Explanation:
Given that,
Capacitor 
Resistor 
Peak voltage = 5.10 V
(A). We need to calculate the crossover frequency
Using formula of frequency

Where, R = resistor
C = capacitor
Put the value into the formula


(B). We need to calculate the
when 
Using formula of 

Put the value into the formula


(C). We need to calculate the
when 
Using formula of 


(D). We need to calculate the
when 
Using formula of 


Hence, This is the required solution.
Galileo Galilei was the first scientist to perform experiments in order to test his ideas. He was also the first astronomer to systematically observe the skies with a telescope.
:)
Im pretty sure that they do
Answer:
The Sun-Earth-Moon system happens to exhibit a striking geometric coincidence, which we examine in the first problem. PROBLEM 1. To an observer on Earth, the Sun and the Moon subtend almost the same angle in the sky. The average angle is 0.52 degrees for the Moon and 0.53 degrees for the Sun.
R = ρ L/A. R= resistance, ρ= resistivity, L= length of the conductor. A = area of the conductor. Resistance is directly proportional to the length of the conductor. So if length of the conductor is decreased, resistance will also decrease. Hence A is the correct option