
<h2><em>calculate</em></h2>
<em>
</em>
<h2><em>reduce </em><em>the </em><em>numbers</em></h2>
<em>
</em>
<h2><em>multiply</em></h2>
<em>
</em>
<h2><em>there </em><em>for </em><em>we </em><em>have </em><em>a </em><em>solution</em><em> to</em><em> the</em><em> </em><em>equation</em></h2>
<em>hope </em><em>it</em><em> helps</em>
<em>#</em><em>c</em><em>a</em><em>r</em><em>r</em><em>y</em><em> </em><em>on</em><em> learning</em>
<em>mark </em><em>me</em><em> as</em><em> brainlist</em><em> plss</em>
The focal length of given concave lens will be -26.85 cm
The height of an image to the height of an object is the ratio that is used to determine a lens' magnification. Additionally, it is provided in terms of object and image distance. It is equivalent to the object distance to image distance ratio.
Given concave lens creates a virtual image at -47.0 cm and a magnification of +1.75.
We have to find focal length
The focal length can be found out by following way:
Magnification = m = +1.75
m = hi/h
hi = -47 cm
1.75 = -47/h
h = -26.85 cm
So the focal length of given concave lens will be -26.85 cm
Learn more about magnification factor here:
brainly.com/question/6947486
#SPJ10
Answer:
You would need to type the numbers in the question or you could have added a picture but you didn't so there is no way to answer this question. Have a nice day!
I didn’t know water has calories
Answer:
E=
Explanation:
We are given that
Charge on ring= Q
Radius of ring=a
We have to find the magnitude of electric filed on the axis at distance a from the ring's center.
We know that the electric field at distance x from the center of ring of radius R is given by

Substitute x=a and R=a
Then, we get




Where K=
Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=