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AVprozaik [17]
3 years ago
8

A horizontal disk rotates about a vertical axis through its center. Point P is midway between the center and the rim of the disk

, and point Q is on the rim. If the disk turns with constant angular velocity, which of the following statements about it are true? (There may be more than one correct choice.)
A. P and Q have the same linear acceleration.
B. The angular velocity of Q is twice as great as the angular velocity of P.
C. The linear acceleration of P is twice as great as the linear acceleration of Q.
D. The linear acceleration of Q is twice as great as the linear acceleration of P.
E. is moving twice Q as fast as P.
Physics
1 answer:
hjlf3 years ago
6 0

Answer:

The answer is explained below.

Explanation:

All the point on the disk has same angular acceleration. Here, the point P is at the midway between the center and the rim of the disk and the point Q is at rim of the disk.

So, the distance of the point Q from the axis is twicee the distance of the point P from the axis.

<em>Rp - R</em>

<em>Rq - 2R</em>

The linear acceleration is

α2 - Rα

So, the linear acceleration of Q is twice as great as the linear acceleration of P.

The speed of the particle when it is in the circular motion depends on the radius of the particle.

In this case, the speed of point Q is twice the speed of point P.

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¿cual es la velocidad de un haz de electrones que marchan sin desviarse cuando pasan a traves de un campo magnetico perpendicula
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Answer:

La velocidad del haz de electrones es 1.78x10⁵ m/s. Este valor se obtuvo asumiendo que el campo magnético dado (3500007) estaba en tesla y que la fuerza venía dada en nN.

Explanation:

Podemos encontrar la velocidad del haz de electrones usando la Ley de Lorentz:

F = |q|vBsin(\theta)     (1)

En donde:

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q: es el módulo de la carga del electron = 1.6x10⁻¹⁹ C

v: es la velocidad del haz de electrones =?

B: es el campo magnético = 3500007 T

θ: es el ángulo entre el vector velocidad y el campo magnético = 90°

Introduciendo los valores en la ecuación (1) y resolviendo para "v" tenemos:

v = \frac{F}{qBsin(\theta)} = \frac{100 \cdot 10^{-9} N}{1.6 \cdot 10^{-19} C*3500007 T*sin(90)} = 1.78 \cdot 10^{5} m/s            

Este valor se calculó asumiendo que el campo magnético está dado en tesla (no tiene unidades en el enunciado). De igual manera se asumió que la fuerza indicada viene dada en nN.

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Espero que te sea de utilidad!                                        

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The concept required to solve this problem is linked to inductance. This can be defined as the product between the permeability in free space by the number of turns squared by the area over the length. Recall that Inductance is defined as the opposition of a conductive element to changes in the current flowing through it. Mathematically it can be described as

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