Connect with experts and get insightful answers on IDNLearn.com. Ask your questions and receive accurate, in-depth answers from our knowledgeable community members.
Sagot :
Answer:
[tex]F_N>F_N'[/tex]
Explanation:
From the question we are told that
Radius of curvature[tex]r=100m[/tex]
Mass [tex]M=300kg[/tex]
initial Speed of Motorcycle [tex]V_1=30m/s[/tex]
Final Speed of Motorcycle [tex]V_2=33m/s[/tex]
Generally the equation Force at initial speed is mathematically given as
[tex]F_N=mg-\frac{mv^2}{R}[/tex]
[tex]F_N=300*9.8-\frac{(300*30)^2}{100}[/tex]
[tex]F_N=240N[/tex]
Generally the equation Force at Final speed is mathematically given as
[tex]F_N'=mg-\frac{mv'^2}{R}[/tex]
[tex]F_N'=300*9.8-\frac{(300*33)^2}{100}[/tex]
[tex]F_N'=-327N[/tex]
Therefore
[tex]F_N>F_N'[/tex]
Following are the calculation to the given question:
Solution:
Using formula:
[tex]\to mg - F^{'}_N=\frac{mv^{2}}{R} \\[/tex]
Calculating the Initial value:
[tex]\to F_N = mg - \frac{mv^2}{R}\\\\[/tex]
[tex]= 300 \times (9.8 - \frac{(30)^2}{100}) \\\\= 300 \times (9.8 - \frac{900}{100}) \\\\= 300 \times (9.8 - 9) \\\\= 300 \times (0.8) \\\\ = 240\ N \\\\[/tex]
Calculating the Final value:
[tex]\to F^{'}_{N}= 300 (9.8 -\frac{33^2}{100})\\\\[/tex]
[tex]= 300 (9.8 -\frac{1089}{100})\\\\= 300 (9.8 - 10.89)\\\\= 300 (- 1.09)\\\\=-327[/tex]
Therefore, the answer is "the new normal force is less than [tex]F_{N}[/tex]" or [tex]\bold{F^{'}_{N}< F_{N}}[/tex].
Learn more:
brainly.com/question/12910909
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your search for answers ends at IDNLearn.com. Thanks for visiting, and we look forward to helping you again soon.
