The average force exerted by the block exists at -375 N.
<h3>
How to find the average force exerted by the block?</h3>
Given:
Initial Velocity, u = 50m/s
Displacement, s = 10cm = 0.10m
Mass, m = 0.03Kg
Using the relation,
![$v^{2}-u^{2}=2 a s](https://tex.z-dn.net/?f=%24v%5E%7B2%7D-u%5E%7B2%7D%3D2%20a%20s)
Substitute the values in the above equation, we get:
![$&0^{2}-50^{2}=2 \mathrm{a}(0.10) \\](https://tex.z-dn.net/?f=%24%260%5E%7B2%7D-50%5E%7B2%7D%3D2%20%5Cmathrm%7Ba%7D%280.10%29%20%5C%5C)
![$ \mathrm{a}= -12500 \mathrm{~ms}^{-2}](https://tex.z-dn.net/?f=%24%20%5Cmathrm%7Ba%7D%3D%20-12500%20%5Cmathrm%7B~ms%7D%5E%7B-2%7D)
We know that, f = ma
Putting the values, we get:
f = (0.03)(-12500)
f = -375 N
The average force exerted by the block exists at -375 N.
To learn more about average force refer to:
brainly.com/question/18652903
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Answer:
y = 3 - 4x
To find the y intercept let x = 0
y = 3 - 4(0)
y = 3 or ( 0 ,3)
Hope this helps.
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