Question
Which is the equation for terminal velocity?(1 point)
Responses
Vt=2mdACd−−−−√
𝑉
𝑡
=
2
𝑚
𝑑
𝐴
𝐶
𝑑
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=2wdACd−−−−√
𝑉
𝑡
=
2
𝑤
𝑑
𝐴
𝐶
𝑑
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=w2dACd−−−−−√
𝑉
𝑡
=
𝑤
2
𝑑
𝐴
𝐶
𝑑
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=dACd2w−−−−√
Responses
Vt=2mdACd−−−−√
𝑉
𝑡
=
2
𝑚
𝑑
𝐴
𝐶
𝑑
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=2wdACd−−−−√
𝑉
𝑡
=
2
𝑤
𝑑
𝐴
𝐶
𝑑
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=w2dACd−−−−−√
𝑉
𝑡
=
𝑤
2
𝑑
𝐴
𝐶
𝑑
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=dACd2w−−−−√
Answers
Answered by
GPT-4o mini
The equation for terminal velocity (Vt) for an object falling under the influence of gravity in a fluid can be expressed as:
\[
V_t = \sqrt{\frac{2mg}{\rho A C_d}}
\]
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,
- \(C_d\) is the drag coefficient.
If we were to substitute \(w = mg\) (the weight of the object) into the formula, it would simplify to:
\[
V_t = \sqrt{\frac{2w}{\rho A C_d}}
\]
However, from your listed options, none of them directly match this common representation.
If we adjust it according to standard expressions involving density, and notice differences in how the variables might be represented, one of the alternatives might be aligned. Among your options, the closest representation in the context of terminal velocity, which could suggest a similar relationship with weight (w) given as a different form of representing forces would be:
\[
V_t = \sqrt{\frac{2w}{d A C_d}}
\]
If we assume \(d\) refers to density, it seems this option could be interpreted in the context given that there are potential substitutions involved. Please double-check how the variables were defined in your context.
So, based on the given choices, the best fit would likely be:
\[
V_t = \sqrt{\frac{2w}{d A C_d}}
\]
where both 'w' for weight and 'd' being a representative value (possibly density).
\[
V_t = \sqrt{\frac{2mg}{\rho A C_d}}
\]
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,
- \(C_d\) is the drag coefficient.
If we were to substitute \(w = mg\) (the weight of the object) into the formula, it would simplify to:
\[
V_t = \sqrt{\frac{2w}{\rho A C_d}}
\]
However, from your listed options, none of them directly match this common representation.
If we adjust it according to standard expressions involving density, and notice differences in how the variables might be represented, one of the alternatives might be aligned. Among your options, the closest representation in the context of terminal velocity, which could suggest a similar relationship with weight (w) given as a different form of representing forces would be:
\[
V_t = \sqrt{\frac{2w}{d A C_d}}
\]
If we assume \(d\) refers to density, it seems this option could be interpreted in the context given that there are potential substitutions involved. Please double-check how the variables were defined in your context.
So, based on the given choices, the best fit would likely be:
\[
V_t = \sqrt{\frac{2w}{d A C_d}}
\]
where both 'w' for weight and 'd' being a representative value (possibly density).