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30 Cards in this Set
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Velocity= displacement/time Average Velocity: (displacement for trip)/time |
Average Velocity= 45m/3s= 15m/s or 1/2(vi+vf)= 1/2(30m/s+0)= 15m/s |
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Acceleration |
rate of change of velocity or direction A= (vf-vi)/(tf-ti) change in velocity/change in time |
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If you are driving on a circular track at a constant speed, is acceleration zero? |
No, because constantly changing direction, so that change is an acceleration |
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Xi = inches vi = velocity in/s a = accecleration in/s2 |
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Newtons first law |
object at rest stays are rest in motion stays in motion |
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Newton's Second Law units for force? Weight? |
Force is proportional to acceleration Force = Mass x Acceleration 1 Newton= force to move 1kg by 1m/s2 Weight is in Newtons= kgx 9.8m/s2 |
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Tennis Ball and bowling ball are dropped. Does one have a greater force, neglecting air resistance? |
Although the acceleration is the same, the bowling ball requires more force (Newtons) to accelerate down because F= Mass kg x Acceleration in m/s2 |
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What is constant acceleration? If a ball is thrown straight up, is it still accelerating when it stops? |
meaning constant force applied F= Mass kg x Acceleration in m/s2 Yes, still accelerating |
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Can object move up or down with decreasing velocity? Can speed be negative? |
Yes No, speed is positive, so if showing a speed curve on a ball thrown up, it slows to zero and speeds back up. |
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Is a ball resting on a table accelerating? |
Yes, table is providing an opposite force. |
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Newtons Third Law |
For every action, there is an equal and opposite reaction.
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Example of a scalar? How do we relate these to vectors? |
Temperature Can only multiply and divide, so act as a coefficient -1X 3N= -3N 2x2N= 4 N |
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Work definition Formula units? |
Force applied over distance W= Force(N) x Displacement (M)= so, Newtonxmeters, or W= MxAxD= (kgxm)/s2 x (m) Joules =Kgxm2/s2 only force in direction of displacement counts towards work |
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Kinetic Energy definition Formula? If you roll a 1kg ball on the floor and exert 6N over 0.2M, What is the kinetic energy? |
Energy associated with the motion of an object KE= Fxd for initial calculation= NxM=J KE= (1/2)(mass)(v2) Work is the kinetic energy, only use other formula to solve for final speed. so KE= W= (kg already factored)6Nx0.2M= 1.2 J |
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0.1 kg ball 6N force over 2M What is the KE? What is the final velocity? |
KE = W= MAd= Fd= 6Nx0.2M= 1.2J KE= 1/2mv2, so 1.2J= (1/2)(0.1kg)(v2)= (1.2J)/(1/2)(0.1kg)= v2 =4.9 m/s |
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How do you calculate final velocity? |
Plug total Joules in as KE for KE=1/2 mv2 and solve for V |
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4kg ball rolling on flat surface at 2 m/s 6N force is applied over 0.5m in opposite direction initial kinetic energy? Work by force? Final Kinetic energy? Final Velocity? |
Initial KE= (1/2)4kgx (2m/s)2= 8J work by forced= 6Nx0.5M= 3J Final KE = 8J-3J= 5J Vf? 5J= (1/2) 4kg (v2)= (5J)/(1/2)4kg= v2= 1.6 m/s |
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Potential Energy |
Same as calculating Work =FxD where distance is height. PE= mgh or mass x acceleration x height |
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Calculate the potential energy of 2kg box on shelf 1.5m high |
PE= mgh PE= (2kg)(9.8m/s2)(1.5m)= 29J |
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Relate the Law of conservation of energy to Potential Energy |
Energy cannot be created or destroyed, therefore 29J of Potential Energy can be converted to 29J of Kinetic Energy |
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What if 10J of energy is used to push a box along a surface? Does it have 10J PE? |
No, 10J was lost in the form of heat energy or friction |
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4kg box 1.5m off the ground. What is its speed when it hits the ground? |
PE= mgh so - (4kg)(9.8m/s2)(1.5m)= 59J Then KE=1/2mv2 59J= (1/2) 4kg v2 =5.4 m/s |
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Skier on top of 30 degree hill, what is speed at bottom? 55kg 100m high |
PE= mgh 55kgx100mx9.8= 53900J= PE which is converted to KE KE= 1/2mv2 53900J=1/2 (55kg) v2 Speed= 44.3 m/s Easier way for velocity with gravity= V= root(2gh) or V=root(2x9.8x100m)= 44.3 m/s |
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What if hill is bumpy or goes up or down |
same speed because energy is conserved |
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55kg skier 100m hill How much energy is lost to "other forces" if final velocity is 10m/s? |
PE= mgh= 55x9.8x100m= 53,900J KE= 1/2mv2 = 1/2 (55kg) (10m/s2)=2750J So for loss, take PE at top- KE at Bottom =51,150J lost to other forces |
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55kg skier already traveling at 10m/s at top of 100m hill What is their speed at bottom assuming no other factors are at play? |
Total energy at top is PE + KE So PE = mgh= 55x9.8x100= 53900 KE= 1/2 55kg x (10)2 = 2750 Total energy at top= 56,650 For speed KE =mv2 or 56650= 1/2(55kg)x (v)2= 45.4 m/s not much different |
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Power Unit Formula? |
The rate that work is done 1J/per second= 1 Watt Joules/second Power=work/time Power = (mgh)/time |
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average power output hiker 55kg who climbs 1300m in 2h |
Power= mgh/t 55x9.8x1300/(2h) need seconds 55x9.8x1300/ 7200= 97.3 Watts or Joules/sec |
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Momentum sign? definition |
Momentum is vector quantitiy signified by little p or p Force = rate of change of momentum |
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momentum p Momentum greater, 100kg man at 2m/s or 2000kg car at 1 m/s? |
p= Mass x Velocity Man p = 100kg x 2 m/s= 200 Car p = 2000kg x 1 m/s = 2000 car is greater |
