I believe it would be orthoclase feldspar, quartz, micas, and amphiboles are the most abundant in granite.<span />
Explanation:
p=mv
p=5.6×75
p= 420
<em>hope</em><em> it</em><em> was</em><em> helpful</em><em> to</em><em> you</em>
The second one is best. Gravity is an attractive force that acts between all objects, whether they're in contact or not.
Answer with Explanation:
We are given that




a.We have to find the expression for the minimum thickness the film can have ,t.
Condition for destructive interference


For minimum thickness m=0
Then, 
b.Substitute the values

Answer:

b)

Explanation:
Let the amplitude of SHM is given as A
so the total energy of SHM is given as

now we know that
a)
kinetic energy is given as

here

so now we have


now its fraction with respect to total energy is given as

b)
Potential energy is given as

so we have

so fraction of energy is given as
