Question

Задача1 Период колебаний контура составляет 10мКс. Чему равен максимальный ток в катушке, если максимальное напряжение на конденсаторе равно 900В? макс энергия электрического поля составляет 0.9 мДж. задача2 Расстояние в воздухе от лампы до поверхности воды равно 1.2м. Наблюдатель находится в воде на глубине 60см. На каком расстоянии от себя он видит лампу? (желательно с подробным описанием

Answer 1

1.
(CU^2)/2=E(конденсатора)- это формула энергии конденсатора
C=(2E)/U^2;-тут мы выразили емкость конденсатора через формулу энергии конденсатора
C=2.2*10^-9 Ф- тут мы вычислили емкость конденсатора
T=2*pi*sqrt(LC)- формула Томсона
 L=T^2/(4pi^2*C)- выразили индуктивность катушки через формулу Томсона
 L=1.13*10^-3 Гн- тут вычислили индуктивность катушки
(L*I^2)/2=E(катушки)- формула энергии катушки
 I=sqrt(2*E/L) выразили ток через катушку
 I=1.25 А нашли искомое значение
Ответ: I=1.25 А
2 не знаю как делать

Answer 2

Problem 1: Maximum Current in the Coil

To find the maximum current in the coil, we need to use the given information about the period of oscillation and the maximum voltage across the capacitor.

The period of oscillation of the circuit is given as 10 ms (milliseconds), which can also be written as 0.01 seconds.

The maximum voltage across the capacitor is given as 900 V.

To find the maximum current, we can use the formula for the maximum current in an oscillating circuit:

I = (2πfC)V

Where: - I is the maximum current - f is the frequency of oscillation (which is the reciprocal of the period) - C is the capacitance of the capacitor - V is the maximum voltage across the capacitor

In this case, we are given the period, not the frequency. So, we need to calculate the frequency first.

The frequency (f) can be calculated as the reciprocal of the period (T):

f = 1 / T

Substituting the given period of 10 ms (0.01 seconds), we have:

f = 1 / 0.01 = 100 Hz

Now, we can substitute the values into the formula for the maximum current:

I = (2π * 100 * C) * 900

However, we don't have the value of capacitance (C) in the given information. Without the value of capacitance, we cannot calculate the maximum current in the coil.

Please provide the value of capacitance (C) so that we can proceed with the calculation.

Problem 2: Distance of Observer from the Lamp

To find the distance at which the observer sees the lamp, we need to consider the refraction of light at the air-water interface.

Given: - Distance from the lamp to the water surface = 1.2 m - Depth of the observer in the water = 60 cm = 0.6 m

Let's assume the observer is looking vertically upwards towards the lamp.

To find the distance at which the observer sees the lamp, we can use Snell's law, which relates the angles of incidence and refraction of light at the interface between two media.

Snell's law states:

n1 * sin(θ1) = n2 * sin(θ2)

Where: - n1 and n2 are the refractive indices of the two media (air and water, in this case) - θ1 is the angle of incidence (between the incident ray and the normal to the interface) - θ2 is the angle of refraction (between the refracted ray and the normal to the interface)

In this case, the observer is looking vertically upwards, so the angle of incidence (θ1) is 90 degrees.

The refractive index of air (n1) is approximately 1, and the refractive index of water (n2) is approximately 1.33.

Let's assume the distance at which the observer sees the lamp is x.

Using Snell's law, we can write:

1 * sin(90) = 1.33 * sin(θ2)

Since sin(90) = 1, the equation simplifies to:

1 = 1.33 * sin(θ2)

To find θ2, we can take the inverse sine (sin⁻¹) of both sides:

θ2 = sin⁻¹(1 / 1.33)

Using a calculator, we find that θ2 is approximately 48.75 degrees.

Now, we can use trigonometry to find the distance x.

In the right-angled triangle formed by the observer, the water surface, and the lamp, the angle between the observer's line of sight and the normal to the water surface is θ2.

Let's call the distance from the observer to the lamp (along the line of sight) as y.

Using the tangent function, we can write:

tan(θ2) = y / 0.6

Substituting the value of θ2, we have:

tan(48.75) = y / 0.6

Solving for y, we find:

y = 0.6 * tan(48.75)

Using a calculator, we find that y is approximately 0.76 m.

Therefore, the observer sees the lamp at a distance of approximately 0.76 meters from themselves.

Please note that the above calculations assume that the light rays from the lamp are not significantly deviated or scattered by the water.