Disregarding friction, how fast will the block move after the spring is released all the way and the block slides away from it?
a 2.5 kg block is launched along the ground by a spring with a spring constant of 56 n/m. the spring is initially compressed 0.75 m.
a.3.5 m/s
b.4.1 m/s
c.13 m/s
d.16 m/s

According to work energy theorem, work done by all the forces is equal to change in kinetic energy.

As in this case only spring exerts force on the block, work done by spring force is considered only.

So, work done by spring = change in kinetic energy

work done by spring =  kx^2
                                       
                                          2

Therefore according to work energy theorem, 

 kx^2              mvi^                  mvf^2
  =            -            ( vi = initial velocity, vf = final 
  2                         2                          2                                              velocity)

Since block was initially at rest, initial kinetic energy is zero

Therefore,

 kx^2              mvf^2
  =      
  2                     2                                           

Therefore, 

Vf ^2 = kx^2/m = 56(0.75)^2 / 2.5 = 12.6

Therefore, v= 3.5 m/s.
Hence option A is correct.
Hope this helps!!

            

The answer is b.the gaseous state is electrically positive, while the plasma state is electrically negative. 

Well the magnitude of its acceleration related to that car starts with the oil change like when you have to change the oil in your car you have to make sure that your precious car is starting really propley and very well