Answer:
forces that are equal in size and opposite in direction. Balanced forces do not result in any change in motion. unbalanced. forces: forces applied to an object in opposite directions that are not equal in size. Unbalanced forces result in a change in motion.
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The answer is A. Plant growth, Plant growth is the only thing that can actually be tested and scientificly measured, hope this helps!!!!
I think u can say thats a constant velocity, but remember if ur turning, or going around a curve, that is also changing velocity. Hope this helps have a great day!
Differentiate the components of position to get the corresponding components of velocity :
![v_x = \dfrac{\mathrm dx}{\mathrm dt} = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) t^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)t](https://tex.z-dn.net/?f=v_x%20%3D%20%5Cdfrac%7B%5Cmathrm%20dx%7D%7B%5Cmathrm%20dt%7D%20%3D%20%5Cleft%281.5%5Cdfrac%7B%5Crm%20m%7D%7B%5Cmathrm%20s%5E3%7D%5Cright%29%20t%5E2%20-%20%5Cleft%284%5Cdfrac%7B%5Crm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t)
![v_y = \dfrac{\mathrm dy}{\mathrm dt} = \left(1\dfrac{\rm m}{\mathrm s^2}\right)t-2\dfrac{\rm m}{\rm s}](https://tex.z-dn.net/?f=v_y%20%3D%20%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dt%7D%20%3D%20%5Cleft%281%5Cdfrac%7B%5Crm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t-2%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D)
At <em>t</em> = 5.0 s, the particle has velocity
![v_x = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) (5.0\,\mathrm s)^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s) = 17.5\dfrac{\rm m}{\rm s}](https://tex.z-dn.net/?f=v_x%20%3D%20%5Cleft%281.5%5Cdfrac%7B%5Crm%20m%7D%7B%5Cmathrm%20s%5E3%7D%5Cright%29%20%285.0%5C%2C%5Cmathrm%20s%29%5E2%20-%20%5Cleft%284%5Cdfrac%7B%5Crm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29%285.0%5C%2C%5Cmathrm%20s%29%20%3D%2017.5%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D)
![v_y = \left(1\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)-2\dfrac{\rm m}{\rm s} = 3.0\dfrac{\rm m}{\rm s}](https://tex.z-dn.net/?f=v_y%20%3D%20%5Cleft%281%5Cdfrac%7B%5Crm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29%285.0%5C%2C%5Cmathrm%20s%29-2%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%20%3D%203.0%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D)
The speed at this time is the magnitude of the velocity :
![\sqrt{{v_x}^2 + {v_y}^2} \approx \boxed{17.8\dfrac{\rm m}{\rm s}}](https://tex.z-dn.net/?f=%5Csqrt%7B%7Bv_x%7D%5E2%20%2B%20%7Bv_y%7D%5E2%7D%20%5Capprox%20%5Cboxed%7B17.8%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%7D)
The direction of motion at this time is the angle
that the velocity vector makes with the positive <em>x</em>-axis, such that
![\tan(\theta) = \dfrac{3.0\frac{\rm m}{\rm s}}{17.5\frac{\rm m}{\rm s}} \implies \theta = \tan^{-1}\left(\dfrac{3.0}{17.5}\right) \approx \boxed{9.73^\circ}](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%29%20%3D%20%5Cdfrac%7B3.0%5Cfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%7D%7B17.5%5Cfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%7D%20%5Cimplies%20%5Ctheta%20%3D%20%5Ctan%5E%7B-1%7D%5Cleft%28%5Cdfrac%7B3.0%7D%7B17.5%7D%5Cright%29%20%5Capprox%20%5Cboxed%7B9.73%5E%5Ccirc%7D)