<span>Assume: neglect of the collar dimensions.
Ď_h=(P*r)/t=(5*125)/8=78.125 MPa ,Ď_a=Ď_h/2=39 MPa
τ=(S*Q)/(I*b)=(40*〖10〗^3*π(〖0.125〗^2-〖0.117〗^2 )*121*〖10〗^(-3))/(π/2 (〖0.125〗^4-〖0.117〗^4 )*8*〖10〗^(-3) )=41.277 MPa
@ Point K:
Ď_z=(+M*c)/I=(40*0.6*121*〖10〗^(-3))/(8.914*〖10〗^(-5) )=32.6 MPa
Using Mohr Circle:
Ď_max=(Ď_h+Ď_a)/2+âš(Ď„^2+((Ď_h-Ď_a)/2)^2 )
Ď_max=104.2 MPa, Ď„_max=45.62 MPa</span>
Answer:
how you can convert electrical energy to mechanical energy
Explanation:
In the Electric Motor lab, what did you attempt to demonstrate with the electric motor?
In the Electric Motor lab, the experiment usually demonstrate energy conversion process, from electrical energy to mechanical energy. This mechanical energy can also be converted back to electrical energy using a generator.
Therefore, what you attempted to demonstrate with the electric motor, is how you can convert electrical energy to mechanical energy
Answer:
an atom can not be broken into smaller pieces
Explanation:
Answer:
a) 
Explanation:
a) the cross-sectional area of the hose would be the square of radius times pi. And since the sectional radius is half of its diameter d. We can express the cross-sectional area A1 in term of diameter d1
Answer:
The mass of unknown object is 8.62Kg
Explanation:
To develop this problem it is necessary to apply the equations related to the Drag force and the Force of Gravity.
For the given point, that is, the moment at which the terminal velocity is reached, the two forces equalize, that is,
By definition we know that the Drag force is defined as
Where,
Drag coefficient
Density
A =Cross-sectional Area
V = Velocity
In the other hand we have,
Where,
Mass of sphere
Mass of unknown object
Equating the two equations we have to
Re-arrange for m_2,
Our values are given by,
Replacing in the equation we have,
<em>Therefore the mass of unknown object is 8.62Kg</em>
