Rock faults are ruptures along which opposite faces of rock have slid past each other. In Fig. 3-35, points A and B coincided before the rock in the foreground slid down to the right. The net displacement A B → is along the plane of the fault. The horizontal component of A B → is the strike-slip AC. The component of A B → that is directed down the plane of the fault is the dip-slip AD. (a) What is the magnitude of the net displacement A B → if the strike-slip is 22.0 m and the dip-slip is 17.0 m? (b) If the plane of the fault is inclined at angle ϕ = 52.0° to the horizontal, what is the vertical component of A B → ? Figure 3-35 Problem 51.
Rock faults are ruptures along which opposite faces of rock have slid past each other. In Fig. 3-35, points A and B coincided before the rock in the foreground slid down to the right. The net displacement A B → is along the plane of the fault. The horizontal component of A B → is the strike-slip AC. The component of A B → that is directed down the plane of the fault is the dip-slip AD. (a) What is the magnitude of the net displacement A B → if the strike-slip is 22.0 m and the dip-slip is 17.0 m? (b) If the plane of the fault is inclined at angle ϕ = 52.0° to the horizontal, what is the vertical component of A B → ? Figure 3-35 Problem 51.
Rock faults are ruptures along which opposite faces of rock have slid past each other. In Fig. 3-35, points A and B coincided before the rock in the foreground slid down to the right. The net displacement
A
B
→
is along the plane of the fault. The horizontal component of
A
B
→
is the strike-slip AC. The component of
A
B
→
that is directed down the plane of the fault is the dip-slip AD. (a) What is the magnitude of the net displacement
A
B
→
if the strike-slip is 22.0 m and the dip-slip is 17.0 m? (b) If the plane of the fault is inclined at angle ϕ = 52.0° to the horizontal, what is the vertical component of
A
B
→
?
determine the magnitude and orientation theta of FB so that the resultant force is directed along the positive y axis and has a magnitude of 1500N.
SOLVE FOR THE RESUL TANT OF A = 2.5 N, 16° NORTH OF EMST
AND B-ION , 40° EAST OF NORTH USING LAW OF SINE AND
COSINE. YOU CAN VERIFY YOuR ANSWER USING THE
COMPONENT METHOD.
Given vector A = xz ax -xy ay +yz az
Transform A into cylindrical coordinate and it's magnitude at point (1,30o,2) is equal to:
Human Biology: Concepts and Current Issues (8th Edition)
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