Answer:
7.5 ft³/min
Step-by-step explanation:
Let x be the depth below the surface of the water. The height, h of the water is thus h = 10 - x.
Now, the volume of water V = Ah where A = area of isosceles base of trough = 1/2bh' where b = base of triangle = 4 ft and h' = height of triangle = 1 ft. So, A = 1/2 × 4 ft × 1 ft = 2 ft²
So, V = Ah = 2h = 2(10 - x)
The rate of change of volume is thus
dV/dt = d[2(10 - x)]/dt = -2dx/dt
Since dV/dt = 15 ft³/min,
dx/dt = -(dV/dt)/2 = -15 ft³/min ÷ 2 = -7.5 ft³/min
Since the height of the water is h = 10 - x, the rate at which the water level is rising is dh/dt = d[10 - x]/dt
= -dx/dt
= -(-7.5 ft³/min)
= 7.5 ft³/min
And the height at this point when x = 8 inches = 8 in × 1 ft/12 in = 0.67 ft is h = (10 - 0.67) ft = 9.33 ft
Answer:
Age 1-NO
Age 2-NO
Age 3-NO
Age 4-NO
Age 5-NO
Age 6-NO
Age 7-NO
Age 8-NO
Age 9-NO
Age 10-NO
Age 11-maybe probably not
Age 12-maybe probably not
Age 13-maybe probably not
Age 14-kinda
Age 15-kinda
Age 16-starting to know more
Age 17-starting to know more
Age 18-pretty sure
Age 19-pretty sure
Age 20-yes
Answer:
2H2S (g) + 3O2 (g) → 2H2O (l) + 2SO2 (g)
Calculate ΔH° from the given data. Is the reaction exothermic or endothermic?
ΔH°f (H2S) = -20.15 kJ/mol; ΔH°f (O2) = 0 kJ/, mol; ΔH°f (H2O) = -285.8 kJ/mol; ΔH°f (SO2) = -296.4 kJ/mol
