Question

Помогите решить задачи! :)1) Предмет находиться на расстоянии 30 см от собирательной линзы, у которой отдалённость от главного фокуса - 60 см. Где(найти расстояние от линзы до изображения) и какое получается изображение?2) На расстоянии 100 м сфотографирована церковь. На фотографии её высота - 35 мм, фокус объектива отдалён на 5 см. Какой высоты церковь?3) Прожектором, у которого фокус объектива отдалён на 0,25 м, на экране получилось в 15 раз увеличенное изображение. На каком расстоянии от объектива находится экран?4) Как изменится частота свободных колебаний в контуре, если его вместимость увеличим в три раза, а индуктивность уменьшим во столько же раз?

Answer 1

№3
$D=\cfrac{1}{F}\\\cfrac{1}{F}=\cfrac{1}{f}+\cfrac{1}{d}\\\cfrac{1}{d}=\cfrac{1}{F}-\cfrac{1}{f}=15-4=12$
Ответ: на расстоянии 12м
№4
$T=2\pi\sqrt{LC}\\\nu=\cfrac{1}{T}\\C_2=3C_1\\L_1=3L_2\\\cfrac{T_1}{T_2}=\cfrac{\nu_2}{\nu_1}=\cfrac{\sqrt{L_1C_1}}{\sqrt{L_2C_2}}=\sqrt{\cfrac{3L_2C_1}{3C_1L_2}}=1$
Ответ: частота не изменится

Answer 2

Task 1: Finding the Image Distance and Image Characteristics

To find the distance from the lens to the image and the characteristics of the image, we can use the lens formula:

1. Distance from the lens to the image (v): - Given: Object distance (u) = -30 cm (since the object is on the same side as the incident light), focal length (f) = 60 cm - Using the lens formula: ``` 1/f = 1/v - 1/u ``` where f is the focal length, v is the image distance, and u is the object distance.

- Calculating the image distance (v): ``` 1/v = 1/f + 1/u 1/v = 1/60 + 1/(-30) 1/v = 1/60 - 1/30 1/v = (1 - 2)/60 1/v = -1/60 v = -60 cm ``` The negative sign indicates that the image is formed on the same side as the object.

2. Characteristics of the Image: - Since the image distance (v) is negative, the image is virtual and upright. - The magnification (M) can be calculated using the formula: ``` M = -v/u M = -(-60)/(-30) M = 2 ``` The positive magnification indicates that the image is upright.

Therefore, the image is virtual, upright, and located 60 cm on the same side as the object.

Task 2: Calculating the Height of the Church

To calculate the height of the church, we can use the concept of similar triangles and the magnification formula:

Given: - Height of the church on the photograph (h') = 35 mm - Focal length of the lens (f) = 5 cm - Actual distance to the church (u) = 100 m

Using the magnification formula: ``` M = h'/h = -v/u ``` where h' is the height of the church on the photograph, h is the actual height of the church, v is the image distance, and u is the object distance.

Rearranging the formula to solve for the actual height of the church (h): ``` h = h' * u / v h = 35 mm * 100 m / (-5 cm) h = 35 * 10 / (-0.05) h = 3500 / (-0.05) h = -70000 m ``` The negative sign indicates that the image is inverted.

Therefore, the actual height of the church is 70,000 meters.

Task 3: Determining the Distance of the Screen from the Lens

To determine the distance of the screen from the lens, we can use the magnification formula for a simple magnifying glass:

Given: - Magnification (M) = 15 - Focal length of the lens (f) = 0.25 m

Using the magnification formula: ``` M = 1 + (d/f) ``` where M is the magnification, d is the distance of the screen from the lens, and f is the focal length of the lens.

Rearranging the formula to solve for the distance of the screen from the lens (d): ``` d = f * (M - 1) d = 0.25 * (15 - 1) d = 0.25 * 14 d = 3.5 m ```

Therefore, the screen is located 3.5 meters from the lens.

Task 4: Effect of Changing Capacitance and Inductance on Frequency of Free Oscillations

When the capacitance in a circuit is increased by a factor of three and the inductance is decreased by the same factor, the frequency of free oscillations in the circuit will change.

The frequency of free oscillations in an LC circuit is given by the formula: ``` f = 1 / (2π√(LC)) ``` where f is the frequency, L is the inductance, and C is the capacitance.

If the capacitance (C) is increased by a factor of three and the inductance (L) is decreased by the same factor, the new frequency (f') can be calculated as: ``` f' = 1 / (2π√(L'/C')) ``` where L' is the new inductance and C' is the new capacitance.

Since the capacitance is increased by a factor of three and the inductance is decreased by the same factor, the new frequency (f') can be calculated as: ``