"Balanced" means that if there's something pulling one way, then there's also
something else pulling the other way.
-- If there's a kid sitting on one end of a see-saw, and another one with the
same weight sitting on the other end, then the see-saw is balanced, and
neither end goes up or down. It's just as if there's nobody sitting on it.
-- If there's a tug-of-war going on, and there are 300 freshmen pulling on one
end of a rope, and another 300 freshmen pulling in the opposite direction on
the other end of the rope, then the hanky hanging from the middle of the rope
doesn't move. The pulls on the rope are balanced, and it's just as if nobody
is pulling on it at all.
-- If a lady in the supermarket is pushing her shopping cart up the aisle, and her
two little kids are in front of the cart pushing it in the other direction, backwards,
toward her. If the kids are strong enough, then the forces on the cart can be
balanced. Then the cart doesn't move at all, and it's just as if nobody is pushing
on it at all.
From these examples, you can see a few things:
-- There's no such thing as "a balanced force" or "an unbalanced force".
It's a <em><u>group</u> of forces</em> that is either balanced or unbalanced.
-- The group of forces is balanced if their strengths and directions are
just right so that each force is canceled out by one or more of the others.
-- When the group of forces on an object is balanced, then the effect on the
object is just as if there were no force on it at all.
Solid particles hope this helped!
Answer:
Explanation:
1 )
Here
wave length used that is λ = 580 nm
=580 x 10⁻⁹
distance between slit d = .46 mm
= .46 x 10⁻³
Angular position of first order interference maxima
= λ / d radian
= 580 x 10⁻⁹ / .46 x 10⁻³
= 0.126 x 10⁻² radian
2 )
Angular position of second order interference maxima
2 x 0.126 x 10⁻² radian
= 0.252 x 10⁻² radian
3 )
For intensity distribution the formula is
I = I₀ cos²δ/2 ( δ is phase difference of two lights.
For angular position of θ1
δ = .126 x 10⁻² radian
I = I₀ cos².126x 10⁻²/2
= I₀ X .998
For angular position of θ2
I = I₀ cos².126x2x 10⁻²/2
= I₀ cos².126x 10⁻²
When we represent what is given to us on a coordinate plane, we have a figure as shown in the attachment.
To find the distance between P and R, we have to find the Net Displacement of the ship (brown arrow in the figure).
For that, we use the rules for Vector addition.
We see that the first displacement 
= 30 km (blue arrow) is along the y-axis, but the second part of the ship's journey 
= 20 km (red arrow) is at an angle with reference to y-axis.
So, we first find the components of the red arrow along X and Y.
Component of along X-axis is given by 
= 10 km
Component of along Y-axis is given by 
= 17.32 km
We now add all the vectors along X and along Y separately.
Net Displacement along X 
Net Displacement along Y 
= 47.32 km
Now that we have the components of the net displacement along X and Y, we make use of Pythagorean Theorem to calculate the 
Therefore, [tex]D_{net} = 48.37 km.
Hence, the distance between the ports P and R is 48 km.
