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Alexeev081 [22]
2 years ago
10

Which energy transformation occurs when a butane lighter is lit?

Physics
1 answer:
UNO [17]2 years ago
7 0
It is electrical energy into connected energy has in relation to the question
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1000 kg car takes travels on a circular track having radius 100 m with speed 10 m/s. What is the magnitude of the centripetal fo
Kryger [21]

Answer:

its a

Explanation:

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2 years ago
The "zone of avoidance" a) Accounts for the absence of any planet between the orbits of Mars and Jupiter Prevents collisions bet
shusha [124]

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b) obscures visual observation through the plane of the Milky Way.

Explanation:

The "zone of avoidance"

Avoidance zone, an area characterized by an apparent lack of galaxies near the Milky Way Galaxy plane and caused by the mysterious impact of interstellar dust. The zone of avoidance is completely a local Milky Way Galaxy effect.

The correct answer is

b) obscures visual observation through the plane of the Milky Way.

7 0
3 years ago
A parallel-plate capacitor has a plate separation of 1.5 mm and is charged to 450 V. 1) If an electron leaves the negative plate
Bumek [7]

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Explanation:

this is the answer to your question

8 0
3 years ago
What is an independent variable?
Natali5045456 [20]

Answer:

The answer is A

Explanation:

Independent variables don't have to depend on other factors of the experiment because they're independent

4 0
2 years ago
Two spherical conductors are separated by a distance much larger than either of their radii. Sphere A has a radius of 11.5 cm an
bonufazy [111]

Explanation:

As the given spheres are connected by a thin wire so, the potential on the spheres are the same.

          \frac{q_{1}}{r_{1}} = \frac{q_{2}}{r_{2}} ......... (1)

Hence, total charge will be as follows.

              q_{1} + q_{2} = Q = -95.5 nC .......... (2)

Using the above two equations, the final equation will be as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

and,    q_{1} = \frac{Qr_{1}}{r_{1} + r_{2}}

Hence, we will calculate the charge on sphere B after the equilibrium is reached as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

                     = \frac{-95.5 \times 74.4 cm}{(11.5 + 74.4) cm}

                     = 82.714 nC

Thus, we can conclude that the charge on sphere B after equilibrium has been reached is 82.714 nC.

                       

5 0
3 years ago
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