Find the edges of the three cubes. The diameter of the base and the height of the cone are 6 cm and 4 cm, respectively. Earth taken out of it has spread evenly all around a width of 3 m it to form an embankment. Calculate the concentration of the dilute solution, c, n(solute) = c(undiluted solution) × V(pipette in L) mol, c(dilue solution) = n(solute) ÷ V(dilute solution in L), moles solute transferred by pipette to volumetric flask, moles solute in volumetric flask after solvent added (dilute solution), The concentration of the second solution is 0.019 mol L. Dilution means to reduce the concentration of a solution. This equation (formula or expression) is very useful because we can rearrange it find: Calculate the new concentration in mol L-1 (molarity) if enough water is added to 100.00 mL of 0.25 mol L-1 sodium chloride solution to make up 1.5 L. Calculate the volume in litres to which 500.00 mL of 0.020 mol L-1 copper sulfate solution must be diluted to make a new solution with a concentration of 0.0010 mol L-1. Therefore, the diameter of the cylindrical vessel is 42 cm. 1. Also, given that the copper sphere is melted and recasted into a right circular cone, Volume of the right circular cone = 1/3 π r2h. Thus, the edges of the three cubes are 6 cm, 8 cm and 10 cm respectively. Stress and anxiety are common experiences for most people. Step 2: Calculate concentration of the stock (undiluted) solution, cstock = n(Na2CO3) ÷ Vstock solution in L, Step 3: Calculate the moles of Na2CO3 in the pipette which is transferred to the second volumetric flask for dilution, n(Na2CO3 pipette) = cstock × Vpipette in L. Step 4: Calculate concentration of dilute solution of Na2CO3 in second volumetric flask. A solution can be diluted by adding more solvent to the stock solution (the starting solution before dilution) in the same vessel. The height and base radius of the cone are also the same. Keep up the great writing. Cancelling out the common part from both sides of the equation we get, Thus, the diameter of the third ball is 5 cm. A vessel in the form of a hollow hemisphere mounted by a hollow cylinder. The radius of the common base is 3.5 cm and the height of the cylindrical and conical portions are 10 cm and 6 cm, respectively. All objects in database are associated to dbo schema, and there is only one login ‘sa’. V1 = volume of stock solution present before dilution in L It is melted and recast into a cone of height 28 cm. It is full of water. Pinal Dave is a SQL Server Performance Tuning Expert and an independent consultant. concentration of a stock solution in mol L-1 = moles of solute ÷ volume of solution in litres, c1 = concentration of stock solution (before dilution) in mol L-1 But now when I disconnect and try to connect using the SA login, it throws the following exception: Login failed for user ‘sa’. The radii of circular bases of a frustum of a right circular cone are 12 cm and 3 cm and the height is 12 cm. What problem is that ??? So, on equating equations (i) and (ii), we get, π × 12 × 14 = π × [(1 + x)2 – (1)2] x 0.4. Eric, I tried running SQL Server Mgmt Studio as an Administrator and got enabled the SA login perfectly. Total height of the vessel = 13 cm = h + r, Inner surface area of the vessel = 2 π r (h + r), Therefore, the inner surface area of the vessel is 572 cm2. Once you learn my business secrets, you will fix the majority of problems in the future. Please let me know how I can resolve this issue or what alteranate I can use. This bucket is emptied on the ground and a conical heap of sand is formed. 24. Let the edges of three cubes (in cm) be 3x, 4x and 5x respectively. Radius of the circular ends of the bucket 5 cm and 15 cm, Let ‘L’ be the slant height of the bucket, = π(5 + 15)26 + π52 = π(520 + 25) = 545π cm2, Therefore, the curved surface area of the bucket = 545π cm2. Therefore, 512 balls can be made of radius 1 cm each with a solid sphere of radius 8 cm. I installed SQL Server 2005 Express Edition, but in my VIEW options, i can not find out where is Solution Explorer and Toolbox windows, help me pls. Note: V2 > V1 and c2 < c1, c1 = concentration of stock solution (before dilution) in mol L-1 The, the radius of the small ball = x/4 cm, Volume of n small balls = Volume of the big ball, Therefore, the number of small balls = 64, Surface area of all small balls/ surface area of big ball = 64 x 4π(x/4)2/ 4π(x)2, Thus, the ratio of the surface area of the small balls to that of the original ball is 4:1. And, the radius of the wire = 2 mm = 0.2 cm, Therefore, the length of the wire = 12150 cm. 23. A cylindrical bucket, 32 cm high and 18 cm of radius of the base, is filled with sand. 3. V2 = total volume of the new dilute solution in L Using the StoPGoPS model for problem solving: What chemical principle will you need to apply? Find the length of pipe. 20. Therefore, the width of the embankment is 5 m. 31. Radius of each circular disc = r = 1.5/2 = 0.75 cm, Height of each circular disc = h = 0.2 cm, Radius of cylinder = R = 4.5/ 2 = 2.25 cm, So, the number of metallic discs required is given by n, n = Volume of cylinder / volume of each circular disc. when i login and connect to object explorer and can see the databases and I it work find. Calculate the height of the cone? 13. Also, find the cost of tin sheet used for making the bucket at the rate of Rs 1.20 per dm2. She primarily focuses on the database domain, helping clients build short and long term multi-channel campaigns to drive leads for their sales pipeline. 