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Tomtit [17]
3 years ago
8

The part of a circuit that increases the electric potential of the electrons is the _____. . . . . conducting wires. resistors.

battery. ammeter
Physics
2 answers:
Readme [11.4K]3 years ago
4 0
<span>The part of a circuit that increases the electric potential of the electrons is the battery. The correct option among all the options that are given in the question is the third option or the penultimate option. I hope that this is the answer that you were looking for and the answer has actually come to your desired help.</span>
masha68 [24]3 years ago
3 0
The part of a circuit that’s increases the electric potential of the electrons is the “battery”.
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A crate is pushed up a ramp at an angle of 30 degree by a 300 N force. How much power is spent in raising the crate to a height
kotegsom [21]

Answer:

You use a force of 150 N to push a 30 kg crate across the floor for a distance of 10 m. If the crate is moving at a speed of 5 m/s…

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The valency of oxygen is 2​
iren2701 [21]

Answer:

Yes the valency of oxygen is 2

5 0
3 years ago
A proton in a cyclotron is moving with a speed of 2.97×107 m/s in a circle of radius 0.568 m. 1.67 × 10−27 kg is the mass of the
vivado [14]

Answer:

B = 0.546 T,  F = 2.59 10⁻¹² N

Explanation:

The magnetic force is

            F = q v x B

We can calculate the magnitude of the force and find the direction by the right hand rule

          F = q v B sin θ

Let's use Newton's second law

         F = m a

Acceleration is centripetal

         a = v² / r

We substitute

       q v B sin θ = m v² / r

The angle between the field and the radius of the circle is 90º so sin 90 = 1

        q B = m v / r

        B = m v / q r

Let's calculate ’

       B = 1.67 10⁻²⁷ 2.97 10⁷ / (1.60 10⁻¹⁹ 0.568)

        B = 0.546 T

The foce is

         F = q v B

         F = 1.60 10⁻¹⁹ 2.97 10⁷ 0.546

         F = 2.59 10⁻¹² N

3 0
3 years ago
78. A film of oil on water will appear dark when it is very thin, because the path length difference becomes small compared with
Hoochie [10]

To solve this problem it is necessary to apply the concepts related to the condition of path difference for destructive interference between the two reflected waves from the top and bottom of a surface.

Mathematically this expression can be described under the equation

\delta = 2nt

Where

n = Refractive index

t = Thickness

In terms of the wavelength the path difference of the reflected waves can be described as

\delta = \frac{\lambda}{4}

Where

\lambda = Wavelenght

Equation the two equations we have that

2nt = \frac{\lambda}{4}

t = \frac{\lambda}{8n}

Our values are given as

\lambda = 380nm \rightarrow Wavelength of light

n = 1.4

t = \frac{380nm}{8*1.4}

t = 33.93nm

Therefore the minimum thickness of the oil for destructive interference to occur is approximately 34.0 nm

4 0
3 years ago
What property of equality can be used to solve the equation. n – 14 = 25
Inga [223]

Do 25-14 and you will get your answer

7 0
3 years ago
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