<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:
Explanation:
In order to solve this problem we need to make a free body diagram of the book and the forces that interact on it. In the picture below you can see the free body diagram with these forces.
The person holding the book is compressing it with his hands, thus exerting a couple of forces of equal magnitude and opposite direction with value F.
Now the key to solving this problem is to analyze the equilibrium condition (Newton's third law) on the x & y axes.
To find the weight of the book we simply multiply the mass of the book by gravity.
W = m*g
W = 1.3[kg] * 9.81[m/s^2]
W = 12.75 [N]
Answer:
d= 1.5 g/cm3
Explanation:
datos
m= 30g
v= 20cm3
d=?
formula
d= m / v
solución
d= 30g / 20cm3 = 1.5g/cm3
The pertinent equation here is F=ma. You haven't shared the mass of the box, so I will use M to represent that mass.
Then F = M(<span>2.3 m/s^2) (answer)</span>
You can mark, lift and clean a ball on the green, but it's a violation to do so when another ball is in motion, as your ball might influence the outcome of that stroke. You can also mark and clean your ball in some instances when it's off the green: cleaning it, for example, just to the point where you can identify it.
