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repository:terminal_velocity [2020/09/01 16:59] porcaro1 [Answer Key] |
repository:terminal_velocity [2021/02/18 19:16] porcaro1 [Activity] |
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- Reflect on your predicted graphs. Explain how they were similar/different than your coding graphs. Revise your graphs as needed. | - Reflect on your predicted graphs. Explain how they were similar/different than your coding graphs. Revise your graphs as needed. | ||
===Code=== | ===Code=== | ||
- | [[https://www.glowscript.org/#/user/david.pastula/folder/MyPrograms/program/TerminalVelocity-MinimalProgram/edit | Link]] | + | [[hthttps://www.glowscript.org/#/user/porcaro1/folder/RepositoryPrograms/program/TerminalVelocity-Incomplete | Link]] |
<code Python [enable_line_numbers="true"]> | <code Python [enable_line_numbers="true"]> | ||
GlowScript 2.8 VPython | GlowScript 2.8 VPython | ||
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===Handout=== | ===Handout=== | ||
==Pre-coding Answers== | ==Pre-coding Answers== | ||
- | - INSERT PICTURE | + | - {{:repository:terminal_velocity_velocity.png?nolink&600|}} |
- | - INSERT PICTURE | + | - {{:repository:terminal_velocity_forces.png?nolink&600|}} |
- | - The equation for terminal velocity is $V=\sqrt{\dfrac{2W}{C\rho A}}$, where | + | - The equation for terminal velocity is $v=\sqrt{\dfrac{2W}{C\rho A}}$, where |
- | - $W$ is the weight of the coin (N) | + | * $W$ is the weight of the coin (N) |
- | - this is calculated by multiplying the mass of the coin by the acceleration of gravity: $W= 0.01135 \text{kg} * 9.81 \text{m/s/s} = 0.111 \text{N}$ | + | * this is calculated by multiplying the mass of the coin by the acceleration of gravity: $W= 0.01135 \text{kg} * 9.81 \text{m/s/s} = 0.111 \text{N}$ |
- | - $C$ is the drag coefficient (1.15, no units) | + | * $C$ is the drag coefficient (1.15, no units) |
- | - $\rho$ is the density of air (1.2754 kg/m³) | + | * $\rho$ is the density of air (1.2754 kg/m³) |
- | - $A$ is the area of the coin | + | * $A$ is the area of the coin |
- | - this is calculated by squaring the radius of the coin and multiplying it by pi: $A= (0.015305 \text{m})^2 * \pi = 0.000736 \text{m²}$ | + | * this is calculated by squaring the radius of the coin and multiplying it by pi: $A= (0.015305 \text{m})^2 * \pi = 0.000736 \text{m²}$ |
- | - $v=\sqrt{\dfrac{2*0.111}{1.15*1.275*0.000736}}=14.3 \text{m/s}$ | + | * $v=\sqrt{\dfrac{2*0.111}{1.15*1.275*0.000736}}=14.3 \text{m/s}$ |
==Post-coding Answers== | ==Post-coding Answers== | ||
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====See Also==== | ====See Also==== | ||
- | * | + | *[[beating_the_train | Beating the Train]] |
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