Answer:
BaF2
Explanation:
since you got the valence numbers just do the scissors move where you:
give the F the 2 and give the Ba the 1 so it be like
BaF2 here is the chemical compound
Answer:
B = 4.059 x 10¹⁵ T
Explanation:
Given,
Number of loop, N = 400
radius of loop, r = 0.65 x 10⁻¹⁵ m
Current, I = 1.05 x 10⁴ A
Magnetic field at the center of the loop
![B = \dfrac{\mu_0NI}{2R}](https://tex.z-dn.net/?f=B%20%3D%20%5Cdfrac%7B%5Cmu_0NI%7D%7B2R%7D)
![B = \dfrac{4\pi\times 10^{-7}\times 400 \times 1.05 \times 10^4}{2\times 0.65\times 10^{-15}}](https://tex.z-dn.net/?f=B%20%3D%20%5Cdfrac%7B4%5Cpi%5Ctimes%2010%5E%7B-7%7D%5Ctimes%20400%20%5Ctimes%201.05%20%5Ctimes%2010%5E4%7D%7B2%5Ctimes%200.65%5Ctimes%2010%5E%7B-15%7D%7D)
B = 4.059 x 10¹⁵ T
Answer:
The factors that affect are depth of the fluid and its density
The angles in the triangle are 91 degrees, 53 degrees and 36 degrees respectively.
<h3>What is the cosine rule?</h3>
From the cosine rule we know that;
c^2 = a^2 + b^2 - 2abcosC
Since;
a = 0.47 m
b = 0.62 m
c = 0.78 m
Then;
(0.78)^2 = (0.47)^2 + (0.62)^2 - 2(0.47 * 0.62)cosC
0.61 = 0.22 + 0.38 - 0.58 cosC
0.61 - ( 0.22 + 0.38) = - 0.58 cosC
0.01 = - 0.58 cosC
C = cos-1(0.01/-0.58)
C = 91 degrees
Using the sine rule;
b/Sin B = c/Sin C
0.62/sinB = 0.78/sin 91
0.62/Sin B = 0.78
B = sin-1 (0.62//0.78)
B = 53 degrees
Angle A is obtained from the sum of angles in a triangle;
180 - (91 + 53)
A = 36 degrees
Learn more about triangle:brainly.com/question/2773823
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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.