Answer:
The final image relative to the converging lens is 34 cm.
Explanation:
Given that,
Focal length of diverging lens = -12.0 cm
Focal length of converging lens = 34.0 cm
Height of object = 2.0 cm
Distance of object = 12 cm
Because object at focal point
We need to calculate the image distance of diverging lens
Using formula of lens
![\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bv%7D%3D%5Cdfrac%7B1%7D%7Bf%7D-%5Cdfrac%7B1%7D%7Bu%7D)
![\dfrac{1}{v}=\dfrac{1}{-12}-\dfrac{1}{-12}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bv%7D%3D%5Cdfrac%7B1%7D%7B-12%7D-%5Cdfrac%7B1%7D%7B-12%7D)
![v=\infty](https://tex.z-dn.net/?f=v%3D%5Cinfty)
The rays are parallel to the principle axis after passing from the diverging lens.
We need to calculate the image distance of converging lens
Now, object distance is ∞
Using formula of lens
![\dfrac{1}{v}=\dfrac{1}{34}-\dfrac{1}{\infty}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bv%7D%3D%5Cdfrac%7B1%7D%7B34%7D-%5Cdfrac%7B1%7D%7B%5Cinfty%7D)
![v=34](https://tex.z-dn.net/?f=v%3D34)
The image distance is 34 cm right to the converging lens.
Hence, The final image relative to the converging lens is 34 cm.
Answer:
b. Cho biết tốc độ của xe là 4m/s. Hãy tính tốc độ góc của điểm trên vành ngoài bánh xe.
Explanation:
#CarryOnLearning
That would be a decomposer
Hope this helps!
<u>Answer</u>
10 feet
<u>Explanation</u>
The Occupational Safety and Health Administration (OSHA) requires one to be not less than 10 feet from overhead power lines that are 0-50,000 v for him or her safety suppose. High voltages above 50 Kv needs a bigger distance that 10 feet.
When current is flowing through a conductor, it produces an electric fields all around it. These fields spreads far from the conductor depending on the current flowing in it. When this fields are strong enough the can cause electrocution.
Complete Question:
Suppose
, where A has the dimensions LT, B has dimensions L²T⁻¹, and C has dimensions LT². Then the exponents n and m have the values
Answer:
The value of n = ¹/₅
The value of m = ³/₅
Explanation:
Given dimensions;
A = LT
B = L²T⁻¹
C = LT²
The values of n and m are calculated as follows;
![LT = [L^2T^{-1}]^n[LT^2]^m\\\\L^1T^1 = [L^{2n}T^{-n}]\times [L^mT^{2m}]\\\\L^1 \times T^1 = [L^{(2n+m)}] \times [T^{(-n +2m)}]\\\\1 = 2n + m -----(1)\\\\1 = -n + 2m ----(2)\\\\from \ (1); \ m = 1-2n, \ \ substitute \ the \ value \ of \ m \ in\ (2)\\\\1= -n +2(1-2n)\\\\1 = -n + 2-4n\\\\1-2 = -5n\\\\-1 = -5n\\\\1= 5n\\\\n = \frac{1}{5} \\\\m = 1 - 2n\\\\m = 1 - 2(\frac{1}{5} )\\\\m = 1- \frac{2}{5} \\\\m = \frac{3}{5}](https://tex.z-dn.net/?f=LT%20%3D%20%5BL%5E2T%5E%7B-1%7D%5D%5En%5BLT%5E2%5D%5Em%5C%5C%5C%5CL%5E1T%5E1%20%3D%20%5BL%5E%7B2n%7DT%5E%7B-n%7D%5D%5Ctimes%20%5BL%5EmT%5E%7B2m%7D%5D%5C%5C%5C%5CL%5E1%20%5Ctimes%20T%5E1%20%3D%20%5BL%5E%7B%282n%2Bm%29%7D%5D%20%5Ctimes%20%5BT%5E%7B%28-n%20%2B2m%29%7D%5D%5C%5C%5C%5C1%20%3D%202n%20%2B%20m%20-----%281%29%5C%5C%5C%5C1%20%3D%20-n%20%2B%202m%20----%282%29%5C%5C%5C%5Cfrom%20%20%5C%20%281%29%3B%20%5C%20m%20%3D%201-2n%2C%20%5C%20%5C%20substitute%20%5C%20the%20%5C%20value%20%5C%20of%20%5C%20m%20%5C%20in%5C%20%20%282%29%5C%5C%5C%5C1%3D%20-n%20%2B2%281-2n%29%5C%5C%5C%5C1%20%3D%20-n%20%2B%202-4n%5C%5C%5C%5C1-2%20%3D%20-5n%5C%5C%5C%5C-1%20%3D%20-5n%5C%5C%5C%5C1%3D%205n%5C%5C%5C%5Cn%20%3D%20%5Cfrac%7B1%7D%7B5%7D%20%5C%5C%5C%5Cm%20%3D%201%20-%202n%5C%5C%5C%5Cm%20%3D%201%20-%202%28%5Cfrac%7B1%7D%7B5%7D%20%29%5C%5C%5C%5Cm%20%3D%201-%20%5Cfrac%7B2%7D%7B5%7D%20%5C%5C%5C%5Cm%20%3D%20%5Cfrac%7B3%7D%7B5%7D)