image is question 12 http://imgur.com/ucmeYmU Two boxes are connected by a cord
ID: 2123573 • Letter: I
Question
image is question 12
http://imgur.com/ucmeYmU
Two boxes are connected by a cord running over a pulley as shown.
Box I of mass 8.0 kg sits on the top of the table.
Box II has a mass of 15.0 kg.
The coefficient of kinetic friction for box I and the table is 0.10.
Draw the free-body diagrams for the two boxes,
a) identifying all of the forces acting on each of the masses.
b- 5 pts) Calculate the acceleration of the system.
c- 5 pts) Calculate the tension in the cord.
question 13
Water is pumped through a neighborhood. It moves with a speed of 1.3 m/s through a
pipe with a diameter 5.0 cm at a pressure of 4.0 atm. (1 atm = 1.01 x 105
Pa)
The pipe narrows to a diameter of 3.0 cm but remains at the same height.
Assume that water is incompressible.
(a What will be the speed of the water in the smaller pipe?
bWhat will be the pressure of the water in the smaller pipe?
Two boxes are connected by a cord running over a pulley as shown. Box I of mass 8.0 kg sits on the top of the table. Box II has a mass of 15.0 kg. The coefficient of kinetic friction for box I and the table is 0.10. Draw the free-body diagrams for the two boxes, identifying all of the forces acting on each of the masses. Calculate the acceleration of the system. Calculate the tension in the cord. Water is pumped through a neighborhood. It moves with a speed of 1.3 m/s through a pipe with a diameter 5.0 cm at a pressure of 4.0 atm. (1 atm = 1.01 x 105 Pa) The pipe narrows to a diameter of 3.0 cm but remains at the same height. Assume that water is incompressible. What will be the speed of the water in the smaller pipe? bWhat will be the pressure of the water in the smaller pipe?Explanation / Answer
Question 14
The forces acting on the beam are :
1) Tension T1 acting upward in the left end..
2) force of 650 N downward at 3 m from the left end ( = 18 - 3 = 15 meter from right end )
3) Weight of 200 N downward at the center i.e. at 9 m from left as well as right end
4) Tension T2 acting upward in the right end
so taking moment of all the forces about the right end...
T1 * 18 = (15*650) + (200 * 9 )
so.. T1 = 641.667 ...
a) Tension in Left cable = T1 = 641.667 N
b) taking moment of all the forces about the Left end...
T2 * 18 = (3*650) + (200 * 9 )
so.. T2 = 208.333 N
Tension in Right cable = T2 = 641.667 N
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