Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
Given,
The initial speed of the car, u=61 km/h
The final speed of the car, v=113 km/h
The drag force on an object is given by,
[tex]F=\frac{1}{2}\rho v^2C_DA[/tex]Where:
• ρ is the density of the air.
,• v is the speed of the object.
,• C_D is the drag coefficient.
,• A is the area of cross-section of the object.
In the given problem, only the speed of the car changes. And density of the air, drag coefficient, and area of the cross section remains the same.
Thus, the initial drag force on the car is
[tex]F_i=\frac{1}{2}\rho u^2C_DA[/tex]The final drag force is given by,
[tex]F_f=\frac{1}{2}\rho v^2C_DA[/tex]The percentage increase in the drag force is given by,
[tex]P=\frac{F_f-F_i}{F_i}_{}\times100[/tex]On substituting,
[tex]\begin{gathered} P=\frac{\frac{1}{2}\rho v^2C_DA-\frac{1}{2}\rho u^2C_DA}{\frac{1}{2}\rho u^2C_DA}\times100 \\ =\frac{\frac{1}{2}\rho C_DA(v^2-u^2)}{\frac{1}{2}\rho u^2C_DA}\times100 \\ =\frac{v^2-u^2}{u^2}\times100 \end{gathered}[/tex]On substituting the values of v and u,
[tex]\begin{gathered} P=\frac{113^2-61^2}{61^2}\times100 \\ =243\% \end{gathered}[/tex]Thus, the percentage increase in the drag force is 243%
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We hope this was helpful. Please come back whenever you need more information or answers to your queries. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
