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IRINA_888 [86]
3 years ago
8

The ball is displaced to the left and then oscillates backwards and forwards between the two plates. The ball touches a plate on

ce every 0.05 s. Every time it touches a plate, a charge of 2.8 × 10−8 C (0.000 000 028 C) is transferred. Calculate the average current produced by the repeated transfer of charge.
Physics
1 answer:
ELEN [110]3 years ago
4 0

Answer:

Average current produced by the repeated transfer of charge is 5.6 × 10⁻⁷ ampere

Explanation:

The formula to be used here is

Q = It

where Q is the quantity of electricity and it is measured coulombs (C); 2.8 × 10⁻⁸ C or 0.000000028 C

I is current and it is measured in ampere (amps or A); unknown

t is time and it is measured in seconds (s); 0.05 s

Since, average current is what is unknown

I =Q/t

I = 0.000000028/0.05

I = 5.6 × 10⁻⁷ A

Average current produced by the repeated transfer of charge is 5.6 × 10⁻⁷ ampere

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From the geometry of the problem, the 20 m-long cable creates the hypotenuse of a right triangle, with the extended of the other two sides of size 20 m * cos(30 deg), which is around 17.3 m. Therefore, the ball has increased by 20 m - 17.3 m = 2.7 m. 

The potential energy will have altered by m*g*h, which is 1400 kg * 9.8 m/s^2 * 1.6 m , or about 37044 joules.

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3 years ago
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Un puente de acero de 100 m de largo a 8° C aumenta su temperatura a 24°C ¿Cuánto medirá su longitud? Valor del coeficiente de d
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La longitud <em>final</em> del puente de acero es 100.018 metros.

Asumamos que la dilatación <em>térmica</em> experimentada por el puente de acero es <em>pequeña</em>, de modo que podemos emplear la siguiente aproximación <em>lineal</em> para determinar la longitud <em>final</em> del puente de acero (L), en metros:

L = L_{o}\cdot [1+\alpha\cdot (T_{f}-T_{o})] (1)

Donde:

  • L_{o} - Longitud inicial del puente, en metros.
  • \alpha - Coeficiente de dilatación, sin unidad.
  • T_{o} - Temperatura inicial, en grados Celsius.
  • T_{f} - Temperatura final, en grados Celsius.

Si tenemos que L_{o} = 100\,m, \alpha = 11.5\times 10^{-6}, T_{o} = 8\,^{\circ}C y T_{f} = 24\,^{\circ}C, entonces la longitud final del puente de acero es:

L = (100\,m)\cdot [1+(11.5\times 10^{-6})\cdot (24\,^{\circ}C - 8\,^{\circ}C)]

L = 100.018\,m

La longitud <em>final</em> del puente de acero es 100.018 metros.

Para aprender más sobre dilatación térmica, invitamos cordialmente a ver esta pregunta verificada: brainly.com/question/24953416

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2 years ago
A neutron star and a black hole are 3.34 x 1012 m from each other at a certain point in their orbit. The neutron star has a mass
m_a_m_a [10]

Answer:

  F=1.65 x 10²⁶ N

Explanation:

Given that

Distance ,R= 3.34 x 10¹² m

Mass m₁= 2.78 x 10³⁰ kg

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we know that gravitational force F given as

F=G\dfrac{m_1m_2}{R^2}

G=Constant

G=6.67 x 10⁻¹¹ Nm²/kg²

Now by putting the values

F=6.67\times 10^{-11}\times \dfrac{2.78\times 10^{30}\times 9.94\times 10^{30}}{(3.34\times 10^{12})^2}\ N

F=1.65 x 10²⁶ N

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Answer:

A.always changing

Explanation:

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Two forces F1 and F2 are acting on a block of mass m=1.5 kg. The magnitude of
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Answer:

Explanation:

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Component of force F₂ in  vertically upward direction = F₁sin37

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b )

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F₂ - 9.58 = 1.5 x 2.5 = 3.75

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