# A hole of radius r is bored through the center of a sphere

a hole of radius r is bored through the center of a sphere The curve in which the plane z = 2 cuts the surface z = . 040 Review Problem - Volume comparisons of two water crystals. Finding the equation. Find the volume of the remaining portion of the sphere. As you can see, the maximum value of the electric field occurs when little r becomes equal to the radius of the distribution and, at that point, the value of electric field is Q over 4 πε 0 big R 2. Now you can draw a right triangle that has as its legs 1) half the axis of the hole, which is 3", starting at the center of the sphere and running to the end of the hole, and 2) the radius of the hole, call it r, at the end of the hole. Volume of a Bore A hole of radius 2 centimeters is bored completely through a solid metal sphere of radius $5 \mathrm{~cm}$. none of the above string to bore a hole through the center of a planet to mine heavy metals in its core. 3 Problem 48E. [Notice that the volume depends only on h, not on r 1 or r 2. Find the volum of the ring-shaped solid that … 🎉 Announcing Numerade's $26M Series A, led by IDG Capital! A cylindrical drill with radius 4 is used to bore a hole through the center of a sphere of radius 5. A cylindrical hole of diameter cm is bored through a sphere of radius cm such that the axis of the cylinder passes through the center of the sphere. First, consider a slice through the sphere that is through the axis of the hole. EXAMPLE 9 The headlight has a hole down the center (Figure 8. cm such that the axis of the cylinder passes through the center of the sphere. Notice that the z-coordinate should depend only on latitude phi and of course the radius of the earth. What is the total surface area of the resulting solid? Express your answer in terms of π. Show that if a body were dropped into the hole it would execute a SHM. I figured I would need that to do this one. A 2 inch by 2 inch by 2 inch block of wood has 1 inch diameter hole drilled. 8 ±0. If the earth were a homegeneous sphere and a straight hole was bored in it through its centre, so when a body is dropped in the hole, it will excutes SHM. the $$x$$-axis) $$r$$ is a fixed number and won’t change . The diameter of the hole is 20. If the sphere has radius and the cylinder has radius find the volume of the spherical ring. 7. gl/JQ8NysFinding the Center-Radius Form of a Circle Given the Endpoints of the Diameter. The 1914 book Tik-Tok of Oz has a tube, that passed from Oz, through the center of the earth, emerging in the country of the Great Jinjin, Tittiti-Hoochoo. Calculate R R R 1 x+y dxdy, where Ris the region bounded by x= 0;y= Use a triple integral in cylindrical coordinates to ﬂnd the volume of the sphere x2 . First let us consider the volume of the entire Sphere, which has radius #2a#. I'm not sure on how to approach . First we sketch the region bounded by the given curves. 2 Problem 70E. Let a a particle of mass m be dropped at time t = 0 into this hole with initial speed zero. 👍. The remaining solid resembles a bead since it has a ﬂat top and bottom with a hole through the middle. Homework Statement The base of S (the solid) is an elliptical region with boundary curve $$9x^2 + 4y^2 = 36$$. 4K] The amount of material removed is the volume of the region within the sphere bounded by the cylinder. Homework Equations x=r*cos(theta) y=r*sin(theta) The Attempt at a Solution i know that the boundaries are: for theta: 0 to 2*pi for r: r1 to r2 You need to specify the diameter/radius of the hole or the sphere. (3 points) A hole of radius ris bored through the center of a sphere of radius R>r. The period of oscillation […] Click here👆to get an answer to your question ️ There are 27 drops of a conducting fluid. (b) Express the volume in part (a) in terms of the height h of the ring. An object of mass m at a distance r from the center of the Earth is pulled toward the center of the Earth only by the mass within the sphere of radius r. A hole of radius is bored through a cylinder of radius at right angles to the axis of the cylinder. After that, from x = 1 to x = 2, each cross-section is a disk with a hole. 4 pi R^2 - 2 pi R (2 R - 2 h) = 4 pi R h. 23-73 A nonconducting solid sphere has a uniform volume charge density ρ. math The volume of the solid of revolution about the x -axis is (2) V = ∫ a b A ( x) d x = ∫ a b π [ R ( x)] 2 d x. Given one or the other, you can perform the rest of the calculations. Notice that the volume depends only on h, not on r 1 or r 2. R! r r R! r A hole of radius is bored through the center of a sphere of radius > r. A cylindrical hole of radius a is bored through a solid right-circular cone of height h and base radius b > a. 2 0 0. Subscribe at http://www. simple oscillatory motion B. I'm having trouble solving this situation. A cylindrical hole of diameter 6 cm is bored through a sphere of radius 5. Preview cm Get help: Video Volume using Integrals. Cylindrical hole of radius a is bored through a sphere of radius 2a. A cylindrical gasoline tank is placed so that the axis of the cylinder is horizontal. So it is proportional to 1 over r2 outside of the sphere, and it is proportional to r inside of the sphere. We know the lengthh(2his the height of the removed cylinder) andnothing else!. 8 meters) per second squared. A round hole of radius √ 3 is bored through the center of a solid sphere of a radius 2. It remains to compute the volume of the two . (a) In terms of M h, find a g at r o. Mass of the body dropped inside the hole = m. (Hint: Set up the integral using washers. 