All categories

3 years ago

A virtual image produced by a lens is always

2 answers:

5 0

The Answer Is D Because When Uu Magnify Large Items The Image Is Reflected Back To The Magnifying Glass Which Makes The Image Appear In The Back

Dmitry_Shevchenko [17]3 years ago

3 0

Option (D) is correct. The virtual image is always firmed in front of the lens or on the same side of the lens as of the object.

Explanation:

The image formed by the lens are categorized in two broad categories.

Real Images
Virtual Images

The Real Image is a image that is formed by the actual meeting of the incoming rays from the object. The real image is an image that can be obtained on the screen.

The Virtual Image is an image that is formed by the the imaginary meeting of the incoming rays from the object. The virtual image cannot be obtained on the screen.

Since the real image is formed by the actual meeting of the rays, the rays would meet on the other side of the lens as that of the object. It means that a real image is formed behind the lens whereas for a virtual image, the diverging rays from object are extended in backward direction and made to meet on the same size as that of the object.

Thus, Option (D) is correct. The virtual image is always firmed in front of the lens or on the same side of the lens as of the object.

Learn More:

1. Nature of the image formed by the mirror brainly.com/question/11743071

2. What type of mirror do dentist use brainly.com/question/997618

3. A small bulb has a resistance of 2 ohm when cold brainly.com/question/10421964

Answer Details:

Grade: Middle School

Subject: Physics

Chapter: Lens

Keywords:

virtual, image, real, object, rays, incoming, converging, diverging, lens, back of lens, front, smaller, larger, screen, obtained.

You might be interested in

Pleaseeeee HELPPPP THIS IS TIMED ALSO,

TEA [102]

Answer:

Friction, normal force, and weight

Explanation:

If the book slows down, it means that there must be friction acting in the opposite direction of the direction the book is moving in.

Weight is caused by the gravitational pull of the Earth on the book, and normal force is the table pushing the book up because the book is pushing down on the table (3rd law.)

Note that weight and normal force is not the 3rd law action-reaction pair. The pair is the force of the book on the table and the force of the table on the book.

8 0

2 years ago

Anyone please help me with this question ASAP

Katyanochek1 [597]

Answer:

increases

Explanation:

it would have to work harder to get to two points together

7 0

3 years ago

A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 7.0 cm has an energy of 14 J. If the

Bingel [31]

Answer:

$E_T= 28J$

Explanation:

The energy of Mass-Spring System the sum of the potential energy of the block plus the kinetic energy of the block:

$E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} mv^2$

Where:

$\Delta x=Amplitude\hspace{3}or\hspace{3}d eformation\hspace{3} of\hspace{3} the\hspace{3} spring\\m=Mass\hspace{3}of\hspace{3}the\hspace{3}block\\k=Constant\hspace{3}of\hspace{3}the\hspace{3}spring\\v=Velocity\hspace{3}of\hspace{3}the\hspace{3}block$

There are two cases, the first case is when the spring is compressed to its maximum value, in this case the value of the kinetic energy is zero, since there is no speed, so:

$E_T=\frac{1}{2} k \Delta x^2\\\\14=\frac{1}{2} k7^2\\\\Solving\hspace{3} for\hspace{3} k\\\\k=\frac{28}{49} =\frac{4}{7}$

The second case is when the block passes through its equilibrium position, in this case the elastic potential energy is zero since $\Delta x=0$ , so:

$E_T=\frac{1}{2} mv^2\\\\14=\frac{1}{2} mv^2\\\\Solving\hspace{3} for\hspace{3} v\\\\v^2=\frac{28}{m}$

Now, let's find the energy of the system when the block is replaced by one whose mass is twice the mass of the original block using the previous data:

$E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} m_2v^2$

Where in this case:

$m_2=New\hspace{3}mass=Twice\hspace{3} the\hspace{3} mass \hspace{3}of\hspace{3} the\hspace{3} original=2m$

Therefore:

$E_T=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{m_2})=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{2m})=14+14=28J$

8 0

3 years ago

What school did Ronald McNair go to and what kind of science did he work in

alina1380 [7]

Answer:

McNair graduated as valedictorian of Carver High School in 1967. In 1971, he received a Bachelor of Science degree in engineering physics, magna cu.m laude, from the North Carolina Agricultural and Technical State University in Greensboro, North Carolina.

5 0

2 years ago

An 8 g bullet leaves the muzzle of a rifle with

Elena-2011 [213]

Answer: 1872 N

Explanation:

This problem can be solved by using one of the Kinematics equations and Newton's second law of motion:

$V^{2}=V_{o}^{2} + 2ad$ (1)