answer:1. **Divide the ring into small elements**: - For a uniformly charged ring with radius ( r = 0.1 , text{m} ), divide it into small differential elements with charge ( Delta Q ). - The total charge on the ring is given by ( Q = sum Delta Q ). 2. **Electric field due to a differential element**: - Consider the contribution of a differential charge element ( Delta Q ) to the electric field at a point on the symmetry axis at a distance ( x ) from the center. - The magnitude of the electric field due to ( Delta Q ) is: [ |Delta E| = k frac{Delta Q}{x^2 + r^2} ] - The component of this field in the direction of the axis (assuming symmetry ensures all other components cancel out) involves ( cos alpha ) where ( cos alpha = frac{x}{sqrt{x^2 + r^2}} ). 3. **Contribution to the axial component**: - Therefore, the contribution to the axial component of the electric field by ( Delta Q ) is: [ Delta E_x = k Delta Q frac{x}{(x^2 + r^2)^{3/2}} ] 4. **Total electric field**: - Summing over all elements, the total electric field at distance ( x ) from the center is: [ E(x) = sum Delta E_x = k frac{x}{(x^2 + r^2)^{3/2}} sum Delta Q = k Q frac{x}{(x^2 + r^2)^{3/2}} ] 5. **Maximizing the electric field**: - Find the value of ( x ) at which the electric field ( E(x) ) is maximized by differentiating ( frac{x}{(x^2 + r^2)^{3/2}} ) and setting it to zero: [ frac{d}{dx} left( frac{x}{(x^2 + r^2)^{3/2}} right) = 0 ] - Differentiating: [ frac{(x^2 + r^2)^{3/2} - x cdot frac{3}{2}(x^2 + r^2)^{1/2} cdot 2x}{(x^2 + r^2)^3} = 0 ] Simplifying: [ (x^2 + r^2)^{3/2} = 3x^2(x^2 + r^2)^{1/2} ] [ (x^2 + r^2) = 3x^2 ] [ r^2 = 2x^2 ] [ x = frac{r}{sqrt{2}} ] - Thus, the maximum is at ( x_0 = frac{r}{sqrt{2}} ). 6. **Calculating the total charge**: - Given that the maximum electric field ( E_{max} = 8 times 10^4 sqrt{3} , text{V/m} ) occurs at ( x_0 = frac{r}{sqrt{2}} ): [ Eleft(frac{r}{sqrt{2}} right) = k Q frac{frac{r}{sqrt{2}}}{left( left(frac{r}{sqrt{2}}right)^2 + r^2 right)^{3/2}} ] [ Eleft(frac{r}{sqrt{2}} right) = k Q frac{frac{r}{sqrt{2}}}{(2r^2/2)^{3/2}} ] [ 8 times 10^4 sqrt{3} = k Q frac{1}{r^2} ] - Using ( k = frac{1}{4 pi epsilon_0} approx 9 times 10^9 ): [ 8 times 10^4 sqrt{3} = 9 times 10^9 Q frac{1}{(0.1)^2} ] [ 8 times 10^4 sqrt{3} = 9 times 10^9 Q times 100 ] [ Q = frac{8 times 10^4 sqrt{3}}{9 times 10^9 times 100} ] [ Q = frac{8 sqrt{3}}{9 times 10^7} ] [ Q = 4 times 10^{-7} , text{C} ] 7. **Electric field at ( x = 0.2 , text{m} )**: - Now calculate the electric field at ( x_1 = 0.2 , text{m} ): [ E(0.2) = k Q frac{0.2}{(0.2^2 + 0.1^2)^{3/2}} ] [ E(0.2) = 9 times 10^9 cdot 4 times 10^{-7} frac{0.2}{(0.04 + 0.01)^{3/2}} ] [ E(0.2) = 3.6 times 10^3 frac{0.2}{0.05^{3/2}} ] [ E(0.2) = 3.6 times 10^3 frac{0.2}{0.01118} ] [ E(0.2) = 3.6 times 10^3 times 17.888 ] [ E(0.2) = 6.44 times 10^4 , text{V/m} ] # Conclusion: [ boxed{Q = 4 times 10^{-7} , text{C}, quad E(0.2 , text{m}) = 6.44 times 10^4 , text{V/m}} ]

answer:Let's calculate the time Max spent on each subject step by step. 1. Biology: 24 minutes 2. History: 1.5 times more than biology History = 24 minutes * 1.5 = 36 minutes 3. Chemistry: 30% less time than biology Chemistry = 24 minutes - (24 minutes * 0.30) = 24 minutes - 7.2 minutes = 16.8 minutes 4. English: Twice as much time as history and chemistry combined English = 2 * (History + Chemistry) English = 2 * (36 minutes + 16.8 minutes) = 2 * 52.8 minutes = 105.6 minutes 5. Geography: Three times more than history and 75% more time than English Geography = 3 * History + (75% of English) Geography = 3 * 36 minutes + (0.75 * 105.6 minutes) Geography = 108 minutes + 79.2 minutes = 187.2 minutes Now, let's add up the time spent on all subjects to find the total time Max spent on his homework. Total time = Biology + History + Chemistry + English + Geography Total time = 24 minutes + 36 minutes + 16.8 minutes + 105.6 minutes + 187.2 minutes Total time = 369.6 minutes Max spent a total of boxed{369.6} minutes on his homework.

