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
→
?
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.
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.
Replace the polar equation r= 8 cos 0 + 4 sin 0 with an equivalent Cartesian equation. Then identify the graph.
The equivalent Cartesian equation is
(Type an equation using x and y as the variables.)
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.