"with the wind" is a tail-wind, and the speeds are added to get the groundspeed.
"against the wind" is a head-wind, and the windspeed is subtracted from the airspeed.
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.
Answer:
v = 3.08 km/s
Explanation:
Given that,
The angular velocity of the satellite = 
A satellite is in orbit 36000km above the surface of the earth.
The radius of the earth is 6400 km
Let v is the velocity of the satellite. It can be calculated as :

So, the velocity of the satellite is 3.08 km/s.
Answer:
0.07756 m
Explanation:
Given mass of object =0.20 kg
spring constant = 120 n/m
maximum speed = 1.9 m/sec
We have to find the amplitude of the motion
We know that maximum speed of the object when it is in harmonic motion is given by
where A is amplitude and
is angular velocity
Angular velocity is given by
where k is spring constant and m is mass
So 

About 5 hours gooood luck