Two soccer players, mary and jane, begin running from approximately the same point at the same time. mary runs in an easterly di
rection at 3.59 m/s, while jane takes off in a direction 42.8 ◦ north of east at 4.67 m/s. what is the magnitude of the velocity of jane relative to mary?
1 answer:
As you illustrate the problem, you may come up with the same diagram as the one shown in the attached picture. The relative velocity is the linear velocity between the two velocities. This is calculated by simply finding the difference of their velocities. Since, the velocity is with respect to Jane relative to Mary, the solution is as follows:
Relative velocity = Velocity of Jane - Velocity of Mary
Relative Velocity = 4.67 m/s - 3.59 m/s = 1.08 m/s
Relative velocity = Velocity of Jane - Velocity of Mary
Relative Velocity = 4.67 m/s - 3.59 m/s = 1.08 m/s
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Height of object is 5 cm
Object distance from a convex lens is 18 cm
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i.e. h = 5 cm
u = -18 cm
f = +10 cm
Let v is distance of the image from the lens. Using lens formula :
The magnification of lens is :
, h' is height of the image
h' = -5.00 cm (in three significant figures)
Answer:
a) 1.248 rad/s
b) 13.728 m/s
c) 0.52 rad/s^2
d) 17.132m/s^2
Explanation:
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r: radius if the trajectory of the astronaut = 11.0m
You replace r and w and obtain:
(c) The tangential acceleration is:
(d) The radial acceleration is: