How to measure the density of a pile of pebbles? (there is no balance)

How to measure the density of a pile of pebbles? (there is no balance)

I believe you should know how to get the volume. The key is the mass. Use the buoyancy formula
F floating = ρ liquid GV discharge
F = g
First, put the small beaker into the large beaker filled with water to measure the overflow water to get the mass of the small beaker, then put the stones into the small beaker, and then put them into the large beaker filled with water to measure the overflow water to get the total mass of the small beaker and the stones, and then get the mass of the stones according to the volume and density of the stones

There is a balance, a beaker and a measuring cylinder to measure the density of large stones. Large stones can be put into the beaker but not into the measuring cylinder

Compared with the beaker, the volume of the measuring cylinder is larger. When measuring larger stones, it is necessary to ensure that the stones can be completely immersed in the water when they are placed in the water. At the same time, it can also measure the change of water level, that is, the volume of stones can be measured. Because the stones are larger, it can be measured with the beaker, but the measuring cup is smaller, and it cannot measure the volume of stones more accurately

Calculation of pressure
It is a service product launched by a cold drink shop. The store provides customers with a juice drink with a density of 950kg / m3 and ice cubes in the drink. The known mass of the drink is 0.6kg. The mass of the ice cubes is 20g, the inner bottom area of the cup is 50cm2, and the depth of the drink in the cup is 10cm (ice cubes are already in the beverage). The total pressure of the drink and ice cubes on the bottom of the cup is calculated. (g = 10N / kg)
Someone tied a rope to an object in the water and tried to salvage it out of the water. The known density of the object is 3 times 10 cubic kilogram per cubic meter, and the volume is 10dm3, g = 10N / kg
How much tension can the rope at least bear to lift the object out of the water?
What is the minimum tension on the rope in the process of lifting an object from the water?

One question: the pressure on the bottom of the cup: ρ = 950kg / m3 * 10N / kg * 0.1M = 950pa, the pressure is PS = 950pa * 50cm2 * 10 to the - 4th power = 4.75n
Second question: to lift an object out of the water, the tension that the rope used can bear is at least the weight of the object, that is, g = mg = ρ VG = 3 times the cubic kilogram of 10 per cubic meter * 10dm3 * 10 - & sup3; * 10N / kg = 300N
In the process of lifting an object from the water, the minimum tension on the rope is the weight of the object in the water minus the weight lost, that is, G-F floating = 300 - ρ water, GV = 300-1000 * 10dm3 * 10 - & sup3; * 10N / kg = 200N

1,1/2 ,1/4 ,1/8 ,1/16 …… The general term formula of

The general formula is 1 / 2 ^ (n-1)

One half, one quarter, one eighth The general term formula of

an＝（－1/2）^n

1、 One half, one quarter, one eighth The general term formula of

An = (- 1 / 2) n-1 (n belongs to positive integer), the power of negative half is n-1

The sum of the first n terms of sequence an is Sn and Sn + half an = 1

That is, 2Sn + an = 2
2s（n-1）+a（n-1）=2
Subtraction of two formulas
2an+an-a（n-1）=0
That is, 3an = a (n-1)
an/ a（n-1）= 1/3
Equal ratio sequence with A1 = 2 / 3 and q = 1 / 3
an= 2/3*（1/3）^(n-1)

Let the sum of the first n terms of an sequence of positive numbers be Sn, Sn = half (an + an), and find the general formula of an

Sn = (1/2)(an + 1/an) (1)
n=1
2(a1)^2 = (a1)^2+a1
a1(a1-1)=0
a1 =1
S(n-1) = (1/2)[a(n-1) +1/a(n-1) ] (2)
(1)-(2)
an = (1/2)(an + 1/an)- (1/2)[a(n-1) +1/a(n-1) ]
1/an - an - a(n-1) -1/a(n-1) =0
a(n-1) - 2an.a(n-1) - an =0
1/an -1/a(n-1) =2
1/an -1/a1 = 2(n-1)
1/an = 2n-1
an = 1/(2n-1)

Given that the sequence an satisfies the n-th power of A1 = half, an times an + 1 = half times a quarter, and N belongs to a positive integer, the general term formula of the sequence an is obtained

According to A1, we get A2 = 1 / 4, and an * an + 1 = (1 / 2) * (1 / 4) ^ n (an + 1) * (an + 2) = (1 / 2) * (1 / 4) * (1 / 4) ^ (n + 1). By comparing the two formulas, we get (an + 2) / an = 1 / 4, so when n is odd, an = A1 * (1 / 4) ^ ((n-1) / 2) = (1 / 2) ^ n, when n is even, an = A2 * (1 / 4) ^ ((n-2) / 2) = (1 / 2) ^ n, in conclusion, an = (1 / 2) ^ n, (half of the nth power)

Given that the first term of sequence {an} is 1 an + 1 = half an + 1, find the general term of sequence {an}
Let {an} satisfy an + 1 of an = n + 1 of N, A1 = 2 to find the general term formula of sequence {an}

(1) To construct an equal ratio sequence:
an+1=1/2an+1
a(n+1)-2=1/2(an-2)
Then the sequence {An-2} is an equal ratio sequence whose initial term is A1-2 = - 1 and common ratio is 1 / 2
So An-2 = - (1 / 2) ^ (n-1)
an=-(1/2)^(n-1)+2.
(2) a(n+1)/an=(n+1)/n
Let n = 1,2,3,4 The results are as follows
a2/a1=2/1
a3/a2=3/2
a4/a3=4/3
a5/a4=5/4
……
an/a(n-1)=n/(n-1)
The result is as follows
an/a1=n
So an = 2n