6 ±0. A cylindrical hole is bored through a steel block, and a cylindrical piston is machined to fit into the hole. About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features Press Copyright Contact us Creators . A cylindrical drill with radius 1 is used to bore a hole throught the center of a sphere of radius 5. A cylindrical drill with radius 5 is used to bore a hole through the center of a sphere of radius 7. 1. Consider a sphere of radius 4 centered at the origin. It predicts a travel time through the tunnel from x=x 0 to x= -x o of- 2539sec 32 3960 5280 2 = ⋅ =π =π τ g R 28(a)A cylindrical drill of radius r 1 is used to bore a hole through the center of a sphere of radius r 2. com Solved: A hole of radius r is bored through the center of a sphere of radius R > r. The hole is bored straight through the centre along its diameter. The cylindrical shell radius is 2 π x 2\pi x 2 π x and the cylindrical shell height is 2 y = 2 r 2 − x 2 . " Its . calculus. 452-15. (a) Show that E(P) = — where P is the point at center of the hole and 90 is the 66. A hole is drilled clear through the center of a solid sphere. Find the period of the simple harmonic motion exhibited by the particle. The sphere contains two spherical holes of radius R1 centered on the y axis at y=+d and y=-d (see figure below). V = (4pi)/3 (R^2-r^2)^(3/2) We basically are being asked to calculate the volume of a spherical bead, that is, a sphere with a hole drilled through it. /X solid is removed, up to the point where \/& reaches 1. Acylindral drill with radius 5 cm is used to bore a hole through the center of a sphere with radius 7 cm. A cylindrical drill with radius {eq}2{/eq} is used to bore a hole through the center of a sphere of radius {eq}3{/eq}. The remaining volume is then the total volume of the sphere minus the volume of the cylindrical hole minus the volume of the two end caps. What charge now rests on the outer surface of the sphere? Sphere can be considered as the solid of revolution. R4 1 R2… 0 Rp z 0 rdrdµdz, (Watch out the order of integration . Set up, but do not evaluate, an integral for the volume cut out. Then the area of the cross section is Thus the volume is Example) A hole of radius is bored through the center of a sphere of radius . a Schwarzschild black hole 1] the event horizon is located at radial coordinate $$R_s=\frac{2GM}{c^2}$$, while . (a) Find the volume of the ring-shaped solid that remains. R 1 0 (R 1−r2 0 dz)rdr]dθ = π 2 (b) A cylindrical hole of radius a is bored through the center of a solid sphere of radius 2a. Find the volume of the ring (two ways), and show that it is independent of the radius of the sphere. 00, and (d) r = R? (e) Sketch a graph of E versus r. Volume of the sphere = 4/3 π r^3 = 4/3 π 13^3 Volume of the new sphere whose radius is 5 = 4/3 π 5^ 3. Drop a i kg mass ball down the hole from the surface. A cylindrical drill with radius 7inches is used to bore a hole through the center of a sphere of radius 9inches. e. While $$r$$ can clearly take different values it will never change once we start the problem. That means that, for each second you fall, you speed up by 32 feet per second — but only near Earth's surface. 039 Review Problem - Bushels of wheat the grain elevator can hold. hole of radius a is bored through the center of a . 2 R2 . Textbook solution for Calculus (MindTap Course List) 8th Edition James Stewart Chapter 5. However, if you are traveling through a tunnel in the Earth, then the gravitational force that you feel is not proportional to the entire mass of the Earth but to the amount of the mass that is contained within a sphere that is centered at the center of mass of the Earth and has a radius equal to the distance you are from the center. Physics This is an application of the napkin ring problem. The height of the spherical end caps at the ends of the cylindrical hole is (R - 3). A six inch long cylindrical hole drilled through a sphere the size of the earth would be shaped as a 6" thick disc of approximately 8000 miles diameter. We insert a +10μC charge at the center of the sphere through a hole in the surface. 🎉 Announcing Numerade's$26M Series A, led by IDG Capital! The radius of the cylindrical hole created is then sqrt(R^2 - 9). A cylindrical drill with radius 2 is used to bore a hole throught the center of a sphere of radius 5. Find the total mass. The solid material of the cylinder has a uniform has a uniform volume charge density ρ.  