Question

Под действием груза массой 500гр пружина динамометра растянется на 4 мм на сколько растянется пружина под действием груза весом 10 ньютон

Answer 1

Жесткость пружины не меняется
k = F1/x1 = F2/x2
x2 = x1 * F2/F1 = x1 * F2/(m1*g) = 4мм * 10Н/(0.500*10)Н = 8мм

Answer 2

1Н = 1 кг*м/с2 
значит 10 Н = 10кг*м/с2 (секунда в квадрате) 
500 грамм = 0,5 кг 

0,5 кг = 4 мм 
10 кг*м/с2 = 80 мм 

10 Н на 80 мм... 

Answer 3

Calculation of Spring Extension

To calculate the extension of the spring under the weight of 10 Newtons, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the extension of the spring. The formula for Hooke's Law is:

F = kx

Where: - F is the force applied (in Newtons) - k is the spring constant (in Newtons per meter) - x is the extension of the spring (in meters)

We can use the given information to calculate the extension of the spring under the weight of 10 Newtons.

Calculation

Given: - Mass of the load = 500 grams = 0.5 kg - Spring extension under this load = 4 mm = 0.004 meters

We can calculate the spring constant using the mass and the spring extension, and then use it to find the extension under the weight of 10 Newtons.

First, let's convert the mass to kilograms: - Mass = 0.5 kg

Now, we can calculate the spring constant using the formula: k = F / x

Where: - F is the force applied (in Newtons) - x is the extension of the spring (in meters)

Given: - Force (weight) = 10 Newtons

We can use the formula to find the spring constant: k = 10 N / 0.004 m

Result

The spring constant, k, is calculated to be 2500 N/m.

Now, we can use this spring constant to find the extension of the spring under the weight of 10 Newtons using Hooke's Law: F = kx

Given: - Force (weight) = 10 Newtons - Spring constant, k = 2500 N/m

We can rearrange the formula to solve for x: x = F / k

Substitute the values: x = 10 N / 2500 N/m x = 0.004 m

Conclusion

The spring will extend by 0.004 meters under the weight of 10 Newtons.