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One of the simplest physical situations to imagine of is a falling object. Thus let's consider a falling object along with mass m and derive a differential equation as, when resolved, will provide us the velocity of the object at any time, t. We will suppose that only gravity and air resistance will act upon the object like it falls. Below is a figure demonstrating the forces which will act upon the object.
Before describing all the terms in such problem we require to set some conventions. We will suppose that forces acting in the downward direction are positive forces whereas forces which act in the upward direction are negative. Similarly, we will suppose that an object moving downward that is a falling object will have a positive velocity.
Here, let's take a look at the forces demonstrated in the diagram above.
FG is the force because of gravity and is provided by FG = mg where g is the acceleration because of gravity. In this type of class I use g = 9.8 m/s2 or g = 32 ft/s2 based on whether we will utilize the metric or British system.
FA is the force because of air resistance and for this illustration we will suppose that this is proportional to the velocity, v, of the mass. Hence the force because of air resistance is then given through FA = -gv , where g > 0. Remember that the "-" is needed to get the correct sign upon the force. Both g and v are positive and the force is acting upward and thus must be negative. The "-" will provide us the correct sign and thus direction for this force.
1000000 divided by 19
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a²+b²=1 a+b
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7(y + 3) - 2(x + 2) = 14, 4 (y - 2) + 3(x - 3) = 2 Ans: 7(y + 3) - 2 (x+ 2) = 14 --------- (1) 4(y- 2) + 3(x - 3) = 2 ----------(2) From (1) 7y +21 -
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