2.2 dm3 of brass is to be drawn into a cylindrical wire of Diameter = 0.25 cm, Volume of cylindrical wire = Volume of brass of 2.2 dm3, Therefore, the length of the cylindrical wire drawn is 448 m. 5. Internal diameter of hollow sphere = 4 cm, So, the internal radius of hollow sphere = 2 cm, External diameter of hollow sphere = 8 cm, So, the external radius of hollow sphere = 4 cm, Volume of the hollow sphere 4/3 π × (43 – 23) … (i), Volume of the cone 1/3 π × 42 × h ….. (ii), As the volume of the hollow sphere and cone are equal. Users member of systemadmin group are can only change the rights of SA user. Therefore, the radius and slant height of the conical heap are 36 cm and 43.26 cm respectively. And the question is solved. Therefore, 450 metallic discs are required. QuickMath allows students to get instant solutions to all kinds of math problems, from algebra and equation solving right through to calculus and matrices. Find the total surface area of the solid. The curved surface area of the cone (S1) = πrL, where L is the slant height of the cone. On comparing equation (i) and (ii) we have. Once you do this and then attempt to enable the SA login, it should work. Please enable javascript and pop-ups to view all page content. So, the radius of the pond = 40/2 = 20 m = r. As the whole view of the pond looks like a hollow cylinder. Please let me know how I can resolve this issue or what alteranate I can use. A solution can be diluted by adding solvent to a given volume of stock solution. Then, Volume of cuboid = (53 x 40 x 15) cm3, External radius of the pipe = 8/2 = 4 cm = R, So, the volume of iron in the pipe = (External Volume) – (Internal Volume). is there a way that i can add solution explorer to the view under the view tab in management studio 2005 express? The diameter of a metallic sphere is equal to 9 cm. And, its volume = 1/3 πr2h = 1/3 π(3.5)2(3), The number of cones = Volume of metallic sphere/ Volume of each cone. (adsbygoogle = window.adsbygoogle || []).push({}); Want chemistry games, drills, tests and more? 22. Subscribe to RSS headline updates from: Powered by FeedBurner, substance which enables the solute to dissolve, determine concentration of dilute solution, determine concentration of stock solution before dilution, determine volume of stock solution used (before dilution), determine concentration of stock solution. Surface area of the solid metallic sphere = 616 cm3, Radius of the solid metallic sphere = 7 cm, Let’s assume r to be the radius of the cone, On equating equations (i) and (ii), we have, Therefore, the diameter of the base of the cone is 14 cm. 27. Let us assume the well to be a solid right circular cylinder, Radius(r) of the cylinder = 3.5/2 m = 1.75 m, Depth of the well or height of the cylinder (h) = 16 m, Now, let the height of the platform be x m, As the earth dug up is spread evenly to form the platform. Find the whole surface and volume of the remaining Cylinder. Depth of the cylindrical vessel = Height of the cylindrical vessel = h = 42 cm (common for both), Inner diameter of the cylindrical vessel = 14 cm, So, the inner radius of the cylindrical vessel = r1 = 14/2 = 7 cm, Outer diameter of the cylindrical vessel = 16 cm, So, the outer radius of the cylindrical vessel = r2 = 16/2 = 8 cm. Height of the cylindrical part (h) = 10 cm, Next, the curved surface area of the cylinder (S3) = 2πrh, Thus, the total surface area (S) = S1 + S2 + S3, Therefore, the total surface area of the solid is 373.38 cm2. A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the area of its whole surface and volume. SQL Server Performance Tuning Practical Workshop is my MOST popular training with no PowerPoint presentations and 100% practical demonstrations. Therefore, the length of the solid cylinder is 79.04 cm. Then, the height of small cone will be = (20 – h) cm, Volume of small cone = 1/125 x volume of big cone. Compare the surface area, of all the smaller balls combined together with that of the original ball. Internal diameter of the hollow sphere = 6 cm, So, the internal radius of the hollow sphere = 6/2 cm = 3 cm = r, External diameter of the hollow sphere = 10 cm, So, the external radius of the hollow sphere = 10/2 cm = 5 cm = R, Volume