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Ivahew [28]
3 years ago
10

What type of image is formed by a lens if m = -1.6?

Physics
2 answers:
IRISSAK [1]3 years ago
7 0

Answer: the image is larger than the object and is real.

Explanation:

anastassius [24]3 years ago
5 0

The image formed from the lens is real, inverted and enlarged

Explanation:

The magnification of a lens is given by:

m=\frac{y'}{y}=-\frac{q}{p} (1)

where:

y' is the size of the image

y is the size of the object

q is the distance of the image from the lens

p is the distance of the object from the lens

For this lens, the magnification is

m = -1.6

We notice the following:

- The magnification is negative

- The magnification is larger than 1

Looking at eq.(1), we can therefore make the following conclusions:

- SInce m=\frac{y'}{y} and |m|>1, it follows that |y'|>|y|, so the image is larger than the object

- Since m=\frac{y'}{y} and m, it follows that y' (since y > 0 for convention): this means that the image is inverted

- Since m=-\frac{q}{p} and m, it follows that q>0 (since p > 0 for convention): this means that the image is real

Learn more about lenses and mirrors here:

brainly.com/question/8737441

#LearnwithBrainly

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Explanation:

We can explain this using the dual nature of magnets.

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We assume that the magnetic lines of forces enters from south pole and leaves from the north pole.

When brought together, like poles repel each other while opposite poles attract each other.

In the picture, the lower two magnets have opposite poles facing each other, hence the force of repulsion is minimum there and the force of attraction is maximum.

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Friction produces heat

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What are the reactants of cellular respiration? Check all that apply. carbon dioxide energy glucose oxygen water
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Answer:

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Explanation:

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A certain frictionless simple pendulum having a length L and mass M swings with period T. If both L and M are doubled, what is t
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The new period is D) √2 T

\texttt{ }

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy and Period of Simple Pendulum formula as follows:

\boxed{E_p = \frac{1}{2}k x^2}

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

\texttt{ }

\boxed{T = 2\pi \sqrt{ \frac{L}{g} }}

where:

<em>T = period of simple pendulum ( s )</em>

<em>L = length of pendulum ( m )</em>

<em>g = gravitational acceleration ( m/s² )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

initial length of pendulum = L₁ = L

initial mass = M₁ = M

final length of pendulum = L₂ = 2L

final mass = M₂ = 2M

initial period = T₁ = T

<u>Asked:</u>

final period = T₂ = ?

<u>Solution:</u>

T_1 : T_2 = 2\pi \sqrt{ \frac{L_1}{g} }} : 2\pi \sqrt{ \frac{L_2}{g} }}

T_1 : T_2 = \sqrt{L_1} : \sqrt{L_2}

T : T_2 = \sqrt{L} : \sqrt{2L}

T : T_2 = 1 : \sqrt{2}

\boxed {T_2 = \sqrt{2}\ T}

\texttt{ }

<h3>Learn more</h3>
  • Kinetic Energy : brainly.com/question/692781
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  • Young Modulus : brainly.com/question/9202964
  • Simple Harmonic Motion : brainly.com/question/12069840

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

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