The residual volume would be the same as the volume of a sphere with radius $\sqrt{5}$. math (a) A cylindrical drill with radius r 1 is used to bore a hole through the center of a sphere of radius r 2. Remember me Not recommended on shared computers. A solid is bounded by z = x2 +y2, z =1andz =4. 2. (a) (10 pts) Set up the integral for the volume. Find step-by-step Calculus solutions and your answer to the following textbook question: A manufacturer drills a hole through the center of a metal sphere of radius R=6, What value of r will produce a ring whose volume is exactly half the volume of the sphere?. what is the volume of the resulting shape - e-eduanswers. / and the hole has radius g = 1. A hole of radius r is bored through the center of a sphere of radius R. 53) A cylindrical hole of diameter 6 cm is bored through a sphere of radius 5 cm such that the axis of the cylinder passes through the center of the sphere. 001R h from the center of a black hole is given by Eq. Given a solid sphere of radius R, remove a cylinder whose central axis goes through the center of the sphere. 8 m/sec>2. Answer: $$\frac{256\pi}{3} \space \text . I arrange them all on a table with the holes A hole of radius is bored through a cylinder of radius at right angles to the axis of the cylinder. The book guides students through the core concepts of calculus and helps them understand how those concepts apply to their lives and the world around them. 00 ± 0. Anyone who's ever had to sit on an airplane for 17 hours, enduring screaming babies, terrible internet, and the constant threat of deep vein thrombosis knows how . Find the volume common to two spheres of radius r r with centers that are 2 h 2 h apart, as shown here. 8 x 0 0. 2 y = 2 r 2 − x 2 . 9. The claim is that the volumeof the remaining solid is 43πh3. A round hole of radius p 3 cms is bored through the center of a solid sphere of radius 2 cms. The disk has radius f= . Find the ratio of the height to the base radius of the cylinder having the largest lateral area. Andrew DeBenedictis. given shell of radius r will have a thickness dr, which gives it a surface area of 4πr2 and a volume of (thickness)(surface area) = 4πr2 dr. All of the . Hi. Apr 24, 2009. Then the points (0, 0, 0), (x, y, 0) and (x, y, z . A hole of radius is bored through the center of a sphere of radius . A right circular cylinder is to be inscribed in a sphere of given radius. Let r be the vector from the center of the sphere to a general point P within the sphere. Describe the motion of the stone –no air friction. Using a Lagrangian nd the equation of motion for the center of mass of the sphere (you can use a Lagrange multiplier but it is not essential). Find the radius when the rates of increase of the surface area and the radius are numerically equal. A cylindrical drill with radius r 1 is used to bore a hole through the center of a sphere of radius r 2. With the hole diameter or radius, you can calculate the sphere radius using Pythagoras, the volume of the enti. It is a direct computation based on already evaluated result/ formula for spherical cap. Use cylindrical shells to find the volume of the solid. If the axis of the hole lies along that of the cone, find the volume of the remaining . 1) A sphere of radius r. . Ball – A sphere with a flow path (hole or tunnel) through the center of it and a connection point for a shaft to rotate it. 038 Review Problem - Circular log with non-uniform cross-section. If the axis of the hole passes through the center of the sphere, find the volume of the metal removed by the drilling. A heavy ball with diameter 10 cm is placed in the bowl and water is poured into the bowl to a depth of h centimeters. Find the ﬂuid force on a circular end of the tank if the tank is half full, assuming that the diameter is Consider the Earth as a homogeneous sphere of radius R and a straight hole is bored in it through its centre. 69. 0 cm. Bore a cylindrical hole through the sphere down the z axis with varying radii “b”. Your task is to determine the volume of the napkin rings for holes bored out of radii 1, 2, and 3. The remaining = 4/3 π 13^3 - 4/3 π 5^3 = 4/3 π(13^3 - 5^3) = 4/3 * π* (2197 - 125) = 4/3*π*2072 New sphere&#039;s radius say r, so volume = 4/3 π r^3. It would leave a very thin ring of material six inches wide and 8000 miles in diameter. The slice is a flat ring or a "washer. The potential of the bigger drop is V2 . 68. You need to specify the diameter/radius of the hole or the sphere. Notice that the volume depends only on h , not on r 1 or r 2 . Napkin ring problem. [Calculus] A cylindrical drill with radius 1 is used to bore a hole through the center of a sphere or radius 5. And when you're very close to the center, you can approximate the mass density of Earth, rho, as constant, so that the mass contained within a sphere of radius r is just the volume (4/3*pi*r 3) times the density: M 1 = rho * 4/3 * pi * r 3. ) 10. 