Question

В бидон с молоком m1=1кг налили молоко,объём которого 5 л.Какую минимальную силу надо приложить , чтобы поднять тбидон с молоком? Плотность молока 1г/см3.g=10H/кг

Answer 1

5л=5000см^3
5000см^*1г/см^3=5000г=5кг
5кг+1кг=6кг
6кг*10Н/кг=60Н.

Answer 2

Calculating the Minimum Force Required to Lift the Milk Can:

To calculate the minimum force required to lift the milk can, we can use the formula for calculating the force needed to lift an object, which is given by the formula:

F = m * g

Where: - F is the force needed to lift the object, - m is the mass of the object, and - g is the acceleration due to gravity.

In this case, the mass of the milk can (m1) is given as 1 kg, and the density of milk is given as 1 g/cm^3. The volume of the milk is 5 liters, which we can convert to cubic centimeters for consistency.

Calculating the Volume of Milk in Cubic Centimeters:

To calculate the volume of milk in cubic centimeters, we can use the conversion factor: 1 liter = 1000 cubic centimeters

So, the volume of milk in cubic centimeters is: 5 liters * 1000 cubic centimeters/liter = 5000 cubic centimeters

Calculating the Mass of the Milk:

The mass of the milk can be calculated using the density formula: Density = Mass / Volume

Rearranging the formula to solve for mass: Mass = Density * Volume

Substituting the given values: Mass = 1 g/cm^3 * 5000 cm^3 = 5000 grams

Converting the mass to kilograms: 5000 grams = 5 kilograms

Calculating the Minimum Force Required:

Now, we can use the mass of the milk (5 kg) and the acceleration due to gravity (10 m/s^2) to calculate the minimum force required to lift the milk can: F = m * g F = 5 kg * 10 m/s^2 F = 50 N

So, the minimum force required to lift the milk can is 50 Newtons.