Respuesta :
Answer:
[tex]\% Change = [\frac{1}{2.8} -\frac{1}{1}]*100 = -64.29\%[/tex]
Explanation:
Assuming that m and v are unchanged.
For this case we have the following formula for the force:
[tex] F = \frac{mv^2}{2d}[/tex]
For this case the new force would be given:
[tex] F_{new}= \frac{mv^2}{2*(1.4 d)}[/tex]
[tex] F_{new}= \frac{mv^2}{2.8 d}[/tex]
And for this case we can calculate the % like this:
[tex] \Change = \frac{\frac{mv^2}{2.8d} -\frac{mv^2}{2d}}{\frac{mv^2}{2d}} *100[/tex]
And doing the algebra we got:
[tex]\% Change = \frac{\frac{mv^2}{2d}}{\frac{mv^2}{2d}} [\frac{1}{2.8} -\frac{1}{1}]*100[/tex]
[tex]\% Change = [\frac{1}{2.8} -\frac{1}{1}]*100 = -64.29\%[/tex]
So then the force decrease 64.29 percent respect the original force.
The change in the average force needed is -64.29%
Percentage difference:
Originally, the force required to stop the puck is given as;
[tex]F=\frac{mv^2}{2d}[/tex]
where d is the stopping distance
Now, if the stopping distance is increased by 40% then the new distance becomes:
[tex]d'=d+\frac{40}{100}d\\\\d'=\frac{7}{5}d[/tex]
So, the force required to stop the puck at a distance d' is given by:
[tex]F_{new}=\frac{mv^2}{2d'}=\frac{5mv^2}{14d}[/tex]
The percentage difference is given by:
[tex]\frac{F_{new}-F}{F}\times100\% = \frac{\frac{5mv^2}{14d}-\frac{mv^2} {2d}}{\frac{mv^2} {2d}}[/tex]
(Fnew - F) / F = -64.29%
Learn more about percentage:
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