<span>Basically just divide the change in position by the change in time. </span>
<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:
s = 23.72 m
v = 21.56 m/s²
Explanation:
given
time to reach the ground (t) = 2.2 second
we know that
a) s = u t + 0.5 g t²
u = 0 m/s
g = 9.8 m/s²
s = 0 + 0.5 × 9.8 × 2.2²
s = 23.72 m
b) impact velocity
v = √(2gh)
v = √(2× 9.8 × 23.72)
v = √464.912
v = 21.56 m/s²
Answer:
what r the questions i can’t see them
Explanation:
