The elastic constant of a spring, also known as the spring constant, is calculated using Hooke's Law, which states that the force needed to extend or compress a spring by some distance is proportional to that distance. The formula is:

F = kx

where:
F is the force applied,
k is the spring constant,
x is the distance the spring is stretched or compressed.

Rearranging the formula to solve for k gives:

k = F/x

Given in the problem, F = 256 N and x = 9.50 cm = 0.095 m (since we need to convert cm to m because the standard unit of distance in physics is meters).

Substituting these values in:

k = 256 N / 0.095 m = 2694.7368421052633 N/m

Rounding this to three significant figures gives k = 2.69 x 10^3 N/m.

So, the elastic constant of the spring is 2.69 x 10^3 N/m. The answer is (a).

Question

The elastic constant of a spring, also known as the spring constant, is calculated using Hooke's Law, which states that the force needed to extend or compress a spring by some distance is proportional to that distance. The formula is:

F = kx

where:
F is the force applied,
k is the spring constant,
x is the distance the spring is stretched or compressed.

Rearranging the formula to solve for k gives:

k = F/x

Given in the problem, F = 256 N and x = 9.50 cm = 0.095 m (since we need to convert cm to m because the standard unit of distance in physics is meters).

Substituting these values in:

k = 256 N / 0.095 m = 2694.7368421052633 N/m

Rounding this to three significant figures gives k = 2.69 x 10^3 N/m.

So, the elastic constant of the spring is 2.69 x 10^3 N/m. The answer is (a).

Knowee AI · Accepted Answer

The elastic constant of a spring, also known as the spring constant, is calculated using Hooke's Law, which states that the force needed to extend or compress a spring by some distance is proportional to that distance. The formula is:

F = kx

where:
F is the force applied,
k is the spring constant,
x is the distance the spring is stretched or compressed.

Rearranging the formula to solve for k gives:

k = F/x

Given in the problem, F = 256 N and x = 9.50 cm = 0.095 m (since we need to convert cm to m because the standard unit of distance in physics is meters).

Substituting these values in:

k = 256 N / 0.095 m = 2694.7368421052633 N/m

Rounding this to three significant figures gives k = 2.69 x 10^3 N/m.

So, the elastic constant of the spring is 2.69 x 10^3 N/m. The answer is (a).

If a force of 256 N is applied to a spring, the spring is stretched a distance of 9.50 cm from its equilibriumposition. What is the elastic constant of this spring?a. 2.69 x 103N/m c. 26:9`N/mb. 2.43 x 103N/m d. 243. N/m

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