60.3° from due south and 5.89 m/s For this problem, first calculate a translation that will put John's destination directly on the origin and apply that translation to Mary's destination. Then the vector from the origin to Mary's new destination will be the relative vector of Mary as compared to John. So John is traveling due south at 6.7 m/s. After 1 second, he will be at coordinates (0,-6.7). The translation will be (0,6.7) Mary is traveling 28° West of due south. So her location after 1 second will be (-sin(28)*10.9, -cos(28)*10.9) = (-5.117240034, -9.624128762) After translating that coordinate up by 6.7, you get (-5.117240034, -2.924128762) The tangent of the angle will be 2.924128762/5.117240034 = 0.57142693 The arc tangent is atan(0.57142693) = 29.74481039° Subtract that value from 90 since you want the complement of the angle which is now 60.25518961° So Mary is traveling 60.3° relative to due south as seen from John's point of view. The magnitude of her relative speed is sqrt(-5.117240034^2 + -2.924128762^2) = 5.893783 m/s Rounding the results to 3 significant digits results in 60.3° and 5.89 m/s
Answer:
Explanation:
Given
-- radius of hydrogen atom
<em>See attachment</em>
Required
Determine the radius of magnesium atom (R)
From the attachment, the ratio of a hydrogen atom to a magnesium atom is:
Simplify
Represent the radius as ratio:
Substitute
Equate both ratios
Express as fraction
Cross Multiply
Hence, the radius of the magnesium atom is: 
Answer:
Its Moving at a constant Speed of 90km/hr
You can convert this speed to Meters/Second or Change it to Km/Second(Because we need our distance in Km)
So 90km/hr x 1/60 x1/60
=90km/3600sec
v=0.025km/sec
When an Object Moves at Constant Speed... Its displacement(distance) is given by
d=vt
given v=0.025km/sec
t=1sec
d=0.025km/sec x 1sec
d=0.025km.
