Answer:
mas of water displaced = 41.4 g
Explanation:
Weight in air = True weight = 45 g
Apparent weight = 3.6 g
Apparent weight = True weight - Buoyant force
Buoyant force = 45 g - 3.6 g = 41.4 g
Weight of water displaced = Buoyant force
Weight of water displaced = 41.4 g dyne
mas of water displaced = 41.4 g
Answer:
(a). The draw-down at a distance 200 m from the well after pumping for 50 hr is 5.383 m.
(b). The draw-down at a distance 200 m from the well after pumping for 50 hr is 6.707 m.
Explanation:
Given that,
Energy 
Transmissivity 
Storage coefficient 
Distance r= 200 m
We need to calculate the draw-down at a distance 200 m from the well after pumping for 50 hr
Using formula of draw-down
Put the value into the formula
We need to calculate the draw-down at a distance 200 m from the well after pumping for 200 hr
Using formula of draw-down
Put the value into the formula
Hence, (a). The draw-down at a distance 200 m from the well after pumping for 50 hr is 5.383 m.
(b). The draw-down at a distance 200 m from the well after pumping for 200 hr is 6.707 m.
Answer:
order d> a = e> c> b = f
Explanation:
Pascal's law states that a change in pressure is transmitted by a liquid, all points are transmitted regardless of the form
P₁ = P₂
Using the definition of pressure
F₁ / A₁ = F₂ / A₂
F₂ = A₂ /A₁ F₁
Now we can examine the results
a) F1 = 4.0 N A1 = 0.9 m2 A2 = 1.8 m2
F₂ = 1.8 / 0.9 4
F₂a = 8 N
b) F1 = 2.0 N A1 = 0.9 m2 A2 = 0.45 m2
F₂b = 0.45 / 0.9 2
F₂b = 1 N
c) F1 2.0 N A1 = 1.8 m2 A2 = 3.6 m2
F₂c = 3.6 / 1.8 2
F₂c = 4 N
d) F1 = 4.0N A1 = 0.45 m2 A2 = 1.8 m2
F₂d = 1.8 / 0.45 4.0
F₂d = 16 m2
e) F1 = 4.0 N A1 = 0.45 m2 A2 = 0.9 m2
F₂e = 0.9 / 0.45 4
F₂e = 8 N
f) F1 = 2.0N A1 = 1.8 m2 A2 = 0.9 m2
F₂f = 0.9 / 1.8 2.0
F₂f = 1 N
Let's classify the structure from highest to lowest
F₂d> F₂a = F₂e> F₂c> F₂b = F₂f
I mean the combinations are
d> a = e> c> b = f
