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3 years ago

8

circuit in Mike’s home is almost at maximum current. If it is a 120 V circuit with about 1/3 of an amp available to safely use,

which bulb would be the maximum wattage Mike could use?

2 answers:

NikAS [45]3 years ago

3 0

Answer:

40 W

Explanation:

The maximum wattage (power) that Mike can use is given by:

$P=VI$

where

V is the voltage

I is the maximum current

In this problem, we have:

$V=120 V$ is the voltage

$I=\frac{1}{3}A=0.33 A$ is the maximum current available

Therefore, the maximum power that could be used is

$P=VI=(120 V)(0.33 A)=40 W$

Free_Kalibri [48]3 years ago

3 0

Answer:

40 W

Explanation:

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3 years ago

Suppose that you have been chosen for a space mission to a distant planet. Due to the length of time you'll be away from Earth y

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I should be active for 15 hours to meet the physical activity requirement.

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4 years ago

Consider a air-filled parallel-plate capacitor with plates of length 8 cm, width 5.52 cm, spaced a distance 1.99 cm apart. Now i

prohojiy [21]

Answer:

The ratio of the new potential energy to the potential energy before the insertion of the dielectric is 0.58

Explanation:

Given that,

Length of plates = 8 cm

Width = 5.52 cm

Distance = 1.99 cm

Dielectric constant = 2.6

Length = 4.4 cm

Potential = 0.8 V

We need to calculate the initial capacitance

Using formula of capacitance

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times8\times5.52\times10^{-4}}{1.99\times10^{-2}}$

$C=1.96\times10^{-12}$

We need to calculate the final capacitance

Using formula of capacitance

$C'=\dfrac{\epsilon_{0}A_{1}}{d}+\dfrac{k\epsilon_{0}A_{2}}{d}$

Put the value into the formula

$C'=(\dfrac{8.85\times10^{-12}}{1.99\times10^{-2}})((4.4\times5.52)+(3.6\times5.52)2.6)\times10^{-4}$

$C'=3.37\times10^{-12}$

We need to calculate the ratio of the new potential energy to the potential energy before the insertion of the dielectric

Using formula of energy

$\dfrac{E}{E'}=\dfrac{\dfrac{1}{2}CV^2}{\dfrac{1}{2}C'V^2}$

Put the value into the formula

$\dfrac{E}{E'}=\dfrac{1.96\times10^{-12}}{3.37\times10^{-12}}$

$\dfrac{E}{E'}=0.58$

Hence, The ratio of the new potential energy to the potential energy before the insertion of the dielectric is 0.58

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