The Sun's gravitational pull keeps our planet orbiting the Sun <span>in a nice nearly-circular orbit.</span>
Answer:
a) 5 N b) 225 N c) 5 N
Explanation:
a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them, acting along the line that joins the charges, as follows:
F₁₂ = K Q₁ Q₂ / r₁₂²
So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N
b) If we reduce the distance, from r, to r/3, as the factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.
So, we will have F₁₂ = 9. 25 N = 225 N
c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:
F₁₂ = 25 N .1/5 = 5 N
The answer you're looking for is: reproductive success.
Hopefully this has helped! :)
To solve the problem it is necessary to take into account the concepts related to simple pendulum, i.e., a point mass that is suspended from a weightless string. Such a pendulum moves in a harmonic motion -the oscillations repeat regularly, and kineticenergy is transformed into potntial energy and vice versa.
In the given problem half of the period is equivalent to 1 second so the pendulum period is,

From the equations describing the period of a simple pendulum you have to

Where
g= gravity
L = Length
T = Period
Re-arrange to find L we have

Replacing the values,


In the case of the reduction of gravity because the pendulum is in another celestial body, as the moon for example would happen that,




In this way preserving the same length of the rope but decreasing the gravity the Period would increase considerably.
Yes but they not supposed to