Answer:
<em>N</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>SI</em><em> </em><em>unit</em><em> </em><em>of</em><em> </em><em>Newton</em>
Equations of the vertical launch:
Vf = Vo - gt
y = yo + Vo*t - gt^2 / 2
Here yo = 35.0m
Vo is unknown
y final = 0
t = 4.00 s
and I will approximate g to 10m/s^2
=> 0 = 35.0 + Vo * 4 - 5 * (4.00)^2 => Vo = [-35 + 5*16] / 4 = - 45 / 4 = -11.25 m/s
The negative sign is due to the fact that the initial velocity is upwards and we assumed that the direction downwards was positive when used g = 10m/s^2.
Answer: 11.25 m/s
Answer:The correct options are:
1. A system is a group of objects analyzed as one unit.
2. Energy that moves across system boundaries is conserved.
Explanation:
A system is defined as group of interrelated or interacting items existing as a single unit or a whole to achieve a specific objective.Energy lost by the system is equal to the energy gained by the surroundings.
Two statements are true about a system:
- A system is a group of objects analyzed as one unit.
- Energy that moves across system boundaries is conserved.
Answer:
The velocity after 2 seconds can be found through:
V = u +a*t
Where V is final velocity, u is initial velocity, a is acceleration and t is time.
V = 0 + 2* 2= 4 meters/second
The distance (s) can be found through:
V^2= u^2 +2*a* s
Where V is final velocity, u is initial velocity, a is acceleration.
4^2= 0^2 + 2 *2*s
16= 0 + 4s
s= 4 meters
Distance (s) can also be found through:
s= ut + 1/2 at^2
s= 0+ 1/2 *2*2^2= 1 *2*2
s= 4 meters
Explanation:
