Which state or states of matter take(s) the shape of its container? A) gas B) solid C) liquid D) liquid and gas
1 answer:
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Answer:
v = 31.3 m / s
Explanation:
The law of the conservation of stable energy that if there are no frictional forces mechanical energy is conserved throughout the point.
Let's look for mechanical energy at two points, the highest where the body is at rest and the lowest where at the bottom of the plane
Highest point
Em₀ = U = m g y
Lowest point
= K = ½ m v²
As there is no friction, mechanical energy is conserved
Em₀ =
m g y = ½ m v²
v = √ 2 g y
Where we can use trigonometry to find and
sin 30 = y / L
y = L sin 30
Let's replace
v = RA (2 g L sin 30)
Let's calculate
v = RA (2 9.8 100.0 sin30)
v = 31.3 m / s
Answer:
Explanation:
Use the one-dimensional equation
Δx = where delta x is the displacement of the object, v0 is the velocity of the object, a is the pull of gravity, and t is the time in seconds. That's our unknown.
Δx = -2 (negative because where it ends up is lower than the point at which it started),
, and
a = -9.8
Filling in:
and simplified a bit:
this should look hauntingly familiar (a quadratic, which is parabolic motion...very important in physics!!). We begin by getting everything on one side of the equals sign and solving for t by factoring:
(the 0 is also indicative of the object landing on the ground! Isn't this a beautiful thing, how it all just works so perfectly together?)
When you factor this however your math/physics teacher has you factoring you will get that
t = 1.3 sec and t = -.31 sec
Since we all know that time can NEVER be negative, it takes the ball 1.3 sec to hit the ground from a height of 2 m if it is rolling off the shelf at 5 m/s.
Answer:
Explanation:
From the position coordinates given , it appears that the ball moves simultaneously along x and y direction.
Displacement along x direction in one second = 4.4 - 1.8 = 2.6 m
So velocity along x direction V_x =
Similarly velocity along y direction V_y(1) =
In the next phase velocity changes both in x and y direction.
velocity in x - direction V_x(2) = [tex]\frac{2}{s}[/tex
Velocity in Y- direction V_y(2) = [tex]\frac{3.1}{s}[/tex
Acceleration in x direction = change of velocity in x direction
= ( 2 - 2.6 ) = -.6 m s⁻²
Acceleration in y direction = ( 3.1 - 2.6 ) = 0.5 m s⁻²
Total acceleration =\sqrt{( -.6 )² + ( .5 )²}
= .78 ms⁻²