<span>Inertia keeps us orbiting because any object with mass has the tendency to resist changes to their direction and speed of movement. Combine that with the interaction of the gravitational attraction of the sun, and that is what keeps Earth in orbit. The sun’s gravitational force is one that is proportional to Earth’s mass, and it acts in a way that is almost exactly perpendicular to Earth’s motion. This keeps Earth from spinning into the sun or far away from it.</span>
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.
Answer:
The answer would be melting point.
Explanation:
Hope this helps. Can you please mark me brainliest
The equation for electrical power is<span>P=VI</span>where V is the voltage and I is the current. This can be rearranged to solve for I in 6(a).
6(b) can be solved with Ohm's Law<span>V=IR</span>or if you'd like, from power, after substituting Ohm's law in for I<span>P=<span><span>V2</span>R</span></span>
For 7, realize that because they are in parallel, their voltages are the same.
We can find the resistance of each lamp from<span>P=<span><span>V2</span>R</span></span>Then the equivalent resistance as<span><span>1<span>R∗</span></span>=<span>1<span>R1</span></span>+<span>1<span>R2</span></span></span>Then the total power as<span><span>Pt</span>=<span><span>V2</span><span>R∗</span></span></span>However, this will reveal that (with a bit of algebra)<span><span>Pt</span>=<span>P1</span>+<span>P2</span></span>
For 8, again the resistance can be found as<span>P=<span><span>V2</span>R</span></span>The energy usage is simply<span><span>E=P⋅t</span></span>