Answer:
The volume of the submerged part of her body is 
Explanation:
Let's define the buoyant force acting on a submerged object.
In a submerged object acts a buoyant force which can be calculated as :
ρ.V.g
Where ''B'' is the buoyant force
Where ''ρ'' is the density of the fluid
Where ''V'' is the submerged volume of the object
Where ''g'' is the acceleration due to gravity
Because the girl is floating we can state that the weight of the girl is equal to the buoyant force.
We can write :
(I)
Where ''W'' is weight
⇒ If we consider ρ =
(water density) and
and replacing this values in the equation (I) ⇒


ρ.V.g = 610N
(II)
The force unit ''N'' (Newton) is defined as

Using this in the equation (II) :



We find that the volume of the submerged part of her body is 
Answer:
Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.
(a) Determine the position of wire 3.
b) Determine the magnitude and direction of current in wire 3
Explanation:
a) 

position of wire = 50 - 1.2
= 48.8cm
b) 

Direction ⇒ downward
Answer:
1.2 × 10^27 neutrons
Explanation:
If one neutron = 1.67 × 10^-27 kg
then in 2kg...the number of neutrons
; 2 ÷ 1.67 × 10^-27
There are.... 1.2 × 10^27 neutrons
I attached a free body diagram for a better understanding of this problem.
We start making summation of Moments in A,



Then we make a summation of Forces in Y,



At the end we calculate the angle with the sin.