of the hollow spherical shell = 4/3 π × (R3 – r3), And given, the length of the solid cylinder = 8/3 cm, Let the radius of the solid cylinder be r cm, Therefore, the diameter of the cylinder is 14 cm. He has authored 12 SQL Server database books, 35 Pluralsight courses and has written over 5200 articles on the database technology on his blog at a https://blog.sqlauthority.com. One such application is to find the surface areas and volumes of different solids and their combinations. A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. Internal diameter of hollow spherical shell = 6 cm, So, the internal radius of hollow spherical shell = 6/2 = 3 cm = r, External diameter of hollow spherical shell = 10 cm, So, the external diameter of hollow spherical shell = 10/2 = 5 cm = R, Let the height of cylinder be taken as h cm, Volume of cylinder = Volume of spherical shell, Therefore, the height of the cylinder = 8/3 cm. Find the width of the embankment? A way of adding the pure solvent to the volumetric flask, for example a glass funnel and a pasteur pipette. A tent is in the form of a cylinder of diameter 20 m and height 2.5 m, surmounted by a cone of equal base and height 7.5 m. Find the capacity of tent and the cost of the canvas at Rs 100 per square meter. 2. is my MOST popular training with no PowerPoint presentations and, Comprehensive Database Performance Health Check, SQL SERVER – Cluster Patching: The RPC Server is Too Busy to Complete This Operation, SQL SERVER – UNION Not Allowed but OR Allowed in Index View – Limitation of the View 6, SQL SERVER – Query Optimization – Remove Bookmark Lookup – Remove RID Lookup – Remove Key Lookup – Part 3, SQL Server Performance Tuning Practical Workshop. Also, find the cost of milk which can completely fill the container, at the rate of Rs.25 per litre. You must be an Administrator on the box to enable the SA user or other SysAdmin users. The solution to be diluted will be in a vessel such as a volumetric flask. Please suggest me with an option to eradicate this issue. The diameter of cylinder is 24 m. The height of the cylindrical portion is 11 m while the vertex of the cone is 16 m above the ground. 14. A solid metallic sphere of radius 5.6 cm is melted and solid cones each of radius 2.8 cm and height 3.2 cm are made. n(Na2CO3) = 0.095 × 0.50 = 0.0475 mol The earth taken out of it has been spread evenly all around it to a width of 4 m to form an embankment. In this chapter, students will be solving problems on finding surface areas and volumes of combinations of solids like a cuboid, cube, right circular cylinder, right circular cone, sphere, hemisphere, and frustum of a cone. I just give brief of my situation, I am running SQL server 2014 r2 in that around 8 database was created, it was running with no problem since 2 years, but recently i m getting issue like suddenly my entire database say any one complete database got deleted that means all tables of that database became blank. God bless all they guys & girls like you who help as out all over the world on a daily basis. Therefore, the height of the embankment so formed is 6.78 m. 32. Find the height of the embankment. We will be linking to this great content on our site. Find the height of the embankment? Excellent article! The diameter of cylinder is 24 m. The height of the cylindrical portion is 11 m while the vertex of the cone is 16 m above the ground. 2.2 cubic dm of brass is to be drawn into a cylindrical wire of 0.25 cm in diameter. 14. Volumes of both, the well and the platform should be the same. As the well is spread evenly to form embankment then the volumes will be same. The radius of the cylindrical base is 20 cm. Find its total surface area. But i noticed that i do not see the solution explorer when I click on the view tab. Three cubes of a metal whose edges are in the ratio 3: 4: 5 are melted and converted into a single cube whose diagonal is 12√3 cm. Came here by searching for domain. 4. [SPname] so it throws the exception. I had abig trouble and need to explrore so much. 1. A solid in the form of a right circular cone mounted on a hemisphere is immersed in tub. The total space between the two vessels is filled with cork dust for heat insulation purposes. The diameters of internal and external surfaces of a hollow spherical shell are 10cm and 6 cm respectively. Find the volume of the bucket if its depth is 12 cm. Find the total surface area and volume of frustum. n(Na2CO3) = 0.0943 × 0.050 = 0.00472 mol, cdilute = n(Na2CO3 pipette) ÷ Vdilute in L Each blog post includes links to relevant AUS-e-TUTE tutorials and problems to solve. 