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examples of distance and displacement problem

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Question:Pls solve the problem with example

Answers:Not clear what's being asked. Do we need to find how far various points move? Surely not, it's too easy, for each point multiply 2*pi*(distance from centre of rotation)*(angle of rotation in degrees) divided by 360.

Question:How do you know when to subtract or add displacement vectors? For ex. if a man walked east +200 m then 100 m west it'd be: d=200 + (-100) = 100 m east. However why wouldn't you use d=df-di? Does it only apply if the final distance is greater than the initial? for ex. a man walked east +200 m then 300 m west. would it be d= (-300) - (+200) = -500? do you only subtract if the final position is a bigger number than the initial? i'm confused. when do we know if we use the equation or not epecially in collinear vectors in displacement?

Answers:Just use common sense. Sometimes it makes no difference which is subtracted from which. It helps if you make a quick sketch on a piece of paper. your example: if a man walked east +200 m then 100 m west it'd be: d=200 + (-100) = 100 m east. is correct. Best if you pick a direction and call that positive displacement, and the opposite negative. Usual is the positive x axis is +, negative x axis is -. your second example: "walked east +200 m then 300 m west" Use the rule I just mentioned. +200 +(-300) = -100, or 100m west. If you make a sketch, you would have seen that 500 is incorrect. Now this all get a lot more complicated when the vectors are no co-inear, so you need to get this down correctly.

Question:I don't know how to do this problem A car is traveling with a constant acceleration, starting from 20m/s [E] to 60/s [E] in 20 seconds. What is the total displacement. If you could give the formula and show how it works, it would be appreciated. Thanks

Answers:Answer --> 800 m How to find it... Given Vi = Initial velocity Vf = Final velocity t = time D = Distance / displacement Values: Vi = 20 m/s Vf = 60 m/s t = 20 s Necessary equation is d = 1/2 ( Vf + Vi ) t D = 1/2 * (60 m/s + 20 m/s) * (20 s) D = 1/2 * (80 m/s) * (20 s) D = 800 m

Question:I've been hearing over and over that what matters in the end is the amount of air displacement you have for how loud it will get. Like, how two 12s would be louder than one 15 because of the larger cone area displacing more air. So would a subwoofer with 0.1 cu. ft. air displacement be twice as loud as one woofer that has 0.05 cu. ft. of displacement? Is this logic correct? (Yes, I am aware that how loud a system is depends on a number of factors, but I am focusing on this one in particular.)

Answers:The air "displacement" you're referring to has to do with the physical size (and shape) of a specific driver and does not affect the SPL of the driver. A subwoofer with 0.1 Ft^3 displacement would simply be more massive than one displacing 0.05 Ft^3. The number is most useful in determining the final size of enclosure in which to mount the subwoofer. The displacement figure is subtracted from the enclosure's "gross" volume (along with anything else that takes up airspace - like a port for example) to arrive at the "net" volume. The volume of air "movement" produced by the subwoofer has a greater effect on the SPL. And air movement is more affected by the total cone area and the distance in and out that the cone is designed to move. "Xmax" specifies the distance the voice coil (and consequently the speaker cone) can move in one direction and generally the greater the Xmax, the more air is moved. But probably more important than either displacement or Xmax, and perhaps the best indication of SPL is the sensitivity, also known as "SPL". A higher sensitivity means the subwoofer is more efficient at turning electrical energy into useful audio. This figure shows the loudness in dB measured with 1 watt of power from a distance 1 meter directly in front of the speaker cone. Each time you double the power, the dB's increase by 3. So if you are comparing competing subwoofers and want to know which is "louder", with the other parameters constant, it's the one with the highest sensitivity.

From Youtube

Distance Formula Problems 2 - Equating Two Distance & Getting Value of X :SkyingBlogger.Com - This is an example of distance formula problems. In this problem we are given two equal distances and by equating them we have to find out the unknown value suppose x as given in above video.

Solving Word Problems in Distance, Rate, and Time Using Quadratics - Example 2 :Solving Word Problems in Distance, Rate, and Time Using Quadratics - Example 2