False
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50 g of liquid X at 10 Celcius and 200 g of liquid Y
mx*cx*(t-tx)+my*cy*(t-ty)=0
cx/cy = - my*(t-ty) : mx*(t-tx) = (my/mx) * (ty - t) / (t-tx)
cx/cy = 200/50*(40-15)/(15-10) = 20
cx/cy = 20
A must be at least 4 full paragraphs probably will need more
Explanation:
Let h is the height of the plane above ground. x is the horizontal distance between the ground and the airport. Let s(t) is the distance between the plane and the airport. So,
...........(1)
Given, h = 4, x = 40 and s(t) = -20 mph
Differentiate equation (1) wrt t


When x = 40, 



So, the speed of the airplane is 241.14 m/s. Hence, this is the required solution.
The speed of tsunami is a.0.32 km.
Steps involved :
The equation s = 356d models the maximum speed that a tsunami can move at. It reads as follows: s = 200 km/h d =?
Let's now change s to s in the equation to determine d: s = 356√d 200 = 356√d √d = 200 ÷ 356 √d = 0.562 Let's square the equation now by squaring both sides: (√d)² = (0.562) ² d = (0.562)² = 0.316 ≈ 0.32
As a result, 0.32 km is roughly the depth (d) of water for a tsunami moving at 200 km/h.
To learn more about tsunami refer : brainly.com/question/11687903
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