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Cloud [144]
3 years ago
5

What an object is placed 8 mm from a concave spherical mirror a clear image can be projected on the screen 16 mm in front of me

to add you to the height of 4 mm the height of the image is
Physics
1 answer:
alexgriva [62]3 years ago
4 0

Concept: The magnification of spherical mirror can be defined by two ways.

(i) In terms of the height of the object and image.

The magnification of the spherical mirror is defined as the ratio of the height of the image'h_{i}' to the height of the object 'h_{o}'. It is denoted by letter 'm'.

Mathematically, it can be written as

m= \frac{h_{i}}{h_{o}}   ------------(1)

(ii) In terms of the object's and image's distances.

The magnification of the spherical mirror is defined as the negative ratio of the image distance'd_{i}' to the object distance 'd_{o}'.

Mathematically, it can be written as

m= - \frac{d_{i}}{d_{o}}   ------------(2)

Now, from equation (1) and (2) we have,

m = \frac{h_{i}}{h_{o}}   = -  \frac{d_{i}}{d_{o}}  -----------(3)

Given: Spherical Concave Mirror,

We will consider positive sign for object's and image's distance because both are in front of the mirror.

Object distance (d_{o}) = + 8 mm.

Image distance (d_{i}) = + 16 mm

Object's height (h_{o}) = + 4 mm

Image's height (h_{i}) =?

Now, apply equation (3)

\frac{h_{i}}{h_{o}}   = - \frac{d_{i}}{d_{o}}

Or,   \frac{h_{i}}{4 mm}   = - \frac{+16 mm}{+8 mm}

Or, hi = - 8 mm

Here; negative sign means, the image will be inverted.

The image's height will be 8 mm.

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Explanation:

When resistors are connected in series, effective resistance is given by

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               V = IR

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               R = 143Ω

Substituting

               V = 0.5 x 143 = 71.5 V

Potential difference is 71.5 V

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The cart will be pulled to the right by the hanging mass, so by Newton's second law, the net force on the cart is

<em>T</em> - 25 N = (8 kg) <em>a</em>

where <em>T</em> is the tension in the rope and <em>a</em> is the acceleration.

The hanging mass has a net force of

(6 kg) <em>g</em> - <em>T</em> = (6 kg) <em>a</em>

where <em>g</em> = 9.8 m/s².

Adding these equations together eliminates <em>T</em>, and we can solve for <em>a</em> :

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