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jekas [21]
3 years ago
6

A car is driven 13 mi east and then certain distance due north, ending up at the position 25 degrees north of east of its initia

l position. (a) the distance traveled by the car due north is (1) less than, (2) equal to, (3) greater than 13 mi. Why? (b) what distance due north does the car travel? Please show work!!
Physics
1 answer:
Kobotan [32]3 years ago
3 0

Answer: Hello mate!

lets define the north as the y-axis and east as the x-axis.

Using the notation (x,y) we can define the initial position of the car as (0,0)

then the car travells 13 mi east, so now the position is (13,0)

then the car travels Y miles to the north, so the position now is (13, Y)

and we know that the final position is 25° degrees north of east of the initial position. This angle says that the distance traveled to the north is less than 13 mi because this angle is closer to the x-axis (or east in this case).

This angle is measured from east to north, then the adjacent cathetus is on the x-axis, in this case, 13mi

And we want to find the distance Y, so we can use the tangent:

Tan(25°) = Y/13  

tan(25°)*13 mi = Y = 6.06 mi.

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Because distance measure from center .

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F=G\frac{m_1m_2}{r^2}

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Using the formula

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Read 2 more answers
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