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
Via half-life equation we have:
Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes. By plugging those values in we get:
There is 6.25 grams left of Ra-229 after 12 minutes.
Answer:
Explanation:
When a ray of light passes through denser medium to rarer medium and then after refraction it comes back into the denser medium, the phenomenon is called total internal refraction.
There are two necessary conditions for the total internal refraction.
(1) The refraction takes place g=from denser medium to rarer medium.
(2) The angle of incidence in the denser medium should be more than the critical angle for that pair of media.
Answer:
The work done by the force on the particle is 29.85 J.
Explanation:
The work is given by:
Where:
x₁: is the lower limit = 0 m
x₂: is the upper limit = 1.9 m
Fₓ: is the force in the horizontal direction = (4.5 + 13.7x - 1.5x²)N
Therefore, the work done by the force on the particle is 29.85 J.
I hope it helps you!
