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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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An insulating sphere is 8.00 cm in diameter and carries a 6.50 ÂµC charge uniformly distributed throughout its interior volume.

Kobotan [32]

Explanation:

(a) Formula to calculate the density is as follows.

$\rho = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}$

= $2.42 \times 10^{-2} C/m^{3}$

Now, calculate the charge as follows.

$q_{in} = \rho(\frac{4}{3} \pi r^{3})$

= $2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}$

= $10.106 \times 10^{-8}$ C

or, = 101.06 nC

(b) For r = 6.50 cm, the value of charge will be calculated as follows.

$q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}$

= 7.454 $\mu C$

7 0

3 years ago

Which of the following is an example of resourcefulness?

tatiyna

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5 0

3 years ago

The length of a certain wire is kept same while its radius is doubled. what is the new resistivity of this wire?

anastassius [24]

The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.

The relationship between the resistance and the resistivity of a wire is
$\rho = \frac{AR}{L}$
where
$\rho$ is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length

the cross-sectional area is given by
$A=\pi r^2$
where r is the radius of the wire. Substituting in the previous equation ,we find
$\rho = \frac{\pi r^2 R}{L}$

For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
$\rho' = \frac{\pi r'^2 R}{L'}= \frac{\pi (2r)^2 R}{L}=4 \frac{\pi r^2 R}{L} = 4 \rho$
Therefore, the new resistivity must be 4 times the original one.

5 0

3 years ago

Read 2 more answers

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UNO [17]

Answer:

160 meters I think

Explanation:

if I'm wrong I'm a troop juSt trying to get points to ask a couple of questions

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