Question

Слиток серебра объём 70 см имеет массу 735 г. 1) Какова масса слитка серебра объёмом 25 см? 2)Каков объём слитка серебра массой 420 г.?

Answer 1

735г : 70см3 = 10,5г - масса 1см3
25см3 * 10,5г = 262,5г - масса слитка 25см3
420г : 10,5г = 40см3 - объём слитка массой 420г

Answer 2

I can help you with your math problem. Here is the solution:

To find the mass of a silver ingot with a volume of 25 cm, we need to use the formula:

$$m = \rho V$$

where $m$ is the mass, $\rho$ is the density, and $V$ is the volume. We can find the density of silver by dividing the mass of the given ingot by its volume:

$$\rho = \frac{735}{70} = 10.5 \text{ g/cm}^3$$

Then, we can plug in the values into the formula and get the answer:

$$m = 10.5 \times 25 = 262.5 \text{ g}$$

The mass of a silver ingot with a volume of 25 cm is 262.5 g.

To find the volume of a silver ingot with a mass of 420 g, we need to use the same formula, but rearrange it to solve for $V$:

$$V = \frac{m}{\rho}$$

We already know the density of silver, so we can plug in the values and get the answer:

$$V = \frac{420}{10.5} = 40 \text{ cm}^3$$

The volume of a silver ingot with a mass of 420 g is 40 cm^3.

I hope this helps. You can also check out these web sources for more information: [1](https://uchi.ru/otvety/questions/slitok-serebra-obemom-70-sm-v-kube-imeet-massu-735-g-1-kakova-massa-slitka-serebra-obemom), [2](https://uchi.ru/otvety/questions/slitok-serebra-obemom-70-sm-v-kube-imeet-massu-735-g-1-kakova-massa-slitka-serebra-obemom-1ba63b83-4c94-4806-95d7-7578b878c595), [3](https://online-otvet.ru/matematika/5cea94b296f4e19a29409a6a).