Answer:
Explanation:
The radius of Earth is 
The density of an object is given by :
The density of Erth is, d = 5515 kg/m³
Where
m is the mass of the Earth
So,
Put all the values,
So, the mass of the Earth is equal to 
.
Answer:
Explanation:
Given that,
First Capacitor is 10 µF
C_1 = 10 µF
Potential difference is
V_1 = 10 V.
The charge on the plate is
q_1 = C_1 × V_1 = 10 × 10^-6 × 10 = 100µC
q_1 = 100 µC
A second capacitor is 5 µF
C_2 = 5 µF
Potential difference is
V_2 = 5V.
Then, the charge on the capacitor 2 is.
q_2 = C_2 × V_2
q_2 = 5µF × 5 = 25 µC
Then, the average capacitance is
q = (q_1 + q_2) / 2
q = (25 + 100) / 2
q = 62.5µC
B. The two capacitor are connected together, then the equivalent capacitance is
Ceq = C_1 + C_2.
Ceq = 10 µF + 5 µF.
Ceq = 15 µF.
The average voltage is
V = (V_1 + V_2) / 2
V = (10 + 5)/2
V = 15 / 2 = 7.5V
Energy dissipated is
U = ½Ceq•V²
U = ½ × 15 × 10^-6 × 7.5²
U = 4.22 × 10^-4 J
U = 422 × 10^-6
U = 422 µJ
I think it may be that of a temperate deciduous forest tho im not sure
thank u for letting me answer and god bless have a good life <3
Answer:
A general solution is 
and a particualr case is mgh, it is just to distance around the radius Earth.
Explanation:
We can use a general equation of the potential energy to understand the particular and general case:
The potential energy is defined as 
, we know that the gravitational force is 
, so we could find the potential energy taking the integral of F.
(1)
We can find the particular case, just finding the gravitational potential energy difference:
. Here Uf is the potential evaluated in r+Δh and Ui is the potential evaluated in r.
Using (1) we can calculate ΔU.
Simplifying and combining terms we have a simplified expression.
(2)
Let's call 
. It is the acceleration due to gravity on the Earth's surface, if r is the radius of Earth and M is the mass of the Earth and we can write (2) as ΔU=mgh, but if we have distance grader than r we should use (2), otherwise, we could get incorrect values of potential energy.
I hope i hleps you!
