Hi, pls help with this question. Thank you.

A lamp is 10% efficient.How much electrical energy must be supplied to the lamp each second if it produces 20 J of light energy

k0ka [10]

If it produces 20J of light energy in a second, then that 20J is the 10% of the supply that becomes useful output.

20 J/s = 10% of Supply

20 J/s = (0.1) x (Supply)

Divide each side by 0.1:

Supply = (20 J/s) / (0.1)

<em>Supply = 200 J/s </em>(200 watts)

========================

Here's something to think about: What could you do to make the lamp more efficient ? Answer: Use it for a heater !

If you use it for a heater, then the HEAT is the 'useful' part, and the light is the part that you really don't care about. Suddenly ... bada-boom ... the lamp is 90% efficient !

6 0

4 years ago

As capacitor was discharging, what did you observe about q on its plates and the motion of charges in the external circuit?

liubo4ka [24]

As capacitor was discharging, The charge on the plate got reversed and the motion of charge is opposite to the flow of current.

The charging contemporary asymptotically processes 0 as the capacitor becomes charged up to the battery voltage.

The capacitor is completely charged when the voltage of the electricity supply is equal to that at the capacitor terminals. that is referred to as capacitor charging; and the charging segment is over when modern-day stops flowing thru the electrical circuit.

A capacitor can be slowly charged to the important voltage and then discharged quick to provide the power wanted. it's far even viable to charge several capacitors to a positive voltage and then discharge them in any such way as to get extra voltage out of the gadget than became installed.

Learn more about capacitor here:-brainly.com/question/14883923

#SPJ4

8 0

2 years ago

How does the scientific process generally begin?

Murljashka [212]

There are several approaches. The most favourable one (in my opinion) is this one:
1. Asking a question
2. Doing a research (how to answer this question)
3. Creating a hypothesis (NOT a thesis!)
4. Experimenting (to prove the hypothesis)
5. Analysing results from the experiment
6. Writing a thesis

5 0

4 years ago

Read 2 more answers

A cat is shot horizontally out of a cannon that is on top of a 125 meter tall cliff. If the cat lands 286 m away from the base o

Elden [556K]

The initial velocity of the cat is 56.6 m/s

Explanation:

The motion of the cat is a projectile motion, consisting of:

- a uniform horizontal motion with constant velocity

- a vertical accelerated motion with constant acceleration (free fall)

We start by analyzing the vertical motion, to find the time it takes for the cat to reach the ground. We use the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where we have (choosing downard as positive direction)

s = 125 m is the vertical displacement of the cat

u = 0 is the initial vertical velocity

t is the time taken to reach the ground

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t, we find

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(125)}{9.8}}=5.05 s$

Now we can analzye the horizontal motion. Since the motion is uniform and the horizontal velocity is constant, it is given by

$v_x =\frac{d}{t}$

where

d = 286 m is the horizontal distance travelled

t = 5.05 s is the time taken

Solving the equation,

$v_x = \frac{286}{5.05}=56.6 m/s$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

3 0

3 years ago

A and B start walking in the same direction at the same time around a circular park of diameter 4200 m. If A walks 10 m more tha

Liono4ka [1.6K]

Answer:

Va = 5000 m / 3600 s = 1.39 m/s

(Va - Vb) 60 = 10

Vb = Va - .167 = 1.22 m/s

(Va - Vb) T = 4200 Π where T is time for A to complete 1 more lap

.17 T = 4200 Π

T = 24700 Π time for A to again catch B

N = 1.39 * 24700 Answer: