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3 years ago

What is the motion of the negative electrons and positive atomic nuclei caused by the external field?

Physics

1 answer:

telo118 [61]3 years ago

3 0

Answer:

The negative electrons moves to the left While the positively charged current moves in the opposite direction to the right when an external field is applied.

The positive atomic nuclei remains almost motionless.

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Una barra de aluminio que esta a 78 GRADOS CENTIGRADOS entra en contacto con una barra de cobre de la misma longitud y área que

stiks02 [169]

Answer:

Al llegar a su equilibrio térmico ambas barran tendrán una temperatura de 53 grados centígrados.

Explanation:

Dado que una barra de aluminio que está a 78 grados centígrados entra en contacto con una barra de cobre de la misma longitud y área que esta a 28 grados centígrados, y posteriormente se lleva acabo la transferencia de energía entre ambas barras llegando a su equilibrio térmico, para determinar la temperatura a la que ambas barras llegarán se debe realizar el siguiente cálculo:

(78 + 28) / 2 = X

106 / 2 = X

53 = X

Por lo tanto, al llegar a su equilibrio térmico ambas barran tendrán una temperatura de 53 grados centígrados.

8 0

3 years ago

Two forces that are not equal in size are

Bumek [7]

They are unbalanced forces ..... Hope this helps :3

4 0

3 years ago

78 million in standard form

Leviafan [203]

Answer:

78 million in standard form is 78,000,000

6 0

3 years ago

What does the Pauli exclusion principle state? . . A.Electrons will be added to an atom at the lowest possible level or sublevel

BARSIC [14]

The Pauli exclusion principle state that : D. Two electrons occupy the same orbital only if they have opposite spins

This happen because he stated that in an atom or molecule, two electrons CANNOT have same four electronic quantum numbers

hope this helps

5 0

3 years ago

You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state

posledela

Answer:

a) y₂ = 49.1 m , t = 1.02 s , b) y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

$v_{y}$ ² = $v_{oy}$ ² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

y-yo = $v_{oy}² /2 g
 y- 0 = 10.0²/2 9.8
 y - 0 = 5.10 m
 
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
 y₂ = 5.1 + 44
 y₂ = 49.1 m
Let's use the other equation to find the time
 [tex]v_{y}$ = $v_{oy}$ - g t

t = $v_{oy}$ / g

t = 10 / 9.8

t = 1.02 s

b) the maximum height

y- 44.0 = $v_{y}$ ² / 2 g

y - 44.0 = 5.1

y = 5.1 +44.0

y = 49.1 m

The time is the same because it does not depend on the initial height

t = 1.02 s

7 0

3 years ago