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(When blocking in football, why does a defending lineman often attempt to get his body under that of his opponent and push upward?) What effect does this have on the friction force between the opposing lineman’s feet and the ground?. Do Now 10.21.10.

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Do now 10 21 10

(When blocking in football, why does a defending lineman often attempt to get his body under that of his opponent and push upward?) What effect does this have on the friction force between the opposing lineman’s feet and the ground?

Do Now

10.21.10

http://cdn-aki.vmixcore.com/825/14811/26/531156701/448/825/5311/fbb4aa1aad4bf20dd9291a6847276a54.jpg


Do now 10 21

10 m/s

C

5 m/s

10 m/s

B

5 m/s

10 m/s

10 m/s

10 m/s

A

10 m/s

10 m/s

Do Now 10/21


Test tuesday

  • Chapters 2, 3, and 6 often attempt to get his body under that of his opponent and push upward?)

    • Newton’s 1st Law (Inertia)

    • Newton’s 2nd Law

    • Friction

    • Air Resistance

Test Tuesday


Objective

  • Explain freefall forces often attempt to get his body under that of his opponent and push upward?)

  • State the two factors that affect air resistance

Objective


6.6 often attempt to get his body under that of his opponent and push upward?) Free Fall Explained

The ratio of weight (F) to mass (m) is the same for the 10-kg cannonball and the 1-kg stone.


6.7 often attempt to get his body under that of his opponent and push upward?) Falling and Air Resistance

The air resistance force an object experiences depends on the object’s speed and surface area.


6.7 often attempt to get his body under that of his opponent and push upward?) Falling and Air Resistance

Terminal Speed

Terminal speed isthe speed at which the acceleration of a falling object is zero because friction balances the weight.


6.7 often attempt to get his body under that of his opponent and push upward?) Falling and Air Resistance

Sky divers reach terminal speed when air resistance equals weight.


6.7 often attempt to get his body under that of his opponent and push upward?) Falling and Air Resistance

  • The terminal speed for a sky diver varies from about 150 to 200 km/h, depending on the weight and orientation of the body.

    • A heavier person will attain a greater terminal speed than a lighter person.

      • The heavier person’s weight “pushes” through the air particles with greater force

    • Body orientation also makes a difference. Dive vs. Spread out.


6.7 often attempt to get his body under that of his opponent and push upward?) Falling and Air Resistance

The flying squirrel increases its area by spreading out. This increases air resistance and decreases the speed of its fall.


6.7 often attempt to get his body under that of his opponent and push upward?) Falling and Air Resistance

A parachute will bring the skydiver’s speed to about 15 km/h to 25 km/h


6.7 often attempt to get his body under that of his opponent and push upward?) Falling and Air Resistance

think!

Which experiences a greater air resistance force, a falling piece of paper or a falling elephant?


6.7 often attempt to get his body under that of his opponent and push upward?) Falling and Air Resistance

think!

Which experiences a greater air resistance force, a falling piece of paper or a falling elephant?

Answer:The elephant! It has a greater frontal area and falls faster than a piece of paper—both of which mean the elephant pushes more air molecules out of the way. The effect of the air resistance force on each, however, is another story!


Practice

Practice


Hw answers newton s 2 nd law ws1

  • 4 m/s acceleration of 6.2 m/s2 and 2 m/s2

  • 3 kg

  • 6 m/s2

  • 1 m/s2

  • 3 m/s2

  • 12 m/s2

  • Half as fast

  • 140,000 N

  • 15 m/s2

  • 7.5 m/s2

HW AnswersNewton’s 2nd Law WS1


Newton s 2 nd law ws2 answers

  • a = 0.25 m/s acceleration of 6.2 m/s2

  • Fnet = 0 N

  • a = 0 m/s2

  • a = 2 m/s2

  • a = 1 m/s2

  • a = 10 m/s2

  • FN = 80 N, m = 8 kg, a = 5 m/s2, Fnet = 40 N

  • FN = 100 N, μ = 0.1, m = 10 kg, a = 1 m/s2, Fnet = 10 N

  • FN = 50 N, Ff = 5 N, Fg = 50 N, a = 1 m/s2, Fnet = 5 N

  • μ = 0.167

  • Fa = 105 N, Fa = 0 N

Newton’s 2nd Law WS2 Answers


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