3.50 per m2, Height of the Cylindrical Portion = Height of the tent – Height of the surmounted Cone, And, given diameter of the cylinder (d) = 36 m, So, its radius (r) of the cylinder = 36/2 = 18 m. Let’s consider L as the slant height of the cone. If the radius of each of the smaller balls is 1/4 of the radius of the original ball, how many such balls are made? Hence, the slant height of the cone (l) = 5 cm, So, the curved surface area of the cone (S1) = πrl, And, the curved surface area of the hemisphere (S2) = 2πr2, Therefore, the curved surface area of the toy is 103.62 cm2. The radii of two of the balls are 1.5 cm and 2 cm. 18. Rain water, which falls on a flat rectangular surface of length 6 m and breadth 4 m is transferred into a cylindrical vessel of internal radius 20 cm .What will be the height of water in the cylindrical vessel if a rainfall of 1 cm has fallen? Nupur Dave is a social media enthusiast and an independent consultant. 0.67 mm which is the thickness of the wire. I’m Hoang Nguyen. Volume of the frustum of a right circular cone = 1/3 π(r22 + r12 + r1 r2 )h, Let ‘L’ be the slant height of cone, then we know that, Total surface area of the frustum = π(r1 + r2) x L + π r12 + π r22. In my Comprehensive Database Performance Health Check, we can work together remotely and resolve your biggest performance troublemakers in less than 4 hours. Volume of the hollow cylinder = π (R2– r2) × h. Therefore, the volume of the hollow cylinder is the required amount of sand needed to spread across to a depth of 20 m. 29. A cylindrical vessel of diameter 14 cm and height 42 cm is fixed symmetrically inside a similar vessel of diameter 16cm and height of 42 cm. Following exception is raised, which is I am not able to resolve: Msg 15165, Level 16, State 1, Procedure sp_refreshsqlmodule_internal, Line 55 Could not find object ‘[DB1].[dbo]. i hv craeted login abc,wen i logged from that login it will not allow me add user db in user mapping show a error 15151, i want to give privigdes to user db by using this login. Consider a cylindrical tub having radius as 5 cm and its length 9.8 cm. A cylindrical vessel having diameter equal to its height is full of water which is poured into two identical cylindrical vessels with diameter 42 cm and height 21 cm which are filled completely. Pinal is also a CrossFit Level 1 Trainer (CF-L1) and CrossFit Level 2 Trainer (CF-L2). Thus, slant height of the cone (L) = 18.3 m, Now, the Curved Surface area of the Cylinder (S1) = 2πrh, And, the Curved Surface area of the cone (S2) = πrL, So, the total curved surface of the tent (S) = S1 + S2, Hence, the total Curved Surface Area (S) = 1533.08 m2, So, 1533.08 m2 of canvas will cost = Rs (3.50 x 1533.08). (adsbygoogle = window.adsbygoogle || []).push({}); © 2006 – 2020 All rights reserved. With this sphere, we have to make spherical balls of radius r = 1 cm, Let’s assume that the number of balls made as n. The volume of the solid sphere = sum of the volumes of n spherical balls. My computer has install windows 7 and SQL Server 2008 Express. n(Na2CO3 in stock) = cstock × Vstock A solution of urea in water contains 16 grams of it in 120 grams of solurion. The diameter of the cylinder (also the same for cone) = 24 m. The height of the Cylindrical part (H1) = 11m, So, Height of the cone part (H2) = 16 – 11 = 5 m, Vertex of the cone above the ground = 11 + 5 = 16 m, Curved Surface area of the Cone (S1) = πRL = 22/7 × 12 × L, Curved Surface Area of Cone (S1) = 22/7 × 12 × 13, Curved Surface Area of Cylinder (S2) = 2πRH1, Thus, the area of Canvas required for tent, S = S1 + S2 = (22/7 × 12 × 13) + (2 × 22/7 × 12 × 11), Therefore, the area of canvas required for the tent is 1320 m2. Therefore, the height of the cylindrical vessel nearly 191 cm. How many cubic cms of the cork dust will be required? 50 circular plates each with diameter 14 cm, As these plates are one above the other, the total thickness of all the plates = 0.5 x 50 = 25 cm, So, the total surface area of the right circular cylinder formed = 2πr × h + 2πr2, Therefore, the total surface area of the cylinder is 1408 cm2. 15. Applications of Mathematics in everyday life are enormous. Its internal diameter is 54 cm and the thickness of the iron sheet used in making roller is 9 cm. Along with 17+ years of hands-on experience, he holds a Masters of Science degree and a number of database certifications. Find the volume largest right circular cone that can be cut out of a cube whose edge is 9 cm. Msg 15151, Level 16, State 1, Line 1 Cannot alter the login ‘BUILTIN\Administrators’, because it does not exist or you do not have permission. 35. The solid cylinder is recast into a hollow cylinder of length 16 cm, external diameter of 20 cm and thickness of 2.5 mm = 0.25 cm, Volume of the solid cylinder = π12h = πh cm3, Let’s assume the length of the solid cylinder as h, Volume of the hollow cylinder = πh(R2– r2), So, the internal radius of the cylinder is 9.75 cm, Volume of the hollow cylinder = π × 16 (100 – 95.0625), Volume of the solid cylinder = volume of the hollow cylinder.
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