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nignag
3 months ago
14

A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through

a potential difference of 2V0, what speed would it gain? A proton is accelerated from rest through a potential difference and gains a speed . If it were accelerated instead through a potential difference of , what speed would it gain? v02√ 4v0 8v0 2v0 SubmitReques
Physics
1 answer:
inna [3.1K]3 months ago
6 0
Answer: The resulting speed is \sqrt{2}v_{0}. Option (a) stands as the correct choice. Explanation: Given the context, the potential difference entails calculations linked to speed assessment. If instead accelerated through a different potential difference, the resulting speed will be computed accordingly.
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Two large insulating parallel plates carry charge of equal magnitude, one positive and the other negative, that is distributed u
Maru [3345]

Answer:

The correct choice is C: points 1, 4, and 5 are equal, followed by 2 and 3 being equal.

Explanation:

Here’s the breakdown:

The electric field from the positive sheets E₁ = б/2E₀

E₂ is from the negative sheet = -б/2E₀

At points 1, 4, and 5, the electric fields created by the sheets oppose each other.

At point 1, the total field is calculated as -E₁ + E₂ = 0.

Similarly, at point A, the total field results in -E₁ - E₂ = 0.

However, at any point in between the plates, the electric field is directed consistently in one way.

At points 2 and 3, the field is directed to the right.

Thus, we have:

E net = E₁ + E₂

= б/2E₀ + -б/2E₀

=б/E₀

Note: Please refer to the attached document for the full question accompanying this solution.

7 0
3 months ago
Three moles of an ideal gas with a molar heat capacity at constant volume of 4.9 cal/(mol∙K) and a molar heat capacity at consta
ValentinkaMS [3465]

Answer:

The amount of heat that enters the gas throughout this two-step process totals 120 cal.

Explanation:

Given that,

Moles present = 3

Heat capacity at volume held constant = 4.9 cal/mol.K

Heat capacity at pressure held constant = 6.9 cal/mol.K

Starting temperature = 300 K

Ending temperature = 320 K

We are tasked with determining the heat absorbed by the gas at constant pressure

Employing the heat formula

\Delta H_{1}=nC_{p}\times\Delta T

Substituting the values into the equation

\Delta H_{1}=3\times6.9\times(320-300)

\Delta H_{1}=414\ cal

Next, we calculate the heat absorbed by the gas at constant volume

Using the corresponding heat formula

\Delta H_{1}=nC_{v}\times\Delta T

Insert the values into the formula

\Delta H_{1}=3\times4.9\times(300-320)

\Delta H_{1}=-294\ cal

Now, it's necessary to evaluate the total heat flow into the gas during both steps

Using the total heat formula

\Delta H_{T}=\Delta H_{1}+\Delta H_{2}

\Delta H_{T}=414-294

\Delta H_{T}=120\ cal

Thus, the heat that transfers into the gas throughout this two-step process amounts to 120 cal.

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