30. Each drop has radius r, and each of them is charged to the same potential V1 . M. A sphere of radius R is uniformly charged. We have step-by-step solutions for your textbooks written by Bartleby experts! r=a = − − 0 + 0 Q 2πa2 Note that when this is combined with (1), the total (free and polarization) charge density is σ tot = σ + σ pol = 0 + 0 Q 2πa2 on either half of the sphere. 👍 Correct answer to the question suppose a hole of radius 1 is bored through a sphere of radius 2. 1 m/sec^2. Ans. Calculate the volume of the remaining solid. The axis of the hole is a distance b from the axis of the cylinder, where a < b < R (as shown in figure). It's pretty easy to show that all the holes are 2*K or 6 inches long. This ring would have the same volume as a six inch diameter sphere with a cylindrical hole of zero diameter. Taking positive r as outward from the center of the Earth: This is the same form as Hooke's Law for a mass on a spring. 6 0. Students also viewed these Calculus questions A hole of radius is bored through a cylinder of radius R > r at right angles to the axis of the cylinder. (b) Does a g at . The cylinder shaped hole has a radius of 2 cm. 8 ms^(-2) Find the volume that remains after a hole of radius 1 is bored through the center of a solid sphere of radius 2. Example 17. Let us ﬂnd the volume cut out (see Figure 3). Book details. For a larger sphere, the band will be thinner but longer. Also find its time period. 6. An ideal string is partially wrapped around the sphere’s equator so that the sphere rotates . 0. Morsel 1. The moment of inertia of a sphere through its center is I= 2 5 ma 2. In the 2012 movie Total Recall, a gravity train called "The Fall" goes through the center of the Earth to commute between Western Europe and Australia. A very long, solid insulating cylinder with radius R has a cylindrical hole with radius a bored along its entire length. The volume of the cylindrical hole is Vh = 6Pi(R^2 - 9). A hole of radius is bored through the middle of a cylinder of radius R > r at right angles to the axis of the cylinder. As a final In Exercises 30–31, find a parameterization for the curve. https://www. A cylindrical hole of radius r is bored through the center of a sphere with radius R > r. 17464525 2. ••22 The radius of a black hole are related by R h and mass M h R h = 2GM h /c 2, where c is the speed of light. the surface of the hole passes through the center of the sp r-ruslan [8. 2 An object occupies the space inside both the cylinder x 2 + y 2 = 1 and the sphere x 2 + y 2 + z 2 = 4, and has density x 2 at ( x, y, z). The acceleration at the surface of the sphere due only to this block would be 3. A hole of radius r is bored through the center of a sphere of radius R > r. I found this to be about 84 min. The remaining volume is pi/6 h^3 = pi/6 * 24√3 = 4√3 pi. series of larger spheres, each radius R, (R and r vary from sphere to sphere) with holes of radius r in them. A 1/4π in B 1/5π in C 1/3π in D 1/8π in E 1/2π in . Suppose that a small hole is drilled straight through the center of the earth, thus connecting two antipodal points on its surface. A solid, uniform sphere of mass M and radius R pivots on a xed, massless, frictionless axle that passes through the center of the sphere. It is obtained by rotating the circle about x-axis. (b) r = 0, (c) r = R/2. Through the center. 3 A right-circular solenoid of ﬁnite length L and radius a has N . If the earth were a homogeneous sphere and a straight hole bored in it through its centre, show that if a body were dropped into the hole it would execute a simple harmonic motion. (a) Show that the electric field at P is given by (Note that the result is independent of the radius of . A golden oldie that usually takes the following form. The small volume we want will be defined by Δ ρ, Δ ϕ , and Δ θ, as pictured in figure 17. Earth is assumed to be a homogeneous sphere. Find the volume of the ring shaped solid that remains. I did a problem similar to this (a cone) and didn't have too much trouble, but i'm kind of stumped on this one. If a hole of radius r is bored through, the hole with generate a circular shape in the sphere. However, if we have a person 50kG and 2 meters tall (yeah, tall and skinny) and person's cg is 1 meter above the surface, or 1. Assume that the gravitational acceleration a g of an object at a distance r o = 1. Area of spherical cap = 2 pi R h where R is sphere radius, h is total cap height, either for a full or for a truncated spherical segment. If a hole of height h is drilled straight through the center of a sphere, the volume of the remaining band does not depend on the size of the sphere. In the diagram below a hole is drilled through the centre of the sphere. R π 2 . Cross-sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in . Its centre is at O and Radius = R. A solid metal cylinder with height of 10 inches and radius of 6 cm has a cylinder-shaped hole drilled through it. Thus the equator lies in the xy-plane (so that our "sphere" is centered at the origin in 3D). 