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3 years ago

Which is a TRUE statement about a series circuit

Physics

1 answer:

Ray Of Light [21]3 years ago

6 0

Answer:

A the charge has only one path through which it can flow.

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Find the momentum of a 25 kg object going 4m/s to the right.

Ronch [10]

X=mass × velocity
x=25×4
=100kgm/s

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4 years ago

Is the answer D?

irga5000 [103]

Nope answer is 500 N downward
draw the freebody diagram to understand

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4 years ago

Which of the following elements is in Group 2?

Rzqust [24]

Answer:

calcium

Explanation:

if you have a periodic table that would be very helpful to you

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3 years ago

Vocabulary word for the total amount of charge in a closed system remains constant

Usimov [2.4K]

Answer:

i think this is it i dont know tho A conservation law stating that the total electric charge of a closed system remains constant over time, regardless of other possible changes within the system. "Conservation of charge." YourDictionary. LoveToKnow

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3 years ago

We can reasonably model a 90-W incandescent lightbulb as a sphere 7.0cm in diameter. Typically, only about 5% of the energy goes

Ronch [10]

Answer:

292.3254055 W/m²

469.26267 V/m

$1.56421\times 10^{-6}\ T$

Explanation:

P = Power of bulb = 90 W

d = Diameter of bulb = 7 cm

r = Radius = $\frac{d}{2}=\frac{7}{2}=3.5\ cm$

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

c = Speed of light = $3\times 10^8\ m/s$

The intensity is given by

$I=\frac{P}{A}\\\Rightarrow I=\frac{90}{4\pi 0.035^2}\\\Rightarrow I=5846.50811\ W/m^2$

5% of this energy goes to the visible light so the intensity is

$I=0.05\times 5846.50811\\\Rightarrow I=292.3254055\ W/m^2$

The visible light intensity at the surface of the bulb is 292.3254055 W/m²

Energy density of the wave is

$u=\frac{1}{2}\epsilon_0E^2$

Energy density is also given by

$\frac{I}{c}=\frac{1}{2}\epsilon_0E^2\\\Rightarrow E=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E=\sqrt{\frac{2\times 292.3254055}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E=469.26267\ V/m$

The amplitude of the electric field at this surface is 469.26267 V/m

Amplitude of a magnetic field is given by

$B=\frac{E}{c}\\\Rightarrow B=\frac{469.26267}{3\times 10^8}\\\Rightarrow B=1.56421\times 10^{-6}\ T$

The amplitude of the magnetic field at this surface is $1.56421\times 10^{-6}\ T$

7 0

3 years ago