An object distance is
presented as s = 5f and we know that the mirror equation relates the image
distance to the object distance and the focal length.
The mirror equation is
1/f = 1/s + 1/s’ where the variable f stands for
the focal length of the mirror. Variable (s)
represents the distance between the mirror surface and the object and the
variable <span>(s’) represents the distance between the mirror surface and
the image. </span>
In addition, a concave mirror
will have a positive focal length (f) and a convex mirror will have a negative
focal length (f).
Now, we then have 1/f = 1/5f
+ 1/s’ which is s’ = 5f/4
Then we get the magnification
ratio that expresses the size or amount of magnification or reduction of the
object or image and to get the magnification, we use this equation: M= s’/s
M= 5f/4x5f
s’ = 1/4s
Therefore, the image height
is one fourth of the object height
Answer:
abiotic
Explanation:
i think but dont take my word for it
Answer:
D. two positively charged objects
Answer:
The height reached is 20m, The time taken to reach 20m is 2 seconds
Explanation:
Observing the equations of motion we can see that the following equation will be most helpful for this question.
![v^{2} = u^{2} + 2as](https://tex.z-dn.net/?f=v%5E%7B2%7D%20%3D%20u%5E%7B2%7D%20%2B%202as)
We are given initial velocity, u
We know that the stone will stop at its maximum height, so final velocity, v
Acceleration, a
And we are looking for the displacement (height reached), s
Substitute the values we are given into the equation
![0^{2} = 20^{2} + 2(10)s](https://tex.z-dn.net/?f=0%5E%7B2%7D%20%3D%2020%5E%7B2%7D%20%2B%202%2810%29s)
Rearrange for s
![0^{2} -20^{2} =20s](https://tex.z-dn.net/?f=0%5E%7B2%7D%20-20%5E%7B2%7D%20%3D20s)
![-400=20s](https://tex.z-dn.net/?f=-400%3D20s)
![\frac{-400}{20} =s](https://tex.z-dn.net/?f=%5Cfrac%7B-400%7D%7B20%7D%20%3Ds)
s = -20 (The negative is just showing direction, it can be ignored for now)
The height reached is 20m
Use a different equation to find the time taken
![s = vt - \frac{1}{2} at^{2}](https://tex.z-dn.net/?f=s%20%3D%20vt%20-%20%5Cfrac%7B1%7D%7B2%7D%20at%5E%7B2%7D)
Substitute in the values we have
![-20=(0)t - \frac{1}{2} (10)t^{2}](https://tex.z-dn.net/?f=-20%3D%280%29t%20-%20%5Cfrac%7B1%7D%7B2%7D%20%2810%29t%5E%7B2%7D)
Rearrange for t
![-20 =0 -5 t^{2}](https://tex.z-dn.net/?f=-20%20%3D0%20-5%20t%5E%7B2%7D)
![\frac{-20}{-5} =t^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B-20%7D%7B-5%7D%20%3Dt%5E%7B2%7D)
![4 = t^{2}](https://tex.z-dn.net/?f=4%20%3D%20t%5E%7B2%7D)
t = 2s
The time taken to reach 20m is 2 seconds