is kg, and the density is kg / M & sup3

M = ρ * V, ρ = 0.2 / 0.4 * 10-3 = 500 kg / M & sup3; the mass of these samples on the moon is 0.2 kg / M & sup3; the mass of the remaining part is 0.1 kg / M & sup3 if cut half

Calculation of physical mass and density
The total mass of a bottle filled with water is 150g. The total mass of a bottle filled with 1.2 × 10 cubic kilogram of certain liquid per cubic meter is 166g. Find the mass and volume of the bottle

Let v be the mass of the bottle
Then M + 1000V = 0.15
m+1200V=0.166
The simultaneous solution is v = 0.00008m & sup3;, M = 0.07kg

Why can we only calculate the mass of the central celestial body by quoting the formula of universal gravitation
For example, in the Earth Moon system, if the moon moves around the earth, then the earth also moves relative to the moon. Can't we consider the moon as the central celestial body?

The relative motion of a circle is different from the relative motion of a straight line. You can say that a straight line a to B goes to the left, which is equivalent to B straight line a goes to the right, but the circle can't. In fact, the moon rotates clockwise (or counterclockwise) around the earth. If you want to change the angle of the earth, it's equivalent to the earth's rotating around the moon

In the formula of universal gravitation f = MW ^ 2R, is m the mass of the central celestial body or the mass of the planetary celestial body?

The circle of celestial bodies
F=mg=mw2r
This is the derivation process. M refers to the orbit around the celestial body
No, please correct me

The calculation formula of universal gravitation

The calculation formula of universal gravitation: F = GMM / (R ^ 2) at the midpoint: when the object is gravitated by F 1 = 2g (m ^ 2) / ((R / 2) ^ 2) is not at the midpoint, let its distance from two stars be a and B respectively, then F 2 = g (m ^ 2) / (a ^ 2) + G (m ^ 2) / (b ^ 2), and a + B = R

Does r in the formula of physical gravitation and centripetal force add the radius of the central celestial body?
Including: F = (GMM) / R ^ 2
F=(mv^2)/R
F = m Ω ^ 2R - [Omega refers to angular velocity]
GM = GR ^ 2 (gold substitution formula)
wait.
Please elaborate on the above three
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`I mean
For example, the moon goes around the earth,
This radius R is the distance from the surface of the moon to the surface of the earth,
Or the distance from the center of the moon to the center of the earth?
Please explain other formulas in this way`
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A celestial body is a moving object that is subject to the universal gravitation of the central celestial body,
It is equivalent to the universal gravitation of the center of the celestial body,
Where R is the distance from the center of the moving celestial body to the center of the central celestial body!

What celestial bodies are similar to the earth in mass and volume?

Mercury, Venus, earth and Mars are called Earth like planets
The reason is that Mercury, Venus and Mars are similar in mass, volume and average density to earth

If the density of a celestial body is the same as that of the earth, near the surface of the celestial body, a small ball with a mass of 4 kg will open from rest at point a
If the density of a celestial body is the same as that of the earth, near the surface of the celestial body, a small ball with a mass of 4kg starts to fall 16m at point a and reaches the surface of the planet, with a speed of 10m / S ^ 2. Ask for:
(1) What is the ratio of the radius of the celestial body to that of the earth?
(2) If the ball is thrown horizontally at point a, and the distance between the landing point and point a is 20m, what is the initial velocity of the ball thrown horizontally?

(1) Gravitation f = g (mm) / R ^ 2
Let the radius of the celestial body and the radius of the earth be r and R respectively; the density of the celestial body and the earth be ρ, and the mass of the object be m
M1=4ρπr^3/3--------------------1
M2=4ρπR^3/3-----------------------2
Acceleration of object on celestial body: a = GM1 / R ^ 2 ------- 3
Earth acceleration: g = GM2 / R ^ 2 -------- 4
Taking 1 and 2 into 3 and 4, we get a = 4 ρ π RG / 3 and g = 4 ρ π RG / 3
a/g=r/R
Acceleration a = VT ^ / 2S = 10 ^ 2 / 2 * 16 = 3.125 M / S
r/R=a/g=3.125/9.8=?
(2) Let v be the initial velocity of the horizontal throw of the ball
20^=V^t^+s^=V^t^+16^
V^t^=144m^---------------------5
And because T ^ = 2as-6 on celestial bodies
The equation 5 and 6 together give v = 3.75m/s

Is the mass, volume and average density of the celestial body close to that of the earth?

Mars, which is why humans are so keen to explore Mars and there are Martians in science fiction movies