Question

Колесо диаметром 14 см делает на некотором расстоянии 28 оборотов. какой должна быть площадь диска колеса, которое делает на таком же расстоянии 56 оборотов?

Answer 1

D=14
1).R(радиус)=14/2=7
2).S= ПR^2. S(1)=7^2П=49П
3). 49П - 28
х - 56
х= 49П*56/28=98П
Ответ: S(2)=98П
примечание: П( число пи )=3, 14

Answer 2

Calculation of the Area of the Wheel's Disk

To calculate the area of the wheel's disk, we need to use the given information about the diameter and the number of revolutions.

The diameter of the wheel is given as 14 cm, and it makes 28 revolutions over a certain distance. We need to find the area of the disk when the wheel makes 56 revolutions over the same distance.

To calculate the area of the disk, we can use the formula for the area of a circle:

Area = π * r^2

where π is a mathematical constant approximately equal to 3.14159, and r is the radius of the circle.

Since the diameter is given, we can find the radius by dividing the diameter by 2:

Radius = Diameter / 2

Let's calculate the radius of the wheel:

Radius = 14 cm / 2 = 7 cm

Now, we can calculate the area of the disk when the wheel makes 28 revolutions:

Area1 = π * (7 cm)^2

To find the area of the disk when the wheel makes 56 revolutions, we can use the concept of proportionality. Since the number of revolutions is doubled, the area of the disk will also be doubled. Therefore:

Area2 = 2 * Area1

Let's substitute the value of Area1 into the equation:

Area2 = 2 * (π * (7 cm)^2)

Now, we can calculate the area of the disk when the wheel makes 56 revolutions.

Please note that the exact value of π is not provided in the search results, so we will use an approximate value of 3.14159 for our calculations.

Calculation:

Area2 = 2 * (3.14159 * (7 cm)^2)

Area2 ≈ 2 * (3.14159 * 49 cm^2)

Area2 ≈ 2 * 153.938 cm^2

Area2 ≈ 307.876 cm^2

Therefore, the area of the disk of the wheel, when it makes 56 revolutions over the same distance, should be approximately 307.876 cm^2.

Please note that the exact value of π may vary depending on the level of precision required for the calculation.