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2 years ago

Eight 7.0W Christmas tree lights are strung in series to a 110 V light source. What is the resistance of each bulb?

Physics

1 answer:

Evgesh-ka [11]2 years ago

8 0

Answer:

Resistance = 1728.57 Ohms

Explanation:

Given the following data;

Power = 7 Watts

Voltage = 110 Volts

To find the resistance of each bulb;

Mathematically, power can be calculated using the formula;

$Power = \frac {Voltage^{2}}{resistance}$

Making resistance the subject of formula, we have;

$Resistance = \frac {Voltage^{2}}{power}$

Substituting into the formula, we have;

$Resistance = \frac {110^{2}}{7}$

$Resistance = \frac {12100}{7}$

Resistance = 1728.57 Ohms

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Answer:

Explanation:

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3 years ago

Does someone know how to do math with that equation

mestny [16]

Answer:

f = 5 cm

Explanation:

using the thin lens equation, given as follows:

$\frac{1}{f} = \frac{1}{d_{o}}+\frac{1}{d_{i}}\\$

where,

f = focal length = ?

do = the distance of object from lens = 20 cm

di = the distance of image from lens = 6.6667 cm

Therefore,

$\frac{1}{f} = \frac{1}{20\ cm}+\frac{1}{6.6667\ cm}\\\\\frac{1}{f} = 0.199999\ cm^{-1}\\\\f = \frac{1}{0.199999\ cm^{-1}}\\\\$

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3 years ago

In terms of newton first law of motion why is it important to wear a seat belt

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Answer:

Seatbelts stop you

Explanation:

Any passengers in the car will also be decelerated to rest if they are strapped to the car by seat belts.

6 0

4 years ago

A bag of sugar weighs 3.50 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration isone-sixth

Alex73 [517]

Answer:

$F_{Earth}$ = 15.57 N

$F_{Moon}$ = 2.60 N

$F_{Uranus}$ = 16.98 N

The mass of the bag is the same on the three planets. m=1.59 kg

Explanation:

The weight of the sugar bag on Earth is:

g=9.81 m/s²

m=3.50 lb=1.59 kg

$F_{Earth}$ =m·g=1.59 kg×9.81 m/s²= 15.57 N

The weight of the sugar bag on the Moon is:

g=9.81 m/s²÷6= 1.635 m/s²

$F_{Moon}$ =m·g=1.59 kg× 1.635 m/s²= 2.60 N

The weight of the sugar bag on the Uranus is:

g=9.81 m/s²×1.09=10.69 m/s²

$F_{Uranus}$ =m·g=1.59 kg×10.69 m/s²= 16.98 N

The mass of the bag is the same on the three planets. m=1.59 kg

5 0

3 years ago

In young’s double slit experiment, the measured fringe width is 0.5 mm for a Sodium light of 589 nm at a distance of 1.5 m. A br

Levart [38]

Answer:

(A). The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

Explanation:

Given that,

Fringe width d = 0.5 mm

Wavelength = 589 nm

Distance of screen and slit D = 1.5 m

Distance of bright fringe y = 1 cm

(A) We need to calculate the order of the bright fringe

Using formula of wavelength

$\lambda=\dfrac{dy}{mD}$

$m=\dfrac{d y}{\lambda D}$

Put the value into the formula

$m=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{589\times10^{-9}\times1.5}$

$m=5.65 = 6$

(B). We need to calculate the width of the bright fringe

Using formula of width of fringe

$\beta=\dfrac{yd}{D}$

Put the value in to the formula

$\beta=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{1.5}$

$\beta=3.33\times10^{-6}\ m$

$\beta=3.33\ \mu m$

Hence, (A). The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

3 0

3 years ago