The physical problem of the first two densities, the physical density problem of the first two years

1.There is a glass bottle with a mass of 10 grams. When the bottle is filled with a certain liquid, the total mass of the bottle and the liquid is 40 grams.

When a metal block is inside the bottle, the total mass of the bottle and the metal block is 90 grams. If the metal block is filled into the bottle and then filled with the above liquid, the total mass of the bottle is 100 grams. The density of the metal block is known to be 8x10 to the power of 3 kg m 3.

Seeking:1The volume of the metal block. 2.The density of the liquid.

2.There is a bottle of quality, filled with water is the total mass, if it is filled with a certain oil, the total mass is, find the density of the oil.

3.Water-saving furniture refers to sanitary ware that consumes less than 6 liters of water per flush. A family installed a new set of water-saving sanitary ware with a water consumption of 5 liters, while the original sanitary ware consumed 9 liters of water at a time.

Q: 1) How many times can 1000kg of water be used to rinse this water-saving fixture?

2) How much kg of water can the household save per month? (Use an average of 10 times a day, 30 days per month).

1. The mass of the metal block is:

m gold = 90-10 = 80g

So the volume of the metal block is:

v gold = m gold gold = 80 8 = 10 cm 3

2. The volume of the glass bottle is V.

When the bottle is filled with a certain liquid, the mass of the liquid is:

m=40-10=30g

If the metal block is filled into the bottle and then filled with the above liquid, the mass of the remaining liquid will be:

mˊ=100-90=10g

The volume of the remaining liquid is:

v = v - v gold = v - 10

Since the density of the liquid does not change, there are:

m v = m v i.e.

30/v=10/（v-10）

Solution: v=15cm3

So the density of the liquid is:

m/v=30/15=2g/cm^3

2. If the volume of water and oil is equal, then there is:

v water = v oil i.e.

800-300) water = (700-300) oil.

Solution: Oil =

3. (1) The volume of 1000kg of water is:

v = m water = 1000 1000 = 1m 3 = 1000l

The number of times this water-saving fixture can be rinsed is:

n = v 5 = 1000 5 = 200 times.

2) The number of litres of water that can be saved per month for the household is:

vˊ=（9-5）×10×30=1200l

The quality of the water saved is:

m = water v = 1000 1200 10 -3 = 1200kg

I pour. Is this a conundrum?

First of all, you have to understand one thing.

Replace the steel with aluminum.

The volume of the material is constant.

This is determined by the design of the aircraft.

Therefore, assuming that the volume of the replacement material is v, aluminum m kg is required to make these aluminum materials, and steel m kg is required to use steel

Then according to the title, there is m-m=104kg

Because m=pv

Therefore, 7900*v-2700*v=104kg can be obtained v=

Therefore m=2700*

The mass is equal, and the volume of A is three times that of B. It can only be concluded that the density of A is only 1 3

The density of B in the back is 4 times that of A. It is also necessary that B is hollow.

A is not hollow I don't know. Pick B

Let the mass of A and B be m

v A = m A.

v B = m B.

Alloy = 2m (v A + v B) = 2m (m A + m B) = 2 A B ( A + B).

Hop = m v

mA+mB) (VA+VB).

2mA (mA, A+mB, B).

2mA [mA(A+B) AB]=2 A-B (A+B)

v A = m A, v B = m B.

Alloy = 2m (v A + v B) = 2m m ( A + B) * A * B = 2 A * B ( A + B).

Let the mass of A and B be m

v A = m A.

v B = m B.

2m (m A + m B).

2 A* B ( A + B).

Doesn't the volume change after synthesis?

There should be a premise here that the alloy volume is equal to the sum of the volumes of the two metals, the volume of the metal = m A.

Ethyl metal volume = m B.

Then the total mass = 2m, the total volume = m A + m B alloy density = 2m (m A + m B) = 2 A * B ( A + B).

1. M solid = M liquid = 900g, V solid = a 3 = (10cm) 3 = 1000cm 3, = M solid V solid = 900g 1000cm 3 = there is a density table to know that the solid may be ice;

2. v=v total-v water=423cm 3-420cm 3=3cm 3, =m v=50g 3cm 3=, so the metal block is not pure gold.

A density is a physical quantity that reflects the characteristics of matter, the characteristics of matter refer to the nature of the substance itself and can be distinguished from each other, people often feel that the dense material is "heavy", the density of the material is "lighter", where the "heavy" and "light" essentially refer to the size of the density.

Mass is the amount of matter contained in an object. The substance contained is reduced, so the mass is reduced. Density is a property of matter, it does not change with the change of mass and volume, and the density of the same substance does not change.

Solution: According to the formula =m v

42g/4c㎥

So this substance may be made of silver.

Go home and read a good book, at least the simplest formula application should be mastered, you can't always rely on others!!

(1) Saline = m v = 60g 50cm 3 = > so, if it does not meet the requirements, water should be added.

2) Let the volume of water that should be added be v water, then brine v brine + water v water = (v brine + v water).

i.e.: water = water].

Solutionable: v water =

When these brine are prepared to meet the requirements, a total of v=v brine + v water = 50l + 50l = 100l is required

The mass of the sample m0 = and the volume v0 =

The density of the sample is the density of the river water =m0 v0=

So the mass of 1m 3 of river water (v total) m total = v total = 1018kg m 3 1m 3 = 1018kg

River water can be seen as a mixture of sand and pure water, so the total mass of river water can be seen as the mass of sand + the mass of pure water.

m total = m sand + m water.

The total volume of river water can be regarded as the volume of sand + the volume of pure water v total = v sand + v water.

Let the volume of sand contained in each cubic meter of river water be v sand, then the volume of pure water is v total - v sand m total = m sand + m water.

m total = sand v sand + water v water.

m total = sand v sand + water (v total - v sand).

1018 = 2500v sand + 1000 (1-v sand) 1018 = 2500v sand + 1000-1000v sand 1500v sand = 18

v sand = m sand = sand v sand = 2500kg m 3