Answer:
the distance from the location of the center of gravity to the location of the center o mass for this system is 1.13m
Explanation:
Given that
m₁=4.4kg
x₁=+1.1m
m₂=3.7kg
x₂=+0.80m
m₃=2.9kg
x₃=+1.6m
The position of the center of mass is
Xcm = [m₁x₁ +m₂x₂ +m₃x₃]/(m₁+m₂+m₃)
= [(4.40kg)(1.1 m)+(3.70 kg)(0.80 m)+(2.90 kg)(1.60 m)]/(4.4 kg + 3.70 kg+2.90 kg)
= 1.13 m
The position of the center of gravity is 1.13m
Therefore, the distance from the location of the center of gravity to the location of the center o mass for this system is 1.13m
Answer:
A quantity that has magnitude but no particular direction is described as scalar. A quantity that has magnitude and acts in a particular direction is described as vector.
Answer:
A)
B)
Explanation:
<u>Given:</u>
Length of the room 
Width of the room 
A) Let A be the area of the room
B)We will calculate uncertainty in each dimension
%uncertainty in length
%uncertainty in width =
The uncertainty in area will be sum of uncertainty in length and width
%uncertainty in Area= %uncertainty in length + %uncertainty in width
%uncertainty in Area
%uncertainty in Area=0.0106
Uncertainty in Area
There Area is
Answer;
-Physical model
A physical representation of a real object, such as a globe of the world, is a physical model.
Explanation;
-A physical model is a simplified material representation, usually on a reduced scale, of an object or phenomenon that needs to investigated.
-The model can be used to simulate the physical conditions involved (temperature, waves, speed etc.) and to predict the particular constraints of the situation.
Answer:
12 m/s
Explanation:
Using the continuity equation, which is an extension of the conservation of mass law
ρ₁A₁v₁ = ρ₂A₂v₂
where 1 and 2 indicate the conditions at two different points of flow, in this case, point 1 is any normal position in the pip and point 2 is the conditions at the restriction.
ρ = density of the fluid flowing; note that the density of the fluid flowing (water) is constant all through the fluid's flow
A₁ = Cross sectional Area of the pipe at point 1 = (πD₁²/4)
A₂ = Cross sectional Area of the pipe at the restriction = (πD₂²/4)
v₁ = velocity of the fluid flowing at point 1 = 3 m/s
v₂ = velocity of the fluid flowing at The restriction = ?
ρ₁A₁v₁ = ρ₂A₂v₂
Becomes
A₁v₁ = A₂v₂ (since ρ₁ = ρ₂)
(πD₁²/4) × 3 = (πD₂²/4) × v₂
3D₁² = D₂² × v₂
But
D₂ = (D₁/2)
And D₂² = (D₁²/4)
3D₁² = D₂² × v₂
3D₁² = (D₁²/4) × v₂
(D₁²/4) × v₂ = 3D₁²
v₂ = 4×3 = 12 m/s
