Answer:
The Sun produces energy by the process of nuclear fusion. Nuclear fusion occurs when lighter nuclei combine to produce a larger, heavier nucleus. In the process, energy is released. Nuclear fusion requires very high temperatures and pressures. Nuclear fusion occurs in the core of the Sun when hydrogen atoms combine to form helium atoms.
Explanation:
I just took the assignment
A. Oysters, clams, and other shellfish are efficient filter feeders that help remove excess nitrogen from waters by incorporating it into their shells and tissue as they grow. ... Adult oysters reportedly can filter up to 50 gallons a day, while large quahogs (hard clams) can clean about 24 gallons of water a day
Answer:
Electric field at a distance of 1.45 cm will be 
Explanation:
We have given the distance d = 1.45 cm = 0.0145 m
And the potential difference 
There is a relation between potential difference and electric field
Electric field at a distance d due to a potential difference is given by
, here E is electric field, V is potential difference and d is distance
So 
Answer:

Explanation:
Given that,
The current in the loop, I = 2 A
The radius of the loop, r = 0.4 m
We need to find the magnetic field at a distance 0.09 m along the axis and above the center of the loop. The formula for the magnetic field at some distance is given as follows :

Put all the values,

So, the required magnetic field is equal to
.
Answer:

is the no. of electrons
Explanation:
Given:
- quantity of charge transferred,

<u>No. of electrons in the given amount of charge:</u>
As we have charge on one electron 
so,


is the no. of electrons
- Now if each water molecules donates one electron:
Then we require
molecules.
<u>Now the no. of moles in this many molecules:</u>

where
Avogadro No.


- We have molecular mass of water as M=18 g/mol.
<u>So, the mass of water in the obtained moles:</u>

where:
m = mass in gram

