Answer:t=0.3253 s
Explanation:
Given
speed of balloon is 
speed of camera 
Initial separation between camera and balloon is 
Suppose after t sec of throw camera reach balloon then,
distance travel by balloon is
and distance travel by camera to reach balloon is
Now
There are two times when camera reaches the same level as balloon and the smaller time is associated with with the first one .
(b)When passenger catches the camera time is 
velocity is given by
and position of camera is same as of balloon so
Position is 
Answer:
Explanation:
Distance travelled by sound in going to target( insect ) and returning back = 2d ,d is distance of target .
time t = .06 s
speed = distance / time
344.9 = 2d / .06
d = 10.35 m
Answer:at 21.6 min they were separated by 12 km
Explanation:
We can consider the next diagram
B2------15km/h------->Dock
|
|
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.
Power is the work done per unit time. Therefore,
Therefore,
The answer is D
Hope this helps
