Yes a kg of hydrogen will have more atoms than a kg of lead, because lead has a higher atomic mass, than hydrogen so it will take more atoms of hydrogen to make a kg than lead
<span>Standard deviation is a calculation. It I used in statistical analysis of a easier job. hoped this helps u </span>
True : <span>There are numerous third-class </span>levers<span> in the human </span>body<span>; one example can be illustrated in the elbow joint</span>
Answer:
it is 2.2 m
Explanation:
because he goes back 2.2 m so 4.4 minus 2.2 equals 2.2
Answer:
a
The x- and y-components of the total force exerted is

b
The magnitude of the force is

The direction of the force is
Clockwise from x-axis
Explanation:
From the question we are told that
The magnitude of the first charge is 
The magnitude of the second charge is 
The position of the second charge from the first one is 
The magnitude of the third charge is 
The position of the third charge from the first one is 


The position of the third charge from the second one is



The force acting on the third charge due to the first and second charge is mathematically represented as

Substituting values



The magnitude of
is mathematically evaluated as

The direction is obtained as

![\theta = tan ^{-1} [-0.63889]](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%20%5E%7B-1%7D%20%5B-0.63889%5D)


