Imagine two people standing at placemark A and placemark E, looking at each other across the fault. Which of the following state
ments correctly describes the relative motion of the land opposite each observer?
a. To both observers, the land opposite them is moving to the right.
b. To both observers, the land opposite them is moving to the left.
c. To the observer at placemark A, the land at placemark Eis moving to the left
d. To the observer at placemark E, the land at placemark A is moving to the right.
a. To both observers, the land opposite them is moving to the right.
b. To both observers, the land opposite them is moving to the left.
c. To the observer at placemark A, the land at placemark Eis moving to the left
d. To the observer at placemark E, the land at placemark A is moving to the right.
1 answer:
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(1) A - B
(2) B - C
(3) - A + B - C
(4) 3A - 2C
(5) - 2A + 3B - C
(6) 2A - 3 (B - C)
Answer:
(1) (3,-5,-4)
(2) (-5, 4, 0)
(3) (-6, 4, 3)
(4) (-3, -2, -11)
(5) (-11, 14, 8)
(6) (17, -12, -6)
Explanation:
A⃗ =(1,0,−3)
B⃗ =(−2,5,1)
C⃗ =(3,1,1)
Vector additions and subtraction are done on a component by component basis, that is, only data from component î can be added to or subtracted from another Vector's component î. And so on for components j and k.
1) (A - B) = (1,0,−3) - (−2,5,1) = (1-(-2), 0-5, -3-1) = (3,-5,-4)
2) (B - C) = (−2,5,1) - (3,1,1) = (-2-3, 5-1, 1-1) = (-5, 4, 0)
3) -A + B - C = -(1,0,−3) + (−2,5,1) - (3,1,1) = (-1-2-3, 0+5-1, 3+1-1) = (-6, 4, 3)
4) 3A - 2C = 3(1,0,−3) - 2(3,1,1) = (3,0,-9) - (6,2,2) = (3-6, 0-2, -9-2) = (-3, -2, -11)
5) -2A + 3B - C = -2(1,0,−3) + 3(−2,5,1) - (3,1,1) = (-2,0,6) + (-6,15,3) - (3,1,1) = (-2-6-3, 0+15-1, 6+3-1) = (-11, 14, 8)
6) 2A - 3 (B - C) = 2(1,0,−3) - 3[(−2,5,1) - (3,1,1)] = (2,0,-6) - 3(-5,4,0) = (2+15, 0-12, -6-0) = (17, -12, -6)
Answer:
44 N/m
Explanation:
The extension, e, of the spring = 2.9 m - 1.4 m = 1.5 m
The work needed to stretch a spring by <em>e</em> is given by
where <em>k</em> is spring constant.
Using the appropriate values,
The force of friction is 19.1 N
Explanation:
According to Newton's second law, the net force acting on the bag is equal to the product between its mass and its acceleration:
where
is the net force
m is the mass
a is the acceleration
The bag is moving at constant speed, so its acceleration is zero:
Therefore the net force is zero as well:
Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:
where
is the horizontal component of the applied force, with
F = 22.5 N
is the force of friction
And solving for , we find
Learn more about friction:
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