In the year 2120, when we have a colony on the moon, an engineer brings an old grandfather clock with her. She knows the clock’s pendulum has a length of 1.0 m and the moon’s gravity is 1.62 m/s^2. If she winds the clock when the time shows 12:00, how many Earth minutes have elapsed when the clock face reads 12:12? Round your answer to 1 decimal place for entry into eCampus. Do not enter units. Example: 12.3

the clock will only read 4.9 min on Earth

Explanation:

This watch can be approximated by a simple pendulum that is a rope with a point mass at its end, the angular speed is

              w = √ g / L

angular velocity is related to frequency and is to period

               w = 2π f = 2π / T

               T = 2π √L /g

let's analyze the situation on the moon,

              T = 2π √(1 / 1,62)

              T = 4.937 s

this indicates that each oscillation corresponds to the time T when the clock has advanced 12 minutes, we can find how many rotten it has made

Let's start by reducing the time to the SI system

            t = 12 min (60s / 1 min) = 720 s

now let's use a direct ratio. If one oscillation in T how many oscillations in t

            #_oscillation = t / T

           #_oscillation = 720 / 4,937

           # _oscillation = 145.8

Let's see how long the same pendulum has on Earth when it gives this number of oscillations

           T_earth = 2π √ L / g

           T_earth = 2π √(1 / 9.8)

          T _earth = 2.01 s

Now we can know the time it uses in the 145.8 oscillations

          t = #_ oscillations T_earth

          t = 145.8 2.01

          t = 292.63 s

let's reduce to minutes

          t = 292.63 s (1min / 60s) = 4.88 min

therefore the clock will only read 4.9 min on Earth

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.