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Steps to Solve

Here's a breakdown of the solutions to the physics questions provided.

32. Dot product and angle The dot product of two vectors $A$ and $B$ is given by $|A||B|\cos(\theta)$, where $\theta$ is the angle between the vectors. If the dot product is zero, then $\cos(\theta) = 0$. This occurs when $\theta = 90^\circ$.

33. Formula: $A \cdot B = |A||B|\cos(\theta)$

34. Set dot product to zero: $0 = |A||B|\cos(\theta)$

35. Solve for $\theta$ when $\cos(\theta) = 0$: $\theta = 90^\circ$

36. Angle with x-axis Given $A_x = 5.6$ and $A_y = -4.7$, we can find the angle $\theta$ using the arctangent function. Note that since $A_x$ is positive and $A_y$ is negative, $\theta$ is in the fourth quadrant.

37. Calculate the angle: $\theta = \arctan(\frac{A_y}{A_x}) = \arctan(\frac{-4.7}{5.6})$

38. Find arctan value* $\theta = \arctan(-0.839) \approx -40^\circ$

39. Convert to positive angle since it is in 4th quadrant: Add 360 degrees

40. Resulting angle: $\theta = -40 + 360 = 320^\circ$

41. Magnitude of sum of vectors Given $A_x = 5.1$, $B_x = -2.6$, $A_y = -5$, and $B_y = -4.3$, we first find the components of the sum $A + B$. Then, we calculate the magnitude of the resulting vector.

42. Find the x component of the resulting vector: $(A + B)_x = A_x + B_x = 5.1 + (-2.6) = 2.5$

43. Find the y component of the resulting vector: $(A + B)_y = A_y + B_y = -5 + (-4.3) = -9.3$

44. Calculate the magnitude of the resulting vector: $|A + B| = \sqrt{(2.5)^2 + (-9.3)^2} = \sqrt{6.25 + 86.49} = \sqrt{92.74} \approx 9.6$

45. Dot product of two vectors $A$ and $B$ Given $A = -3i + 6j - 5k$ and $B = -2i + 3j + k$, the dot product is calculated as follows:

46. Dot product formula: $A \cdot B = A_x B_x + A_y B_y + A_z B_z$

47. Calculation: $A \cdot B = (-3)(-2) + (6)(3) + (-5)(1) = 6 + 18 - 5 = 19$

48. Difference between vectors Given $A = -3i + 6j - 5k$ and $B = -2i + 3j + k$, the difference $A - B$ is:

49. Subtract vectors: $A - B = (-3 - (-2))i + (6 - 3)j + (-5 - 1)k$

50. Simplify: $A - B = (-3 + 2)i + (6 - 3)j + (-5 - 1)k = -1i + 3j - 6k = -i + 3j - 6k$

51. Magnitude of the Sum Given $A = -3i + 6j - 5k$ and $B = -2i + 3j + k$, the sum $A + B$ is:

52. Add vectors: $A + B = (-3 + (-2))i + (6 + 3)j + (-5 + 1)k$

53. Simplify: $A + B = (-5)i + (9)j + (-4)k = -5i + 9j - 4k$

54. Find magnitude: $|A + B| = \sqrt{(-5)^2 + (9)^2 + (-4)^2} = \sqrt{25 + 81 + 16} = \sqrt{122} \approx 11.0$

55. Torque and angular velocity When the torque on a rotating body decreases, its angular velocity decreases. Torque is directly proportional to angular acceleration, and angular acceleration is the rate of change of angular velocity.

56. Area under acceleration-time graph The area under the acceleration-time graph represents the change in velocity in that time interval. This is because acceleration is the rate of change of velocity with respect to time.

57. Momentum of objects in opposite direction The momentum of two objects moving with the same speed but in opposite directions, upon collision, is zero, assuming they have the same mass. Before collision the total momentum is $p_1 + p_2 = mv - mv = 0$. After collision, due to conservation of momentum, the final momentum is also 0.