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:
Timbre also known as color or tone quantity which is used for music notes sound and tone are mostly know for guitars or pianos .
Does this help .
time=distance/speed
1.6/100 secs = 0.016secs=16millisecs
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Answer:
354 m/s
Explanation:
For the second overtune (Third harmonic) of an open pipe,
λ = 2L/3................................ Equation 1
Where L = Length of the open pipe, λ = Wave length.
Given: L = 1.75 m.
Substitute into equation 1
λ = 2(1.75)/3
λ = 1.17 m.
From the question,
V = λf.......................... Equation 2
V = speed of sound in the room, f = frequency
Given: f = 303 Hz.
Substitute into equation 2
V = 1.17(303)
V = 353.5
V ≈ 354 m/s
Hence the right answer is 354 m/s
