Answer:
2.23 × 10^6 g of F- must be added to the cylindrical reservoir in order to obtain a drinking water with a concentration of 0.8ppm of F-
Explanation:
Here are the steps of how to arrive at the answer:
The volume of a cylinder = ((pi)D²/4) × H
Where D = diameter of the cylindrical reservoir = 2.02 × 10^2m
H = Height of the reservoir = 87.32m
Therefore volume of cylindrical reservoir = (3.142×202²/4)m² × 87.32m = 2798740.647m³
1ppm = 1g/m³
0.8ppm = 0.8 × 1g/m³
= 0.8g/m³
Therefore to obtain drinking water of concentration 0.8g/m³ in a reservoir of volume 2798740.647m³, F- of mass = 0.8g/m³ × 2798740.647m³ = 2.23 × 10^6 g must be added to the tank.
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You need to add a picture or answers!
Answer:
3: I can´t see the text/image, but it depend on the mass and the force applied to the ball, if both are too high, it will be harder to make a home run. (Second law)
4:It would be easier to make a home run because there is no interruption between the ball and the space the same travels. (Third law)
Explanation:
Answer:
ΔK = 24 joules.
Explanation:
Δ
Work done on the object
Work is equal to the dot product of force supplied and the displacement of the object.
* Δ
Δ can be found by subtracting the vectors (7.0, -8.0) and (11.0, -5.0), which is written as Δ = (11.0 - 7.0, -5.0 - -8.0) which equals (4.0, 3.0).
This gives us
* 
= 
= 
J
Answer:
B
endothermic: heat taking in
exothermic: heat given out
