If the spring constant is 12.6N/m and a spring is stretched by 0.25m, how much force has been applied?

The magnitude of the applied force is 3.15 N

Explanation:

Hooke's law states that for small deformations, the change in dimensions, 'x', of an object is directly proportional to the force, 'F', of the acting load acting that bring about the deformation

Mathematically, Hooke's Law can be expressed as follows;

F = k·x

Where;

F = The applied load

k = The force constant of the object

x = The amount of deformation of the object

The given parameters are;

The spring's force constant, k = 12.6 N/m

The amount of stretching in the spring, x = 0.25 m

Therefore, we have;

The force applied, F = k·x = 12.6 N/m × 0.25 m = 3.15 N.

the best answer is d because the strong cohesion is caused by the ability of water molecules to undergo hydrogen bonding.

a is incorrect because when water solidifies, it becomes   ice, which expands and occupies a bigger volume for the same mass of water in its liquid state.

b is incorrect because water has high surface tension. besides bonding to itself it hydrogen bonds to other molecules at the top decreasing possibility for breaking of attractive forces.

c is incorrect because the water molecule has oxygen which is highly electronegative and so pulls the electron pair towards itself creating a dipole moment, making the molecule polar.

e is incorrect because water has strong adhesion, ability to stick to other molecules such as glass because of this molecules having stronger hydrogen bonding than water.

the answer is a for sure mark brainliest i had got this right on usatestprep