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nikklg [1K]
4 years ago
15

a small negatively charged sphere with a mass of 5.4*10^-5 is suspended between two parallel plates. the potential difference is

360V to hold the sphere stationary. what is the sphere's charge?

Physics
1 answer:
labwork [276]4 years ago
3 0
Here, Fe = Fg
q.E = m.g
We have: E = 360 V
m = 5.4 × 10⁻⁵
g = 9.8 m/s²   [ constant value for earth system ]

Substitute their values into the expression:
q (360) = 5.4 × 10⁻⁵ × 9.8
q = 52.92 × 10⁻⁵ / 360
q = -1.47 × 10⁻⁶  [ negative sign represents the nature of charge ] 

So, Your Final answer would be 1.47 × 10⁻⁶

Hope this helps!
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Answer:

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⇒displacement=16+9−−−−−√=25−−√=5m

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3 years ago
The most abundant climate type over the face of the earth is
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It is Humid Continental but i could be wrong
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Por una tubería de 0.06 m de diámetro circula agua con una velocidad desconocida, al llegar a la parte estrecha de la tubería de
Vesnalui [34]

Answer:

La velocidad con la que se desplaza el agua antes de llegar a la parte estrecha de la tubería es 1.156 \frac{m}{s}

Explanation:

La ecuación de continuidad es simplemente una expresión matemática del principio de conservación de la masa.  Este principio establece que la masa de un objeto o colección de objetos nunca cambia con el tiempo.

La ecuación de continuidad es la relación que existe entre el área y la velocidad que tiene un fluido en un lugar determinado y dice que el caudal de un fluido es constante a lo largo de un circuito hidráulico.

En otras palabras, la ecuación de continuidad se basa en que el caudal (Q) del fluido ha de permanecer constante a lo largo de toda la conducción. Cuando un fluido fluye por un conducto de diámetro variable, su velocidad cambia debido a que la sección transversal varía de una sección del conducto a otra.

Entonces, siendo el caudal es el producto de la superficie de una sección del conducto por la velocidad con que fluye el fluido,  en dos puntos de una misma tubería se cumple:

Q1=Q2

A1*v1= A2*v2

donde:

  • A es la superficie de las secciones transversales de los puntos 1 y 2 del conducto.
  • v es la velocidad del flujo en los puntos 1 y 2 de la tubería.

Siendo A=pi*r^{2} =pi*(\frac{D}{2} )^{2} =\frac{pi*D^{2} }{4} , donde pi es el número π, r es el radio del conducto y D el diámetro del conducto, entonces:

\frac{pi*D1^{2} }{4}*v1=\frac{pi*D2^{2} }{4}*v2

En este caso:

  • D1: 0.06 m
  • v1: ?
  • D2: 0.04 m
  • v2: 2.6 m/s

Reemplazando:

\frac{pi*(0.06m)^{2} }{4}*v1=\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}

Resolviendo:

v1=\frac{\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}}{\frac{pi*(0.06m)^{2} }{4}}

v1=\frac{(0.04m)^{2} }{(0.06m)^{2}  }*2.6\frac{m}{s}

v1= 1.156 \frac{m}{s}

<u><em>La velocidad con la que se desplaza el agua antes de llegar a la parte estrecha de la tubería es 1.156 </em></u>\frac{m}{s}<u><em></em></u>

8 0
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Answer: The magnitude of the acceleration is 11.25 m/s²
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The frequency of the other wave is 613 Hz or 607 Hz.

The difference between the frequencies of two waves is called the beat frequency.

Here, one wave has a frequency 610 Hz and the beat frequency is 3 beats per second.

Which has a higher frequency is not mentioned. Therefore, there are two possibilities.

Δf = | 610 - 613 | = 3

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Therefore, the frequency of the other wave is 613 Hz or 607 Hz.

Learn more about beat frequency here:

brainly.com/question/14157895

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