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telo118 [61]
4 years ago
15

Is gravity air resistance?

Physics
1 answer:
kolezko [41]4 years ago
7 0

The engineering principles that this project is based upon are gravity and air resistance. Gravity and air resistance are fundamental concepts of many forms of engineering. After we demonstrate the physical principles of gravity and air resistance through various experiments, the children are to pair off into groups of 2 to apply these concepts to an activity. The demonstrations will include dropping a parachute, a small and large feather, and a maple seed and explaining how air resistance affects their movement.


Using only paper towels, paper, scotch tape, scissors and string as construction materials, the children are to brainstorm, design and create a contraption that will make a pencil or a piece of pasta, depending on their age, fall as slowly as possible. This activity is designed for children ages 6-16. The level of difficulty can be adapted with different materials and different falling objects. The older children will be given heavier objects and a limited amount of material.


By participating in this activity, the children will gain a better understanding of how gravity and air resistance affects falling objects. They will have to work cooperatively with team members in order to reach the goal. While each group tests their innovation, the others will compare their ideas to the one that is being tested and learn the strengths and weaknesses of different designs.


As part of a teachers demonstration our group constructed three parachutes with different characteristics out of nylon. One of the parachutes has one large hole in the center of it. Another parachute has several holes throughout. The third parachute has no holes in it. The idea is that the parachute with the most holes will fall the fastest, where as the one without any hole will fall the slowest, and the one with only one hole will fall between the other two. This will demonstrate how the surface area of a parachute causes the air resistance and makes objects fall slowly.


Our activity meets all of the specified requirements. The children are to solve a problem using physical principals that they have learned, thus engaging in an engineering activity. This activity appeals to all types of children; everyone loves to drop things. Components to the kit are safe, durable and do not take up much room. The activity description is very open-ended and so the children are free to design an original contraption anyway they please. The materials are all cheap and easily accessible at home and at school. Given the time it will take for children to brainstorm ideas, design, construct and test a device, it will take the full length of a class period. The whole kit will fit in a plastic storage tub easily because none of the objects in the kit are longer than one foot.

<span>Discovering Gravity & Air ResistanceBy Jason, Allison, and Dave</span><span> <span>If two objects, an elephant and a feather, are dropped off of the side of a building, which would experience a greater force of air resistance during the fall? Which would hit the ground first?<span>  Note: The above graphic requires Flash.</span>If two balls, one made of foam, and one made of metal, are dropped simultaneously from a tall building, which will fall the fastest?</span></span> <span> <span>What's included in the kit:(1) Teachers Manual(3) Demo Parachutes(1) Roll of Paper Towels(1) Pack of Pencils (24)(1) Spool of string(2) Rolls of Scotch Tape(10) Pairs of Safety Scissors(100) Sheets of Paper(1) Box of Paper Clips(1) Box of Pasta(2) Large Feathers(2) Small Feathers(2) Toy Parachutes   </span></span>

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At which of the following temperature and pressure levels would a gas be most likely to follow the ideal gas law? A. 0 K and 100
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The Ideal Gas Law makes a few assumptions from the Kinetic-Molecular Theory. These assumptions make our work much easier but aren't true under all conditions. The assumptions are,

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So, in analyzing the four choices given, we look for low P and high T.

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4 years ago
What is the force per unit area at this point acting normal to the surface with unit nor- Side View √√ mal vector n = (1/ 2)ex +
Mumz [18]

Complete Question:

Given \sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] at a point. What is the force per unit area at this point acting normal to the surface with\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z   ? Are there any shear stresses acting on this surface?

Answer:

Force per unit area, \sigma_n = 28 MPa

There are shear stresses acting on the surface since \tau \neq 0

Explanation:

\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]

equation of the normal, \b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z

\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]

Traction vector on n, T_n = \sigma \b n

T_n =  \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]

T_n = \left[\begin{array}{ccc}\frac{23}{\sqrt{2} }\\0\\\frac{27}{\sqrt{33} }\end{array}\right]

T_n = \frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z

To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.

\sigma_n = T_n . \b n

\sigma \b n = (\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z) . ((1/ \sqrt{2} ) \b e_x + 0 \b  e_y +(1/ \sqrt{2}) \b e_z)\\\\\sigma \b n = 28 MPa

If the shear stress, \tau, is calculated and it is not equal to zero, this means there are shear stresses.

\tau = T_n  - \sigma_n \b n

\tau =  [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau =  [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau =  \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z

\tau = \sqrt{(-5/\sqrt{2})^2  + (27/\sqrt{2})^2 + (5/\sqrt{2})^2} \\\\ \tau = 19.74 MPa

Since \tau \neq 0, there are shear stresses acting on the surface.

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notka56 [123]

Answer:

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