"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.
Answer:
3 seconds
Explanation:
Applying,
Applying,
v = u±gt................ Equation 1
Where v = final velocity, u = initial velocity, t = time, g = acceleration due to gravity.
From the question,
Given: v = 0 m/s ( at the maximum height), u = 30 m/s
Constant: g = -10 m/s
Substitute these values into equation 1
0 = 30-10t
10t = 30
t = 30/10
t = 3 seconds
Answer:
Part a)
Part b)
North of East
Explanation:
Speed of train towards East = 60 km/h
displacement towards East is given as
now it turns towards 50 degree East of North
so its distance is given as
then finally it moves towards west for 50 min
Now the total displacement of the train is given as
now total time duration of the motion is given as
now average velocity is given as
Part a)
magnitude of the average velocity is given as
Part b)
Direction of the velocity is given as
North of East
Answer:
5.1*10^3 J/m^3
Explanation:
Using E = q/A*eo
And
q =75*10^-6 C
A = 0.25
eo = 8.85*10^-12
Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]
= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]
= 5.1*10^3 J/m^3
