The component of the crate's weight that is parallel to the ramp is the only force that acts in the direction of the crate's displacement. This component has a magnitude of
<em>F</em> = <em>mg</em> sin(20.0°) = (15.0 kg) (9.81 m/s^2) sin(20.0°) ≈ 50.3 N
Then the work done by this force on the crate as it slides down the ramp is
<em>W</em> = <em>F d</em> = (50.3 N) (2.0 m) ≈ 101 J
The work-energy theorem says that the total work done on the crate is equal to the change in its kinetic energy. Since it starts at rest, its initial kinetic energy is 0, so
<em>W</em> = <em>K</em> = 1/2 <em>mv</em> ^2
Solve for <em>v</em> :
<em>v</em> = √(2<em>W</em>/<em>m</em>) = √(2 (101 J) / (2.0 m)) ≈ 10.0 m/s
Answer:
10 km at 45° south..........
Answer:
A chemical reaction
Explanation:
In a physical change the chemical composition of a matter stays the same. One example is a change in state of matter. Consider water, you can heat it to vapour and cool it to form ice but you can get the exact same mass and colour after any physical change happens.
To form a two or more substance an internal arrangement of atoms must happen. For that chemical change should occur as only in chemical reactions atoms are rearrange to form new substances.
Just ignore the horizontal component
if you have a vertical displacement of 15m, 0ms^1 initial velocity, end velocity is ignored, we know the acceleration due to gravity as 9.81ms^2 so we can work out the time using SUVAT
S=15
U=0
V=?
A=9.81
T=?
S=UT + 0.5 AT^2
UT=0
therefore,
S=0.5AT^2
rearrange to:
T=SQR (2S/A)
T = 1.75 seconds
