Question

Очень срочно нужна помощь с задачей по физике!!!! Какой радиус r должен иметь алюминиевый шарик,натертый парафином, чтобы он "плавал" в воде, погрузившись ровно наполовину?

Answer 1

"Подъемная" сила от натяжения в таком случае- 2*pi*r*sigma , где sigma - пов. натяжение для воды.
Сложите с весом шарика (со знаком минус) и архимедовой силой и решите уравнение.

Answer 2

Calculation of the Radius of the Aluminum Ball

To determine the radius of the aluminum ball that will float in water, submerged halfway, we need to consider the buoyant force acting on the ball and the weight of the ball.

The buoyant force is the upward force exerted on an object submerged in a fluid and is equal to the weight of the fluid displaced by the object. In this case, the buoyant force acting on the aluminum ball is equal to the weight of the water displaced by the submerged portion of the ball.

The weight of the ball can be calculated using the formula:

Weight of the ball = Density of aluminum x Volume of the ball x Acceleration due to gravity

The volume of the ball can be calculated using the formula:

Volume of the ball = (4/3) x π x r^3

where r is the radius of the ball.

Since the ball is floating halfway in water, the weight of the ball is equal to the buoyant force acting on it. Therefore, we can equate the weight of the ball to the buoyant force:

Weight of the ball = Buoyant force

Substituting the formulas for weight and buoyant force, we get:

Density of aluminum x Volume of the ball x Acceleration due to gravity = Density of water x Volume of water displaced x Acceleration due to gravity

Canceling out the common terms and rearranging the equation, we get:

Density of aluminum x Volume of the ball = Density of water x Volume of water displaced

Since the ball is submerged halfway, the volume of water displaced is equal to half the volume of the ball:

Volume of water displaced = (1/2) x Volume of the ball

Substituting this into the equation, we get:

Density of aluminum x Volume of the ball = Density of water x (1/2) x Volume of the ball

Canceling out the common terms, we find:

Density of aluminum = (1/2) x Density of water

Now, we can substitute the known values for the densities of aluminum and water:

Density of aluminum = (1/2) x 1000 kg/m^3 (density of water is approximately 1000 kg/m^3)

Simplifying the equation, we find:

Density of aluminum = 500 kg/m^3

To find the radius of the aluminum ball, we can rearrange the formula for the volume of the ball:

Volume of the ball = (4/3) x π x r^3

Substituting the known value for the density of aluminum, we can solve for the radius:

500 kg/m^3 = (4/3) x π x r^3

Simplifying the equation, we find:

r^3 = (3/4) x (500 kg/m^3) / π

Taking the cube root of both sides, we find:

r = (3/4) x (500 kg/m^3) / π)^(1/3)

Calculating this expression, we find that the radius of the aluminum ball should be approximately 0.252 meters.

Please note that this calculation assumes a uniform density for the aluminum ball and neglects any other factors that may affect the buoyancy of the ball, such as surface tension or the presence of the paraffin coating.