Final answer:
The statement that a similarity transformation mapping one circle to another must include a rotation is false. Similarity transformations can be scaling or translation without requiring rotation. Other true/false statements provided also explore vector addition, the Pythagorean theorem, wave-particle duality, and angular momentum.
Step-by-step explanation:
The statement 'A similarity transformation that maps one circle onto another must include a rotation' is false. Similarity transformations include scaling (dilation), translation, and/or rotation. To map one circle onto another, only a dilation is necessary if the circles are concentric. A translation would be required if the circles are not at the same position. A rotation is not a mandatory component unless the circles need to be aligned in a particular orientation.
Now let's consider some of the other true/false statements provided:
- 60. True - A vector can form the shape of a right angle triangle with its x and y components.
- 36. True - We can use the Pythagorean theorem to calculate the length of the resultant vector obtained from the addition of two vectors which are at right angles to each other.
- 37. True - The direction of the resultant vector depends on both the magnitude and direction of added vectors.
- 26. False - wave-particle duality exists only at the quantum level, not for objects on the macroscopic scale.
- 41, 83, 84, GRASP CHECK - The statements about vectors and their resultant angles or magnitudes require specific information, such as magnitudes and directions of the vectors. Without full data, these cannot be determined as true or false in general.
- 19. c. They're equal - If both system A and system B have disks with the same mass and magnitude of angular velocity, the angular momentum for both systems would be equal, as angular momentum is a product of moment of inertia and angular velocity, and for similar mass and angular velocity, the larger radius in system B compensates for the opposite rotation direction of one disk.