answer:1. **Understanding the Problem:** We have a cube with side length 100 cm. Point P starts at vertex A and moves towards vertex C_1 at a constant speed of 3 cm/s. Point Q starts at the same time from vertex A_1 and moves towards vertex B_1 at a constant speed of 2 cm/s. We need to find the minimum distance between P and Q during their motion, precise to 0.0001 cm. 2. **Coordinate System Setup:** First, set up a coordinate system for the cube: - Let A = (0, 0, 0). - The vertices of the cube have coordinates given by: - B = (100, 0, 0), - C = (100, 100, 0), - D = (0, 100, 0), - A_1 = (0, 0, 100), - B_1 = (100, 0, 100), - C_1 = (100, 100, 100), - D_1 = (0, 100, 100). 3. **Parameterize the Motion of Points:** Let P be moving along the line AC_1 which is the space diagonal: - ( A = (0, 0, 0) ) - ( C_1 = (100, 100, 100) ) The position of P as a function of time t (in seconds) is: [ mathbf{P}(t) = (3t, 3t, 3t) ] since it travels along (100, 100, 100) at 3 cm/s. Let Q be moving along the edge A_1B_1: - ( A_1 = (0, 0, 100) ) - ( B_1 = (100, 0, 100) ) The position of Q as a function of time t is: [ mathbf{Q}(t) = (2t, 0, 100) ] since it travels along (100, 0, 100) at 2 cm/s. 4. **Distance Between P and Q:** The distance d(t) between points P and Q is given by: [ d(t) = sqrt{(3t - 2t)^2 + (3t - 0)^2 + (3t - 100)^2} ] Simplify the expression inside the square root: [ d(t) = sqrt{(t)^2 + (3t)^2 + (3t - 100)^2} ] [ d(t) = sqrt{t^2 + 9t^2 + (3t - 100)^2} ] [ d(t) = sqrt{t^2 + 9t^2 + 9t^2 - 600t + 10000} ] [ d(t) = sqrt{19t^2 - 600t + 10000} ] 5. **Minimizing the Distance:** To find the minimum distance, take the derivative of d(t): [ d'(t) = frac{d}{dt}left(sqrt{19t^2 - 600t + 10000}right) = frac{1}{2sqrt{19t^2 - 600t + 10000}} cdot (38t - 600) ] Set the derivative equal to zero to find the critical points: [ 0 = frac{1}{2sqrt{19t^2 - 600t + 10000}} cdot (38t - 600) ] Solve for t: [ 38t - 600 = 0 ] [ t = frac{600}{38} = frac{300}{19} approx 15.7895 text{ seconds} ] 6. **Substitute t Back Into d(t):** Calculate d(t) at t approx 15.7895: [ dleft( frac{300}{19} right) = sqrt{19left(frac{300}{19}right)^2 - 600left(frac{300}{19}right) + 10000} ] [ = sqrt{19 cdot frac{90000}{361} - frac{180000}{19} + 10000} ] [ = sqrt{frac{1710000}{361} - frac{3420000}{361} + 10000} ] [ = sqrt{frac{1710000 - 3420000 + 3610000}{361}} ] [ = sqrt{frac{1900000}{361}} ] [ = frac{sqrt{1900000}}{19} ] [ = frac{100sqrt{190}}{19} ] [ approx frac{817}{2450} , text{cm} ] 7. **Final Answer:** Therefore, the minimum distance between P and Q during their motion is (boxed{frac{817}{2450}}).

answer:For option A, y= dfrac {x^{2}-1}{x-1}=x+1(xneq 1), its domain is different from y=x+1(x∈R), so they are not the same function. For option B, y=lg x(x > 0), its domain is different from y= dfrac {1}{2}lg x^{2}=lg |x|(xneq 0), and their corresponding relationships are also different, so they are not the same function. For option C, y= sqrt {x^{2}}-1=x-1(xgeqslant 0), its domain is different from y=x-1(x∈R), so they are not the same function. For option D, y=x(x∈R), its domain is the same as y=log _{a}a^{x}=x(x∈R), and their corresponding relationships are also the same, so they are the same function. Therefore, the answer is boxed{D}. The solution involves determining whether two functions are the same by checking if their domains and corresponding relationships are the same. This is a basic problem.