Answer:at 21.6 min they were separated by 12 km
Explanation:
We can consider the next diagram
B2------15km/h------->Dock
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B1 at 20km/h
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V
So by the time B1 leaves, being B2 traveling at constant 15km/h and getting to the dock one hour later means it was at 15km from the dock, the other boat, B1 is at a distance at a given time, considering constant speed of 20km/h*t going south, where t is in hours, meanwhile from the dock the B2 is at a distance of (15km-15km/h*t), t=0, when it is 8pm.
Then we have a right triangle and the distance from boat B1 to boat B2, can be measured as the square root of (15-15*t)^2 +(20*t)^2. We are looking for a minimum, then we have to find the derivative with respect to t. This is 5*(25*t-9)/(sqrt(25*t^2-18*t+9)), this derivative is zero at t=9/25=0,36 h = 21.6 min, now to be sure it is a minimum we apply the second derivative criteria that states that if the second derivative at the given critical point is positive it means here we have a minimum, and by calculating the second derivative we find it is 720/(25 t^2 - 18 t + 9)^(3/2) that is positive at t=9/25, then we have our answer. And besides replacing the value of t we get the distance is 12 km.
p=mv so wouldn't u multiply them?
Answer:
Objects with mass exert forces on each other via the force of gravity. This force is proportional to the mass of the two interacting objects, and is inversely proportional to the square of the distance between them. The factors G, M, and r are the same for all masses at the surface of the Earth.
5 second fall starting at 0 m/s
ball strikes ground at a speed = 49 meters per second.
For a 50 kg person receives an absorbed dose of gamma radiation of 20 millirads, the total energy absorbed is mathematically given as
E=0.1457J
<h3>What is the total energy absorbed?</h3>
Generally, the equation for the total energy absorbed is mathematically given as
E=mass*gamma radiation
Therefore
E=50*20*19^{-3}
E=0.1457J
In conclusion, the total energy absorbed
E=0.1457J
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