D: radio waves
Explanation:
radio wves are use to carry satellite signals
The answer is c. the degree of wetness of the paper towels.
Answer:
(a) 25 m
(b) 75 m
Explanation:
Given that the jogger runs at a constant rate of 10.0 m every 2.0 seconds.
So, the speed of the jogger,
Let d be the distance covered by him in time, t s.
As distance=(speed) x (time)
So, 
From equation (i)
As the jogger starts from origin, so, the distance, 
, also represents the position of the jogger at the time 
s.
The position-time graph has been shown.
(a) From equation (ii), for t=5.0 s
So, the jogger is at a distance of 25 m from the origin.
(b) Similarly, for t=15.0 s
So, the jogger is at a distance of 75 m from the origin.
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
|
|
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.
