A person with normal vision can focus on objects as close as a few centimeters from the eye up to objects infinitely far away. T

Answer:

        h = 12.8 cm

Explanation:

The initial parameters are as follows:

distance = 6.4 cm

• when the object descends, its weight matches the spring's force

        weight = spring force

         mg = ky... equation 1

• potential energy stored in a stretched spring = work done by the spring

        mgh = 0.5 x k x h^{2}....equation 2

• Substituting from equation 1 into equation 2

                kyh =  0.5 x k x h^{2}

                y =  0.5 x h

                2y = h

• where y is 6.4, yielding the maximum elongation as

          h = 2 x 6.4 = 12.8 cm

Answer:

0.018 J

Explanation:

The work required to bring the charge from infinity to the point P is equal to the change in its electric potential energy. This can be expressed as

where

represents the charge's magnitude

and signifies the potential difference between point P and infinity.

After substituting into the formula, we arrive at

Answer:

a = 18.28 ft/s²

Explanation:

the values provided are:

duration of force application, t= 10 s

Work done = 10 Btu

mass of the object = 15 lb

acceleration, a =? ft/s²

1 Btu = 778.15 ft.lbf

thus, 10 Btu = 7781.5 ft.lbf

m = 0.466 slug

So,

the work is equivalent to the change in kinetic energy

The acceleration of the object is therefore

         a = 18.28 ft/s²

the constant acceleration of the object is calculated to be 18.28 ft/s²