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OverLord2011 [107]
3 years ago
12

The wattage of a bulb is 24 W when it is connected to a 12 V battery. Calculate its effective wattage if it operates on a 6 V ba

ttery.
Physics
2 answers:
posledela3 years ago
4 0

The power dissipated by a resistance ' R '
powered by a battery ' V ' is                            
                                                 P = V² / R .

Before we play with actual numbers, look at that formula, and
notice that the power is proportional to the square of the voltage. 
So if the voltage suddenly drops by half, then the power that's
dissipated by the resistance drops to 1/2² = 1/4 of where it started out.

Now that we have the pure Science worked out, we're ready to do
the Engineering, and apply our findings to some actual technology.

We have a light bulb that's designed to dissipate energy at the rate
of 24W when you connect it to a source of 12V .  But all we have
in the lab today is a 6V battery.  What power will our bulb dissipate
if we connect it to the 6V that we have, instead of the 12V that it's
designed for ?

We're all standing around scratching our beards, when the new hot-shot
post-doc lab assistant reminds us of the formula up in the first paragraph,
and shows us that since we're running it at 1/2 of the design voltage, we
expect it to dissipate 1/4 of design power = 6 watts.

It'll give us a bit of light so we can still walk around the lab without bumping
into things, but it won't get hot enough to warm our lunches.  We'll have to
do that over a Bunsen burner or in the microwave oven today, and one of
you junior members of the lab staff will have to stop at Lady-o'-Rack and
pick up either a new battery or a new bulb on your way in to work tomorrow.
tresset_1 [31]3 years ago
3 0
It would be 12W because: 6v is half of 12v so half of 24w would be 12w
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Answer:

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Explanation:

5 0
3 years ago
Read 2 more answers
You fill two balloons with gas, one with hydrogen and one with carbon dioxide. You hold a match to each balloon. The hydrogen ba
miv72 [106K]

Answer:E. Hydrogen was able to participate in an exergonic reaction and carbon dioxide couldn't

Explanation:

An exergonic reaction releases energy to the environment. The combustion of hydrogen contained in the balloon is a chemical reaction. The reaction can take place because hydrogen combines with oxygen in air, that is, the gas is combustible. CO2 does not support combustion, it does not combine with oxygen in air and it is also denser than air, hence does not participate in the exergonic reaction.

4 0
3 years ago
Constants A capacitor is connected across an ac source that has voltage amplitude 59 0 V and frequency 77 0 Hz Part C What is th
Vladimir79 [104]

Answer:

C = 1.77 \times 10^{-4} F

Explanation:

As we know that in AC circuit we have

V = i x_c

here we have

V = 59 V

i = 5.05 A

so we will have

x_c = \frac{59}{5.05}

x_c = 11.68 ohm

also we know that

x_c = \frac{1}{\omega C}

here we will have

11.68 = \frac{1}{(2\pi 77)C}

C = 1.77 \times 10^{-4} F

7 0
3 years ago
A certain digital camera having a lens with focal length 7.50 cmcm focuses on an object 1.70 mm tall that is 4.70 mm from the le
Crank

Answer:

a. 7.62cm

b. Real and inverted

c. 2.76 cm

d. 3450

Explanation:

We proceed as follows;

a. the lens equation that relates the object distance to the image distance with the focal length is given as follows;

1/f = 1/p + 1/q

making q the subject of the formula;

q = pf/p-f

From the question;

p = 4.70m

f = 7.5cm = 0.075m

Substituting these, we have ;

q = (4.7)(0.075)/(4.7-0.075) = 0.3525/4.625 = 0.0762 = 7.62 cm

b. The image is real and inverted since the image distance is positive

c. We want to calculate how tall the image is

Mathematically;

h1 = (q/p)h0

h1 = (7.62/4.70)* 1.7

h1 = 2.76 cm

d. We want to calculate the number of pixels that fit into this image

Mathematically:

n = h1/8 micro meter

n = 2.76cm/8 micro meter = 2.76 * 10^-2/8 * 10^-6 = 3450

3 0
3 years ago
Please answer both questions and not just one. Thanks!
Margarita [4]
It would be d and c hoped i helped!
8 0
3 years ago
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