We consider a falling particle whose mass is given by m. we assume that the forces acting on the particle are both the air resistance, which is proportional to the velocity of the particle, and the gravitation. the position of the particle at time t is given by x(t) and the direction of the gravitation is negative x direction. the equation of motion of the particle is given by d2r(t) dr(t) (1) mg a dt m dt2 11 where a is a positive constant, and g is the magnitude of gravitational acceleration we assume that the initial position of the particle is given by x(0) = 0 and the initial velocity of the particle is 0. answer the following questions. as the differential equation of the velocity v(t) of the particle (1) rewrite eq. (1) (2) when t oo, the acceleration becomes 0, and the velocity reaches the terminal velocity v(oo). express the terminal velocity using m, g and a. (3) we introduce a new variable u(t) v(t) - v(o0). rewrite eq. (1) as the differential equation of u(t) and express u(0) using m, g and a (4) express ut) as a function of t. (5) express v(t) as a function of t (6) select the correct statement from the following choices: (a) the mechanical energy of the particle increases with increasing time (b) the mechanical energy of the particle decreases with increasing time. (c) the mechanical energy does not change during the fall

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).

photocells convert light energy to electricity

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j