It's 12.1 m/s, assuming that's the launch velocity that's given.
For projectile motion, velocity's y-component is parabolic/quadratic. It's x-component is constant, so you don't need to know it.
Hydro then electric or solar power
I think its B or D, most likely D.
Explanation:
solution: mass m = 5g = 0.005kg; extension e = 7cm = 0.07m; force f = 70 N; velocity = ?; using: work done in elastic material w = 1/2 fe = 1/2 ke2 = 1/2 mv2 - the kinetic energy of the moving stone. 1/2 fe =...
Answer:
By a factor of 1/4.
Explanation:
The impulse force that applies to an object undergoing rapid deceleration just before coming to a stop on the ground is given by the following formula,
in which 
, 
represent the change in momentum and the time taken for that change.
If one increases the time that is taken for the momentum change (which remains constant for this situation) by a factor 4 and if that new force is represented by 
, the following manipulation confirms the answer to this question.
![\begin{aligned}\\\small F_1 &=\small \frac{\Delta (mV)}{4\Delta t}\\\\&=\small \frac{1}{4}\times\bigg[\frac{\Delta (mV)}{\Delta t}\bigg]\\\\&=\small \frac{1}{4}F\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5C%5C%5Csmall%20F_1%20%26%3D%5Csmall%20%5Cfrac%7B%5CDelta%20%28mV%29%7D%7B4%5CDelta%20t%7D%5C%5C%5C%5C%26%3D%5Csmall%20%5Cfrac%7B1%7D%7B4%7D%5Ctimes%5Cbigg%5B%5Cfrac%7B%5CDelta%20%28mV%29%7D%7B%5CDelta%20t%7D%5Cbigg%5D%5C%5C%5C%5C%26%3D%5Csmall%20%5Cfrac%7B1%7D%7B4%7DF%5Cend%7Baligned%7D)
Here 
is the force that was applied to the object previously.
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