PART A)
By Snell's law we know that

here we know that



now from above equation we have


so it will refract by angle 39.3 degree
PART B)
Here as we can see that image formed on the other side of lens
So it is a real and inverted image
Also we can see that size of image is lesser than the size of object here
Here we can use concave mirror to form same type of real and inverted image
PART C)
As per the mirror formula we know that



so image will form at 30 cm from mirror
it is virtual image and smaller in size
Answer:

Explanation:
As we know that magnitude of two vectors is given as

here we know that
A = magnitude of vector A
B = magnitude of vector B
= angle between two vectors
so here we know that
A = 30 units
B = 40 units
angle = 90 degree
so we have



Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.
Given the data in the question;
Hubble's constant; 
Age of the universe; 
We know that, the reciprocal of the Hubble's constant (
) gives an estimate of the age of the universe (
). It is expressed as:

Now,
Hubble's constant; 
We know that;

so
![1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m](https://tex.z-dn.net/?f=1%5C%20Million%5C%20light%5C%20years%20%3D%20%5B9.46%20%2A%2010%5E%7B15%7Dm%5D%20%2A%2010%5E6%20%3D%209.46%20%2A%2010%5E%7B21%7Dm)
Therefore;

Now, we input this Hubble's constant value into our equation;

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.
Learn more: brainly.com/question/14019680
The electric force is small in comparison because charges are VERY small but the
gravitational force has to do with the mass of the Earth and Moon.