Answer:
The magnification of the image is ![M = -1.5](https://tex.z-dn.net/?f=M%20%20%3D%20-1.5)
Explanation:
From the question we are told that
The object distance is ![u = 14 cm](https://tex.z-dn.net/?f=u%20%3D%2014%20cm)
The image distance is ![v = 21 \ cm](https://tex.z-dn.net/?f=v%20%3D%2021%20%5C%20cm)
The magnification of the image is mathematically represented as
![M = - \frac{v}{u}](https://tex.z-dn.net/?f=M%20%20%3D%20-%20%20%5Cfrac%7Bv%7D%7Bu%7D)
substituting values
![M = - \frac{21}{14}](https://tex.z-dn.net/?f=M%20%20%3D%20-%20%20%5Cfrac%7B21%7D%7B14%7D)
![M = -1.5](https://tex.z-dn.net/?f=M%20%20%3D%20-1.5)
The negative value means that the image is real , inverted and enlarged
Answer:
true
Hoped this helped(⁄ ⁄>⁄ ▽ ⁄<⁄ ⁄)
I’m pretty sure the answer would be :
9.8m/s2
Explanation: gravity =9.8m/s2
Answer:
Explanation:
Let the internal resistance be r .
Since in open circuit the volt is 1.55 V , this will be the source voltage .
Source voltage = 1.55
If external resistance be R .
1.55 / (R + r ) = .500
R + r = 3.1 ohm
So sum of internal resistance and external resistance will be 3.1 ohm.