Question

Бутылка, заполненная газом, при нормальном атмосферном давлении и температуре 27 °С плотно закрыта пробкой диаметром 2,5 см. Чтобы вытащить пробку из бутылки, её нагрели на 100 0С. Чему равна сила трения, удерживающая пробку в бутылке?

Answer 1

Ответ:

F≈16,36Н.

Объяснение:

Решение на фотках. Конечно, я всё очень подробно рассчитал, можно было упростить, если вместо формулы для площади пробки сразу написать S.

Answer 2

Calculation of Frictional Force

To calculate the frictional force holding the cork in the bottle, we can use the equation:

Frictional force = Normal force * Coefficient of friction

The normal force is the force exerted by the cork on the bottle due to gravity. At normal atmospheric pressure and temperature, the cork is tightly sealed in the bottle, so the normal force is equal to the weight of the cork.

The coefficient of friction depends on the materials in contact. Since the question does not specify the materials, we will assume a typical value for the coefficient of static friction between cork and glass, which is around 0.6.

Let's calculate the frictional force step by step:

1. Calculate the weight of the cork: - The weight of an object can be calculated using the equation: Weight = Mass * Acceleration due to gravity - The mass of the cork is not given, so we cannot calculate the exact weight. However, we can assume a typical mass for a cork, which is around 5 grams (0.005 kg). - The acceleration due to gravity is approximately 9.8 m/s^2. - Therefore, the weight of the cork is: Weight = 0.005 kg * 9.8 m/s^2 = 0.049 N

2. Calculate the frictional force: - The frictional force is equal to the product of the normal force and the coefficient of friction. - The normal force is equal to the weight of the cork, which we calculated to be 0.049 N. - The coefficient of friction between cork and glass is assumed to be 0.6. - Therefore, the frictional force is: Frictional force = 0.049 N * 0.6 = 0.0294 N

So, the frictional force holding the cork in the bottle is approximately 0.0294 N.