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svetlana [45]
3 years ago
8

Which of the following represents a virtual image? ​

Physics
1 answer:
Sauron [17]3 years ago
7 0

Answer:

A

Explanation:

A virtual image is on the same side as the object, which by convention is drawn on the left side.  So -di would represent a virtual image.

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A circular area with a radius of 6.50 cm lies in the xy-plane. What is the magnitude of the magnetic flux through this circle du
White raven [17]

Answer:

a. \phi _B=3.0528\times10^-^3T\ m^2\\ \\b. \phi_B=1.83299\times10^-^3T\ m^2\\\\c.\phi_B=0

Explanation:

#Consider a circular area of radius R=2.98cm in the xy-plane at z=0. This means all the are vector points toward the +ve z-axis.

a. first, find the magnetic flux if the magnetic field has a magnitude of B=0.23T and points toward the +ve z-axis. The angle between the magnetic field and the area is \theta=0. Hence the magnetic flux:-

\phi _B=\int {\bar B} . d\bar A \\=\int BdAcos(\theta)=BAcos(0)=BA\\\\=\pi R^2B=\pi(6.50\times10^-^3m)^2(0.230T)\\=3.0528\times10^-^3 T\ m^2

Hence flux magnitude in +z direction is 3.0528\times10^-^3T \ m^2

b. We now find the magnetic flux when the field has a magnitude of <em>B=0.230T</em> and points at an angle of \theta=53.1\textdegree from the +z direction.

Magnetic flux is calculated as:

\phi _B=\int\bar B \bar dA\\=\int BdAcos (\theta)=BAcos(0)=BA\\=\pi R^2B=\pi(6.50\times 10^-^2m)^2(0.230T)\\=1.83299\times 10^-^3 T \ m^2

Hence the flux at an angle of 53.1\textdegree is 1.83299\times 10^-^3T \ m^2

c. We now need to find the magnetic flux if the field has a magnitue of B=0.230T and points in the direction of +y-direction. As with the previous parts, the magnetic flux will be calculated as:

\phi_B= \int\bar B \times d\bar A\\=\int BdAcos(\theta)\\=BAcos(90\textdegree)\\=0

Hence the magnetic flux in the +y-direction is zero.

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A force F moves a load from the bottom of a slope to the top
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