Question
Which is the equation for terminal velocity?(1 point)
Responses
Vt=w2dACd−−−−−√
upper V subscript t baseline equals Start Root Start Fraction w over 2 d upper A Upper C subscript d baseline End Fraction End Root
Vt=2wdACd−−−−√
upper V subscript t baseline equals Start Root Start Fraction 2 w over d upper A Upper C subscript d baseline End Fraction End Root
Vt=2mdACd−−−−√
upper V subscript t baseline equals Start Root Start Fraction 2 m over d upper A Upper C subscript d baseline End Fraction End Root
Vt=dACd2w−−−−√
upper V subscript t baseline equals Start Root Start Fraction d upper A Upper C subscript d baseline over 2 w End Fraction End Root
Responses
Vt=w2dACd−−−−−√
upper V subscript t baseline equals Start Root Start Fraction w over 2 d upper A Upper C subscript d baseline End Fraction End Root
Vt=2wdACd−−−−√
upper V subscript t baseline equals Start Root Start Fraction 2 w over d upper A Upper C subscript d baseline End Fraction End Root
Vt=2mdACd−−−−√
upper V subscript t baseline equals Start Root Start Fraction 2 m over d upper A Upper C subscript d baseline End Fraction End Root
Vt=dACd2w−−−−√
upper V subscript t baseline equals Start Root Start Fraction d upper A Upper C subscript d baseline over 2 w End Fraction End Root
Answers
Answered by
GPT-4o mini
The correct equation for terminal velocity (\(V_t\)) is:
\[
V_t = \sqrt{\frac{2mg}{\rho A C_d}}
\]
Where:
- \(m\) is the mass of the object,
- \(g\) is the acceleration due to gravity,
- \(\rho\) is the density of the fluid,
- \(A\) is the cross-sectional area of the object, and
- \(C_d\) is the drag coefficient.
However, from your options, none of them exactly match this equation, but they seem to be rearrangements or forms connected to the terminal velocity equation. The closest to a correct formula using given variables (considering \(w\) as weight which is \(mg\)) would be simplified as:
\[
V_t = \sqrt{\frac{w}{\frac{1}{2} \rho A C_d}} \quad \text{(if you consider weight instead of mass)}
\]
Among the options you provided, the form:
\[
V_t = \sqrt{\frac{2w}{\rho A C_d}}
\]
If we were to adjust the notation accordingly, and assuming \(d\) correlates to density, area \(A\), and drag coefficient \(C_d\), the first option could represent this.
If you'd prefer a specific alignment to your selections, then please specify or clarify further!
\[
V_t = \sqrt{\frac{2mg}{\rho A C_d}}
\]
Where:
- \(m\) is the mass of the object,
- \(g\) is the acceleration due to gravity,
- \(\rho\) is the density of the fluid,
- \(A\) is the cross-sectional area of the object, and
- \(C_d\) is the drag coefficient.
However, from your options, none of them exactly match this equation, but they seem to be rearrangements or forms connected to the terminal velocity equation. The closest to a correct formula using given variables (considering \(w\) as weight which is \(mg\)) would be simplified as:
\[
V_t = \sqrt{\frac{w}{\frac{1}{2} \rho A C_d}} \quad \text{(if you consider weight instead of mass)}
\]
Among the options you provided, the form:
\[
V_t = \sqrt{\frac{2w}{\rho A C_d}}
\]
If we were to adjust the notation accordingly, and assuming \(d\) correlates to density, area \(A\), and drag coefficient \(C_d\), the first option could represent this.
If you'd prefer a specific alignment to your selections, then please specify or clarify further!