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MaRussiya [10]
1 year ago
12

compared to a solid wire of the same gauge, stranded wire? select one: a. has a slightly larger diameter b. will carry more curr

ent c. is less susceptible to interference d. is more susceptible to interference
Physics
1 answer:
Sedbober [7]1 year ago
5 0

compared to a solid wire of the same gauge, stranded wire is more susceptible to interference.

A solid wire consists of a solid metal core while stranded wires are made of a quantity of thinner wires that are twisted together into an organized bundle.

But because of the bundle feature of the stranded wires, they have more surface areas in comparison to the solid wires, so there is more chances to have dissipation in this type of wires, specially when high frequency current is being passed.

Due to more dissipation it is more likely to have interference.

To learn more about stranded wire, visit here

brainly.com/question/28987082

#SPJ4

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Answer:   13.0m/s

Explanation:

26sin(30)=v in y-direction or vertical direction

v=26m/s*sin(30)=26m/s*(1/2)=13.0m/s

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Can someone help me with this science question like uhm!??!? anyone
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Answer:

mechanical layers of the earth

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7 0
2 years ago
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
5. During the annual shuffleboard competition, Renee gives her puck an initial speed of 9.32 m/s. Once leaving her stick, the pu
abruzzese [7]
V=0 v²=0, A=v-u/t. T=v-u/a. T= 0-9.32/-4.06 therefore time = 2.296 seconds
3 0
2 years ago
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