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.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol