All planets revolve around the sun counterclockwise,and most also rotate on their axis counterclockwise true or false
1 answer:
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Answer:
<em>mass of the ice is 254980463.8T kg</em>
<em>where T is the value of the thickness omitted in the question.</em>
Explanation:
The ice on Walden Pond is .......... thick. The area of the pond is approximately equal to the area of a circle with radius 297 m. Find the mass of the ice. Answer in kg.
<em>The value of the thickness of the ice T is omitted, but I will show the solution, and the real answer can be gotten by multiplying the final calculated answer here by the thickness of the ice omitted.</em>
Given the radius of the equivalent circle of the ice = 297 m'
the area of the ice can be gotten from area A = =
= 277152.678 m^2
recall that the density of ice p ≅ 920 kg/m^3
also,
density of ice p = (mass of ice, m) ÷ (volume of ice, v)
i.e p = m/v
and,
m = pv
substituting the value of the density of water p into the equation, we have,
mass of the ice, m = 920v ....... equ 1
The volume of the ice above will be = (area of the ice, A) x (thickness of the ice, T)
i.e v = AT
substituting the value of area A into the equation, we have
v = 277152.678T ......equ 2
substitute value of v into equ 1
mass of the ice, m = 920 x (277152.678T)
mass of the ice, m = 254980463.8T kg
where T is the thickness of the ice
NB: To get the mass, multiply this answer with the thickness T given in the question.
Answer:
v₂ = 15.24 m / s
Explanation:
This is an exercise in fluid mechanics
Let's write Bernoulli's equation, where the subscript 1 is for the factory pipe and the subscript 2 is for the tank.
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
They indicate the pressure in the factory P₁ = 140000 Pa, the velocity
v₁ = 5.5 m / s and the initial height is zero y₁ = 0
the tank is at a pressure of P2 = 2000 Pa and a height of y₂ = 6.0 m
P₁ -P₂ + ρ g (y₁ -y₂) + ½ ρ v₁² = ½ ρ v₂²
let's calculate
140,000 - 2000 + ρ 9.8 (0- 6) + ½ ρ 5.5² = ½ ρ v₂²
138000 - ρ 58.8 + ρ 15.125 = ½ ρ v2²
v₂² = 2 (138000 /ρ - 58.8 + 15.125)
v₂ =
In the exercise they do not indicate what type of liquid is being used, suppose it is water with
ρ = 1000 kg / m³
v₂ =
v₂ = 15.24 m / s