| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.89 |
| Score | 0% | 58% |
Drag is a type of:
kinetic energy |
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work |
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potential energy |
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friction |
Drag is friction that opposes movement through a fluid like liquid or air. The amount of drag depends on the shape and speed of the object with slower objects experiencing less drag than faster objects and more aerodynamic objects experiencing less drag than those with a large leading surface area.
Concurrent forces:
pass through a common point |
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act in a common plane |
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act in a common dimension |
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act along the same line of action |
Collinear forces act along the same line of action, concurrent forces pass through a common point and coplanar forces act in a common plane.
A truck is using a rope to pull a car. Tension in the rope is greatest in which of the following places?
near the car |
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near the truck |
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in the middle |
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tension is equal in all parts of the rope |
Tension is a force that stretches or elongates something. When a cable or rope is used to pull an object, for example, it stretches internally as it accepts the weight that it's moving. Although tension is often treated as applying equally to all parts of a material, it's greater at the places where the material is under the most stress.
| 10 | |
| 7 | |
| 1.43 | |
| 0.7 |
The mechanical advantage of a wheel and axle is the input radius divided by the output radius:
MA = \( \frac{r_i}{r_o} \)
In this case, the input radius (where the effort force is being applied) is 7 and the output radius (where the resistance is being applied) is 10 for a mechanical advantage of \( \frac{7}{10} \) = 0.7
| 8.59 lbs. | |
| 34.38 lbs. | |
| 11.46 lbs. | |
| 11 lbs. |
To balance this lever the torques on each side of the fulcrum must be equal. Torque is weight x distance from the fulcrum so the equation for equilibrium is:
Rada = Rbdb
where a represents the left side of the fulcrum and b the right, R is resistance (weight) and d is the distance from the fulcrum.Solving for Ra, our missing value, and plugging in our variables yields:
Ra = \( \frac{R_bd_b}{d_a} \) = \( \frac{55 lbs. \times 5 ft.}{8 ft.} \) = \( \frac{275 ft⋅lb}{8 ft.} \) = 34.38 lbs.