Adquisición de conocimiento, de Aplicación, de Metacognición y Actividad Integradora
Enviado por MarioRico3009 • 26 de Octubre de 2017 • Prácticas o problemas • 2.578 Palabras (11 Páginas) • 223 Visitas
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Knowledge Acquisition Activity
In this activity you will be able to recall a topic seen in previous leering unit, which we explored a little more in this unit. It consist of 3 phases that you will be solving as you advance in contest of this unit. And the end, integrate all in a documented and give it to your teacher for a review. Once your teacher cheeked it, include it in your portfolio of evidence.
Phase 1
- - Make a team with 4 classmates (or follow the instruction of your teacher) and respond the following questions that were see in the previous learning unit.
- Outline that differences between vector and scalar quantities.
Points out the differences between scalar and vector quantities.
Scalar quantities: whose single magnitude is sufficient to represent it.
Vector Quantities: is fully specified for a magnitude and a difference
B) What are the meanings of concurrent and non-concurrent vectors?
Concurrent: when the direction of the vectors intersect at some point.
Non-competitors: the lines that determine their action on a straight line are not cut.
C) How does the magnitude of the resultant between two vectors change if the angle between them decreases?
Let the angle between the two vectors decrease that each time they are more closely attached to each other.
D) What is the sum of three times vectors that form a closed triangle?
Two concurrent vectors are taken as representing, lines are drawn parallel to the vectors obtained a logarithm.
E) What is the procedure to add 3 concurrent vectors?
From a vector to another vector is translated to form a triangle the resulting vector is in the line that forms the triangle and its point of application will match the origin of the first point and its vector.
F) if 2 force vectors have magnitudes of 8N and 10N, for what angle is the magnitude of the resultant maximum and for which the minimum?
Maximum
R = 8N + 10N = 18N
(18) 2 = (82) + (10) 2 -2 (8) (10) -cos (a)
324 = 64 + 100-160 cos (a)
Minimal
R = 10N-8N = 2N
(2) (2) (2) (2) (2) (2) (8) cos (a)
(A) = 0
H) under what condition is the magnitude of a vector equal to that of one of its components?
Said vector has no or no angle with any Cartesian coordinate axis.
I) if a vector F is drawn in an xy plane, what interval can the angle of the vector itself take if ...
Its components x and y are positive
0
Its component x is positive and y is negative -90 (270)
The components x and y are negative 90
If its component x is negative and y is positive 180
2.- After answering the questions, performed tougher with your teacher a guide discussion to cooperate your answer with those of the others teams. Modify your responses if necessary; Integrates questions in a document.
Phase 2
1.-identif with the help of your teacher the cardinal points EAST, WEST, NORTH AND SOUTH
2.- Take as reference your classroom estimates the vector position of
Magnitude | Direction | |
The school administration | 10 m | West |
The soccer field | 40 m | South |
The main entrance | 35m | |
Your hose | 18 km | |
Main square of your neighbor | 17 km | West |
Prefecture | 10 m | South |
3. - Integrate the chart above to the same document where you answer the previous questions
Phase 3
Solve the following vetor sums using the method that your teacher indicates
Problem | Procedure | Interpretation |
Find the resutan force for the sum f1=90N and F2= 35N if two of them form an angle of 60 | 90cos60= 45 90sen60= 77.94 35cos60= 30.31 Fr= 124.99 Cos=59.99 | [pic 10] |
Find the position Final (relative to its initial position) of a person if it travels 6km to the north 8km to the east and 7km to the south. | 6km + 8km - 7km= 7 | [pic 11] |
Find the resultant force acting on a body to which F1 = 40N is applied at 30 F2 = 50N at 250 and F3 = 10N at 300 | 40cos 30 = 34.64 40sen 30 = 20 50cos 250 = -17.10 50sen 250 = -46.98 70cos 300 = 300 70sen 300 = 35 FR= 53.14 sen= 8.6 | [pic 12] |
Find the resultant force acting on a body to which F1 = 80N is applied to 130 F2 = 50N to 250 and F3 = 70N to 300. | 80cos130= -51.42 80sen130= 61.28 50cos250= -17.10 50sen250= -46.98 70cos300= 35 70sen300= -60.62 Fr= 57.17 Sen= 54.10 | [pic 13] |
Find the final position of a car if it travels 15km to the North, 16km to the South and 17km to the West. | Fx = -32 50cos90° = 0 38cos360°= 38 70cos180°=-70 Fy= 50 50sen90° =50 38sen360°= 0 70sen180° =0 Fr= [pic 14] Fr= 59 [pic 15] [pic 16] | [pic 17] |
Find the final position of a car, if it travels 50km to the NE, 38km to the South and 70km to the West. | Fx=-1 15cos90° =0 16cos360°=16 17cos180°=-17 Fy= 15 15sen90° =15 16sen360°= 0 17sen180°=0 Fr=[pic 18] Fr=15.03 | [pic 19] |
Find the resultant force acting on a body to which it is applied F1 = 90N at 160 ° and F2 = 100N at 230 | Fx=-148.7 90cos160°= -84.5 100cos230°= -64.2 Fy= -46.6 90sen160° =30.7 100sen230°=-76.6 Fr=[pic 20] Fr= 149 |
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