answer:**Analysis** This problem examines the knowledge of the diagonal formula of a rectangular solid and the calculation of the surface area of a sphere, which are basic topics. **Solution** Given that PA, PB, and PC are equal to each other and the base is an equilateral triangle, the tetrahedron P-ABC is a regular tetrahedron. The projection of P on the base plane is the center O of the base equilateral triangle. Since BC perp PO, BC perp AO, it follows that BC perp plane PAO, which implies BC perp PA. Also, given PA perp PB, it is known that PA perp plane PBC, so it can be concluded that PA perp PC, meaning PA, PB, and PC are mutually perpendicular. Since it is a circumscribed sphere of the tetrahedron, the regular tetrahedron P-ABC can be considered as a corner cut from a cube that shares the same circumscribed sphere with the original cube. Therefore, 4R^2 = 3 times (3^2) = 27, and V = frac{4pi R^3}{3} = frac{27sqrt{3}pi}{2}. Thus, the answer is boxed{frac{27sqrt{3}pi}{2}}.

answer:Solution: The radian measure equivalent to 60^circ is pi cdot frac{60^circ}{180^circ} = frac{pi}{3}. Therefore, the correct option is: boxed{D}. Based on the fact that pi radians equal 180^circ, we find the radian value for 60^circ. This question primarily tests the method of converting degrees to radians and is considered a basic question.

answer:(1) Since the set A={x|3leq x<7}, then complement_R A={x|x<3 text{ or } xgeq7}, Since B={x|2<x<10}, then (complement_R A) cap B={x|2<x<3 text{ or } 7leq x<10}. (2) Since the non-empty set B={x|x=log_{2}m}, if B subseteq {1,2}, then log_{2}m=1 or log_{2}m=2, thus m=2^1 or m=2^2. Therefore, the values of the real number m are boxed{2 text{ or } 4}.

answer:We start with the number 3 displayed on the calculator. Each time the x^2 key is pressed, the displayed number is replaced by its square. We need to determine how many times we must press the key until the displayed number exceeds 1000. 1. **Initial Display**: 3 2. **First Press**: 3^2 = 9 3. **Second Press**: 9^2 = 81 4. **Third Press**: 81^2 = 6561 After the third press, the displayed number is 6561, which is greater than 1000. Therefore, we need to press the x^2 key **three times** to produce a displayed number greater than 1000. The problem makes sense because starting from 3 and applying the squaring operation repeatedly yields numbers that are squared each iteration, and the iterations achieve a result beyond 1000 before counting became unfeasible or ambiguous. Hence, the answer is text{C}. The final answer is boxed{C) 3}