A pressure that will support a column of Hg to a height of 256 mm would support a column of water to what height? The density of
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The question is incomplete. Here is the complete question.
Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.
(a) What is the total mass of the three boxes?
(b) What is the mass of each box?
Answer: (a) Total mass = 2384.5kg;
(b) m1 = 915kg;
m2 = 605kg;
m3 = 864.5kg;
Explanation: The image of the boxes is described in the picture below.
(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:
Total mass of the system of boxes is 2384.5kg.
(b) For each mass, analyse each box and make them each a free-body diagram.
<u>For </u><u>:</u>
The only force acting On the box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.
= 915kg
<u>For </u><u>:</u>
There are two forces acting on : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:
= 605kg
<u>For </u><u>:</u>
= 864.5kg
Hi there!
We can use the kinematic equation:
vf = Final velocity (? m/s)
vi = initial velocity (0 m/s, dropped from rest)
a = acceleration (due to gravity, 9.8 m/s²)
d = distance (9.8 m)
Simplify the equation to solve for vf:
Substitute in the given values: