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Anna [14]
3 years ago
5

NEED HELP ASAP!!!! problem is shown in the attached image

Physics
1 answer:
Lena [83]3 years ago
5 0

Answer:

im sorry but what is the question please tell me in the comments s i can help you

Explanation:

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ΔF=125.22 %

Explanation:

We know that drag force on the car given as

F_D=\dfrac{1}{2}\rho C_DA v^2

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All other quantity are constant so we can say that drag force and velocity can be given as

\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}

Now by putting the values

\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}

\dfrac{F_D_1}{F_D_2}=\dfrac{50^2}{75^2}

\dfrac{F_D_1}{F_D_2}=0.444

Percentage Change in the drag force

\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100

\Delta F=\dfrac{F_D_2-0.444F_D_2}{0.444F_D_2}\times 100

\Delta F=\dfrac{1-0.444}{0.444}\times 100

ΔF=125.22 %

Therefore force will increase by 125.22  %.

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