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Novosadov [1.4K]
3 years ago
14

There is a 12 v potential difference between the positive and negative ends of the jumper cables, which are a short distance apa

rt. an electron at the negative end ready to jump to the positive end has a certain amount of potential energy. on what quantities does this electrical potential energy depend? view available hint(s)
Physics
1 answer:
steposvetlana [31]3 years ago
4 0
The electric potential energy of the electron depends on the potential difference applied between the two ends of the cable. Indeed, the electric potential energy of a charge is given by
U=q \Delta V
where q is the magnitude of the charge, while \Delta V is the potential difference applied. So, U depends on \Delta V.
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A 0.20 mass on a horizontal spring is pulled back a certain distance and released. The maximum speed of the mass is measured to
olga55 [171]

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d

Explanation:

Ya gon find the Kenitic Energy first

K=½mv²===> K=½×0.2×(0.2)²===> 0.1(0.04)===> 0.004

and now the replacement:

0.004=½×0.4V²====> v²=0.02===> V=0.14m/s

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In 2.5 S, a Car increases its speed from 20.0m/s to 25.0 m/s what is the acceleration of the car​
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2 m/s

Explanation:

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3 years ago
Will give correct answer brainliest
sineoko [7]

200 \times 80\%

<h2><em>calculate</em></h2>

<em>200 \times  \frac{80}{100}</em>

<h2><em>reduce </em><em>the </em><em>numbers</em></h2>

<em>2 \times 80</em>

<h2><em>multiply</em></h2>

<em>= 160</em>

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<em>hope </em><em>it</em><em> helps</em>

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6 0
3 years ago
Read 2 more answers
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 car changes velocity at a constant acceleration of 2.5m/s to reach 43.7m/s in 2.7 s how fast was the car moving when it began
Montano1993 [528]

The formula we can use in this case is:

v = v0 + a t

where v is final velocity, v0 is initial velocity, a is acceleration and t is time

So finding for v0:

v0 = v – a t

v0 = 43.7 – (2.5) 2.7

v0 =  36.95 m/s

8 0
3 years ago
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