It would be a neutral,
The acidity of water is 7 as well
Yes, <span> the moon fall partly into earth's shadow when it is in its full size</span>
<span>Here I think you have to find the velocity in x and y components where x is east and y is north
So as air speed indicator shows the negative speed in y component and adding it in
air speed while multiplying with the direction component we will get the velocity as velocity is a vector quantity so direction is also required
v=-28 m/s y + 18 m/s (- x/sqrt(2) - y/sqrt(2))
solving
v= -12.7 m/s x-40.7 m/s y
if magnitude of velocity or speed is required then
speed= sqrt(12.7^2 + 40.7^2)
speed= 42.63 m/s
if angle is asked
angle = arctan (40.7/12.7)
angle = 72.67 degrees south of west</span>
That depends on a few things that you haven't told us about the setup.
So I'm going to assume one of them, and then give you the answer
in terms of another one:
-- Assume a Class-I lever . . . the fulcrum is between the load and the effort.
-- Then the effort needed to lift the load is
(the weight of the load) x (13 / the distance between the fulcrum and the effort)
Answer:
1.5 m
Explanation:
a = v/t
a = 2/1,5 = 4/3
x = 1/2 a t² + vt + x
x = 1/2 × 4/3 × 9/4 = 3/2 =1.5 m
