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I am having trouble in physics?
1) A quarterback takes the ball from the line of scrimmage, runs backward for 6 yards, and then runs sideways parallel to the line of scrimmage for 4 yards. At this point, he throws a 34 yard forward pass straight down the field. What is the magnitude of the football's resultant displacement?
2) A submarine dives 88 m at an angle of 11° below the horizontal. What are the two components?(x and y)
3) A person walks 38° north of east for 3.10 km. How far would another person walk due north and due east to arrive at the same location?(km due north, km due east)
I have no idea how to solve these so any help is appreciated. Please explain how to solve them not just give the answer. Thanks
1) The first question is simply the addition of vectors. But because these are vectors, you must take direction into account. Do not simply add 32 + 4 - 6.
The simplest way to understand this is to draw a very simple diagram. Get a piece of paper, imagine football field and from the starting point on the field, draw a line 6 units backwards, then 4 units sideways (it doesn't matter which direction, as the question didn't specify), then another one 34 units forward. Label each line with their lengths in yards.
The aim here is to find the distance between the starting point of the first line, and the end point of the last line. Draw a dotted line straight between these two points, and then another dotted line from the staring point straight across till it meets the 34 yard line. You should now have a tringle, and from that, you should be able to calculate the lenths of the two dotted lines you just drew.
The width is easy. It's the same length as the 4 yard line behind it. The other side of the triangle can be found by doing 34 - 6, leaving you with 28. So you now have a triangle with sides 4 and 28 and you need to find the hypotenuse. You can do this with Pythagoras theorem easily enough.
2) This question will use trigonometry. Draw another diagram for this, with the path of the submarine following a path from the water's surface, at an angle of 11˚ down to a depth of 88m. With that line as the hypotenuse, make that into a triangle by drawing drawing a vertical line from the surface of the water to the end of the submarine's path. Mark the angle between the water surface and submarine path as 11˚
You're told that the depth is 88m, so that's your y component. To find the x component, all you need to know is that:
tan θ = opposite / adjacent
Or, in this particular case
tan(11˚) = y / x = 88 / x
Just rearrange that equation to solve for x.
3) This is very similar to the last question. Draw another triangle, label the known side and angle, and then use trigonometry to find the lengths of the sides. The only difference is this time you know the length of the hypotenuse, and so to find the other two sides, you'll have to use sin and cos.
sin θ = opposite / hypotenuse
cos θ = adjacent / hypotenuse
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