Answer:
Explanation:
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A child walks due east on the deck of a ship at 2 miles per hour. the ship is moving north at a speed of 18 miles per hour. find
the speed and direction of the child relative to the surface of the water.
1 answer:
Refer to the figure shown below.
The velocity of the child and the velocity of the ship should be added vectorially to find the speed and direction of the child relative to the water surface.
The magnitude of the child's velocity is
v = √(2² + 18²) = 18.11 mph
The direction of the child's speed is
θ = tan⁻¹ (18/2) = tan⁻¹ 9 = 83.7° north of east or counterclockwise from the eastern direction.
Answer:
The magnitude is 18.1 mph.
The direction is 84° north of east.
The velocity of the child and the velocity of the ship should be added vectorially to find the speed and direction of the child relative to the water surface.
The magnitude of the child's velocity is
v = √(2² + 18²) = 18.11 mph
The direction of the child's speed is
θ = tan⁻¹ (18/2) = tan⁻¹ 9 = 83.7° north of east or counterclockwise from the eastern direction.
Answer:
The magnitude is 18.1 mph.
The direction is 84° north of east.
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Answer:
1.25 focal lengths
Explanation:
The lens equation states that:
where
f is the focal length
p is the object distance
q is the image distance
In this problem, the image is 4 times as far from the lens as is the object: this means that
If we substitute this into the lens equation and we rearrange it, we get
so, the object distance measured in focal lengths is
1.25 focal lenghts