Answer:
The centripetal acceleration is 6.95 m/s²
Explanation:
Given;
angular displacement of the blade, θ = 90.08⁰
duration of motion of the blade, t = 0.4 s
radius of the circle moved by the blade, r = 0.45 m
The angular speed of the blade in radian is calculated as;
The centripetal acceleration is calculated as;
a = ω²r
a = (3.93)² x 0.45
a = 6.95 m/s²
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.