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IceJOKER [234]
4 years ago
12

What is the difference between a series circuit and a parallel circuit?

Physics
2 answers:
fenix001 [56]4 years ago
3 0
Hey....................

In a series circuit, the current through each of the components is the same, and the voltage across the circuit<span> is the sum of the voltages across each component.

 </span>In a parallel circuit<span>, the voltage across each of the components is the same, and the total current is the sum of the currents through each component.</span>
HACTEHA [7]4 years ago
3 0
In a series circuit, the voltage across the circuit<span> is the sum of the voltages across each component. </span>In a parallel circuit<span>, the total current is the sum of the currents through each component.</span>
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b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul
Nat2105 [25]

Answer:

b) Betelgeuse would be \approx 1.43 \cdot 10^{6} times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is \approx 1.37 \cdot 10^{-5} } the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

B = \displaystyle{\frac{L}{4\pi d^2}} (1)

where B is the brightness, L is the star luminosity and d, the distance from the star to the point where the brightness is calculated (measured). Thus:

b) B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}} and B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}} where L_{Sun} is the Sun luminosity (3.9 x 10^{26} W) but we don't need to know this value for solving the problem. ly is light years.

Finding the ratio between the two brightness we get:

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is 1.581 \cdot 10^{-5}\ ly. Then

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

4 0
3 years ago
A fine-grained sedimentary rock is known as _____.
Yakvenalex [24]
The finer‐grained<span> s</span>edimentary rocks<span> are called shale, siltstone, and mudstone</span>
5 0
4 years ago
A car’s bumper is designed to withstand a 4.0-km/h (1.1-m/s) collision with an immovable object without damage to the body of th
poizon [28]

Answer:

The force bumper at 0.200m

F=2722.5 N

Explanation:

Using the energy theorem of work

W=K_{f}- K_{i}

W=ΔK

W=F*d

ΔK=\frac{1}{2}*m*v_{f} ^{2} -\frac{1}{2}*m*v_{i} ^{2}

v_{i} =0

ΔK=F*d=\frac{1}{2}*m*v_{f} ^{2} -\frac{1}{2}*m*v_{i} ^{2}

F*d=\frac{1}{2}*m*0 ^{2}- \frac{1}{2}*m*v_{i} ^{2}\\F*d=-\frac{1}{2}*m*v_{i} ^{2}=\frac{1}{2}*900kg*(1.1\frac{m}{s})^{2}\\F*d=544.5\frac{kg*m}{s^{2}}\\ F=\frac{544.5\frac{kg*m}{s^{2}}}{0.2m} \\F=2722.5 N

3 0
3 years ago
The energy lost to friction is destroyed.TrueFalse
leva [86]
False!!!!!!!!!!!!!!!!!!
5 0
3 years ago
What is the Law of Conservation of Energy? A) Energy cannot be created, but can be destroyed as it transforms. B) Energy can be
ivann1987 [24]

the answer is d because energy cannot be created nor destroyed, only changed

4 0
3 years ago
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