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>
We only need to look at Newton's 2nd law of motion:
Net force = (mass) x (acceleration)
If the net force on an object is zero, then either its mass or its acceleration must be zero. If it's called an "object", then its mass isn't zero, so its acceleration is zero.
Given there are three blocks of masses , and (ref image in attachment)
When all three masses move together at an acceleration a, the force F is given by
F = ( + + ) *a ................(equation 1)
Also it is given that does not move with respect to , which gives tension T is exerted on pulley by only, Hence tension T is
T = *a ..........(equation 2)
There is also also tension exerted by . There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by
T = ................(equation 3)
From equation 2 and 3, we get
*a =
Squaring both sides we get
* = * (+)
* = ( * )+ ( *)
( - ) * = *
= */( - )
Taking square root on both sides, we get acceleration a
a = *g/()
Hence substituting the value of a in equation 1, we get