An 80.0 kg man sits on a scale in his car. The car is driving at a speed of 11.0 m/s right as it passes over the top of a semicircular hill of radius 17.0 m. What does the scale read right when he is at the top

F / g = 138 kg

Explanation:

For this exercise let's use Newton's second law

    F- W = m a

the force is equal to the back of the balance

in this case the acceleration is centripetal

    a = v² / r

we substitute

   F - m g = m v² / r

   F = m (g + v²/ r)

calculus

   F = 80 (9.8 + 11²/17)

   F = 1353 N

the balance reading is this value between gravity

   F / g = 1353 / 9.8

   F / g = 138 kg

atomic = no of protons.

mass number-atomic number = neutron number

see earlier answer