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3 years ago

In concave mirror, the size of image depends upon

Physics

2 answers:

irina1246 [14]3 years ago

8 0

Answer:

The distance of the object placed on the principal axis from the concave mirror.

Explanation:

In a concave mirror, the nature of the image formed formed by the object placed in front of the mirror depends on the position of the object placed in from of the mirror. It all depends on the distance between the mirror and the object placed on the principal axis.

The closer the object is to the lens, the more larger or magnified the image formed will be. For example an object placed between the focal point and the pole of a concave produces a much larger image than an object placed beyond the centre of curvature of such mirror.

Dmitrij [34]3 years ago

4 0

Answer:

The location of the object relative to the vertex, V, the focus, F center of curvature, C and infinity

Explanation:

The location of the object relative to the vertex, V, the focus, F center of curvature, C and infinity such that;

The image of an object placed at V, touching the mirror is the upright and the same size as the object and virtual

The image of an object placed between V and F is upright and magnified also virtual

The image of an object placed at the focus is at infinity

The image of an object placed between C and F is real, magnified and inverted

The image of an object placed at C is real, same size and inverted

The image of an object placed between C and infinity is real, diminished and inverted

The image of an object placed at infinity is size zero and real.

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viktelen [127]

lol what is this question

5 0

2 years ago

The wavelength of a certain portion of an electromagnetic wave is 314 nm. what is its frequency and classification?

nadezda [96]

The frequency of the electromagnetic wave is 9.55 × 1014 Hz and it is classified as ultraviolet.

<h3>What is meant by electromagnetic waves?</h3>

Electromagnetic waves are forms of energy that are invisible and travel throughout the universe. However, some of the effects of this energy are visible. The light that we see is a component of the electromagnetic spectrum.

Electromagnetic waves, or EM waves, are produced by vibrations between an electric field and a magnetic field. In other words, electromagnetic waves are made up of oscillating magnetic and electric fields.

<h3>How do you calculate the speed of an electromagnetic wave?</h3>

The wavelength and frequency of any periodic wave are used to calculate its speed. v = λf.

In free space, the speed of any electromagnetic wave is equal to the speed of light, c = 3 $*$ 108 m/s.

The frequency of the electromagnetic wave is 9.55 × 1014 Hz and it is classified as ultraviolet.

To learn more about electromagnetic wave refer to:

brainly.com/question/25847009

#SPJ4

5 0

1 year ago

Two long, parallel wires separated by 3.50 cm carry currents in opposite directions. The current in one wire is 1.55 A, and the

vaieri [72.5K]

Answer:

Therefore,

The magnitude of the force per unit length that one wire exerts on the other is

$\dfrac{F}{l}=2.79\times 10^{-5}\ N/m$

Explanation:

Given:

Two long, parallel wires separated by a distance,

d = 3.50 cm = 0.035 meter

Currents,

$I_{1}=1.55\ A\\I_{2}=3.15\ A$

To Find:

Magnitude of the force per unit length that one wire exerts on the other,

$\dfrac{F}{l}=?$

Solution:

Magnitude of the force per unit length on each of @ parallel wires seperated by the distance d and carrying currents I₁ and I₂ is given by,

$\dfrac{F}{l}=\dfrac{\mu_{0}\times I_{1}\times I_{2}}{2\pi\times d}$

where,

$\mu_{0}=permeability\ of\ free\ space =4\pi\times 10^{-7}$

Substituting the values we get

$\dfrac{F}{l}=\dfrac{4\pi\times 10^{-7}\times 1.55\times 3.15}{2\pi\times 0.035}$

$\dfrac{F}{l}=2.79\times 10^{-5}\ N/m$

Therefore,

The magnitude of the force per unit length that one wire exerts on the other is

$\dfrac{F}{l}=2.79\times 10^{-5}\ N/m$

7 0

3 years ago

Read 2 more answers

Using hooke's law find the elastic constant of a spring that stretches 2 cm when 4newton force is applied to it

erica [24]

<u>Answer:</u>

2N/cm

<u>Step-by-step explanation:</u>

According to the Hooke's Law, the force required to extend or compress a spring is directly proportional distance you can stretch it, which is represented as:

$F=kx$