Answer:
the period of the 16 m pendulum is twice the period of the 4 m pendulum
Explanation:
Recall that the period (T) of a pendulum of length (L) is defined as:
where "g" is the local acceleration of gravity.
SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:
therefore the period of the 16 m pendulum is twice the period of the 4 m pendulum.
Answer:
q = 3.6 10⁵ C
Explanation:
To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth
r = 6 , 37 106 m
E = k q / r²
q = E r² / k
q =
q = 4.5 10⁵ C
Now let's calculate the charge on the planet with E = 222 N / c and radius
r = 0.6 r_ Earth
r = 0.6 6.37 10⁶ = 3.822 10⁶ m
E = k q / r²
q = E r² / k
q =
q = 3.6 10⁵ C
The concept required to solve this problem is quantization of charge.
First the number of electrons will be calculated and then the total mass of the charge.
With these data it will be possible to calculate the percentage of load in the mass.
Here Q is the charge, n is the number of electrons and e is the charge on the electron
Replacing,
According to the quantization of charge the charge is defined as product of the number of electron and the charge on the electron
The total mass of the charge is
Here,
m = Mass of the charge
n = Number of electrons
= Mass of the electron
Replacing we have