1 Answer

Ask a Question

Kemi · Answer 1 · 2023-05-14T10:00:00+0000

0.495 m/s

Step-by-step explanation

the formula for the terminal velocity is given by:

$\begin{gathered} v=\sqrt[]{(2mg)/(\sigma AC)} \\ \text{where} \\ \end{gathered}$

m is the mass

g is 9.81 m/s²

ρ is density

A is area

C is the drag coefficient

then

Step 1

Let's find the mass

$\begin{gathered} \sigma=(m)/(v) \\ m=\sigma\cdot v \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(0.01m)^3 \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(1\cdot10^(-6)) \\ \text{mass}=2\cdot10^(-3)\operatorname{kg} \\ \text{mass}=0.002\text{ kg } \\ \text{Area}=(0.01\text{ m}\cdot0.01m)=0.0001m^2 \end{gathered}$

now, replace

$\begin{gathered} v=\sqrt[]{(2mg)/(\sigma AC)} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }(m)/(s^2))}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452(m^2)/(s^2)} \\ v=0.495\text{ m/s} \end{gathered}$

hence, the answer is 0.495 m/s

A cube has a drag coefficient of 0.8. What would be the terminal velocity of a sugar-example-1

0 Comments

Your comment on this question:

Your answer

1 Answer

0 Comments

Your comment on this answer:

Other Questions