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ivolga24 [154]
3 years ago
9

Assume that two of the electrons at the negative terminal have attached themselves to a nearby neutral atom. There is now a nega

tive ion with a charge -2e at this terminal. What are the electric potential and electric potential energy of the negative ion relative to the electron?
The electric potential and the electric potential energy are both twice as much.
The electric potential is twice as much and the electric potential energy is the same.
The electric potential is the same and the electric potential energy is twice as much.
The electric potential and the electric potential energy are both the same.
The electric potential is the same and the electric potential energy is increased by the mass ratio of the oxygen ion to the electron.
The electric potential is twice as much and the electric potential energy is increased by the mass ratio of the oxygen ion to the electron.
Physics
1 answer:
devlian [24]3 years ago
5 0

Answer: its negative

Explanation: becase it is

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zzz [600]

Answer:

See attached document

Explanation:

Entire process for deriving the asked expression dV across the bridge as function of dP is illustrated in the attachment below.

The document gives a step-by step process for arriving at the expression. However, manipulation of algebraic equations is skipped for the conciseness of the document.

It also gives the expression for the case when all resistors have different nominal values.

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6 0
3 years ago
A proton that has a mass m and is moving at +164 m/s undergoes a head-on elastic collision with a stationary carbon nucleus of m
Irina18 [472]
The concept of this problem is the Law of Conservation of Momentum. Momentum is the product of mass and velocity. To obey the law, the momentum before and after collision should be equal:

m₁ v₁ + m₂v₂ = m₁v₁' + m₂v₂', where
m₁ and m₂ are the masses of the proton and the carbon nucleus, respectively,
v₁ and v₂ are the velocities of the proton and the carbon nucleus before collision, respectively,
v₁' and v₂' are the velocities of the proton and the carbon nucleus after collision, respectively,

m(164) + 12m(0) = mv₁' + 12mv₂'
164 = v₁' + 12v₂'  --> equation 1

The second equation is the coefficient of restitution, e, which is equal to 1 for perfect collision. The equation is

(v₂' - v₁')/(v₁ - v₂) = 1
(v₂' - v₁')/(164 - 0) = 1
v₂' - v₁'=164 ---> equation 2

Solving equations 1 and 2 simultaneously, v₁' =  -138.77 m/s and v₂' = +25.23 m/s. This means that after the collision, the proton bounced to the left at 138.77 m/s, while the stationary carbon nucleus move to the right at 25.23 m/s.
7 0
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Martina seems to have several different personalities. In one, she is 7 years
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Answer:Dissociative Identity Disorder

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mass of remaining cake=0.7549kg

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