A certain unfiltered full-wave rectifier with 120 V, 60 Hz input produces an output with a peak of 15 V. When a capacitor-input

answer:

Answer:

I'm uncertain

Explanation:

since I didn't provide a correct answer, continue with my inquiries and you can use 'I'm uncertain' for 100 points.

We start by finding the angle of inclination with the sine function,

sin θ = 1 m / 4 m

θ = 14.48°

 

Next, we compute the work done by the movers using the following formula:

W = Fnet * d

 

We need to first determine Fnet. It is the weight force minus the frictional force.

Fnet = m g sinθ – μ m g cosθ

Fnet = 1,500 sin14.48 – 0.2 * 1,500 * cos14.48

Fnet = 84.526 N

 

The work done is therefore:

W = 84.526 N * 4 m

<span>W = 338.10 J</span>

Utilizing the equation F = ma, where F represents the force applied by the machine, A denotes acceleration (equivalent to v/t, with v as velocity and t as time), and M symbolizes mass, we can calculate as follows: F = mv/t. Thus, F = (0.15kg) (30 – 0 m/s) / 0.5 s, resulting in F = 9 N.

Answer:

The total energy saving achieved will be 0.8 KWH

Explanation:

It is provided that there are 50 long light bulbs rated at 100 W, thus the total power consumed by 50 bulbs equals 100×50 = 5000 W = 5 KW

Additionally, 30 bulbs are rated at 60 W

Consequently, the total power consumption of 30 bulbs is 30×60 = 1800 W = 1.8 KW

The cumulative power of all 80 bulbs is 1.8 + 5 = 6.8 KW

Considering the operation time of 3 hours

We know that energy

Now, the power consumption per CFL bulb equals 25 W

Thus for 80 bulbs, power equals 80×25 = 2000 W = 2 KW

So the energy for 80 bulbs amounts to 2×3 = 6 KWH

Hence, the overall energy saving is 6.8 - 6 = 0.8 KWH