Answer:
B
Explanation:
Friction acts in a direction opposite to the motion of an object
Coastal areas experience two low tides and two high tides every lunar day, or 24 hours and 50 minutes. The two tidal bulges caused by inertia and gravity will rotate around the Earth as the moons position changes. These bulges represent high tides while the flat sides indicate low tides.
Over the course of 1 day, the position of the moon does not change very much compared to the rotation of the earth. As the earth rotates below the moon, one point on the earth will go through all levels of tide as the day passes by. Strongly attracted, middling, weakly attracted, and then middleagain. From our perspective it looks like "high, low, high, low." Or equivalently you can think about how the points below the moon and opposite the moon will be high tide, and as the earth rotates, those areas will change.
Answer:
If the object is moving with an acceleration of +4 m/s/s (i.e., changing its velocity by 4 m/s per second), then the slope of the line will be +4 m/s/s.
Explanation:
<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>