8k points) oscillations Click here👆to get an answer to your question ️ If this earth were a homogenous sphere and a straight hole bored in it trough its centre. 037 Review Problem - Amount of material the factory chimney contain. Imagine that a hole is drilled through the center of the Earth to the other side. asked Sep 23, 2020 in Oscillations by Raghuveer01 ( 50. Radius of sphere = R, radius of hole = r, length of hole = 2L [so L=3], height of cap = h. Set up, but do not evaluate, an integral for the length of the curve y = ex cosx on the interval [0;ˇ=2]. If the axis of the hole passes through the center of the sphere, find the volume of the metal . On the surface, all of Earth's matter lies below your feet but, as you fall, more . Find the frequency of small oscillations of the sphere. Find the mass in Cylindrical coordinates of the solid that remains with density 5 ( x, y, 2) - (1 - 2 ) grams/ in And so you'd make a round-trip, back through all the various layers of mantle, outer core, and inner core, to the center of the Earth, and back once again through the northern layers. At Earth's surface, gravity pulls on us at 32 feet (9. 4 xx 10^(5) m and g=9. ] Problem 11. We have step-by-step solutions for your textbooks written by Bartleby experts! Problem 19 Easy Difficulty. We pick the positive x-axis to pass through the Prime Meridian on the equator, and the positive y-axis to pass through 90 degrees West longitude there. Find the volume V of the remaining portion of the sphere. ~ VOLUMES BY CYLINDRICAL SHELLS y FIGURE 2 Some volume problems are very difficult to handle by the methods of the preceding section. Use cylindrical shells to compute the volume V of a napkin ring of height 5 h created by drilling a hole with radius r through the center of a sphere of radius R and express the answer in terms of h . A hole of radius r is bored through the center of a sphere of radius R > r. A bowl is shaped like a hemisphere with diameter 30 cm. 13-11 (it is, for large black holes). (a) A cylindrical drill with radius r 1 is used to bore a hole through the center of a sphere of radius r 2 . Spheres for n > 2 are sometimes called hyperspheres. Rotating the region (bounded by y= p R2 x2 and y= r) about the x-axis . (b) Express the volume in part ain terms of the height of the ring. 3. [math]V . 5 kg. com/kisonecat A hole of radius r is bored through the center of a sphere of radius R. plane of the plate, passing through the point on the outer rim of the plate closest to the center of the hole. the volume of a sphere of radiush! r0 R A cylindrical drill with radius r1 is used to bore a hole through the center of a sphere of radius r2. What is the volume of the remaining cylinder with the hole But when you're closer to the center of the Earth, M 1 is less than M Earth. (a) A cylindrical drill with radius r_{1} is used to bore a hole through the center of a sphere of radius r_{2} . #4. A hole of radius r is bored through the center of a sphere of radius R. (12. That’s a famous problem that I first saw in Scientific American many years ago. A round hole of radius sqrt(3) ft is bored through the center of a solid sphere of a radius 2 ft. You might also look up the napkin ring problem, where it is shown that if a hole of length h is drilled through a sphere, the remaining volume is independent of the radius of the sphere! In this case, h/2 = R^2 - r^2 = √3, so h = 2√3. where K is 3 inches in all cases. hole. Geometry. S 1: a 1-sphere is a circle of radius r; S 2: a 2-sphere is an ordinary sphere; S 3: a 3-sphere is a sphere in 4-dimensional Euclidean space. A Charged Spherical Shell with a Hole The figure below shows a circular hole of radius b (white) bored through a spherical shell (gray) with radius R and uniform charge per unit Area cr. 11. If the flux of the electric field through a closed surface that includes all the sphere is f, find: a) (4pts) The charge density r in terms of the quantities above and fundamental constants. The arrangement is kept on a horizontal table (the surface of concave mirror is frictionless and sliding not rolling). Valve body – A pressure vessel that contains the components needed to control or shut off the flow through a pipe. 3k+ A cylindrical hole of radius a is bored through a solid right-circular cone of height h and base radius b > a. y = sqrt (r^2 - x^2) and then rotated the semi circle around . The mass of the sphere with a radius the same as the distance the ball is from the centre at any stage is the only thing which contributes a value to the net force on the ball – all the matter further away has a net force of zero on the ball (if you’re finding it hard to visualise this, don’t forget there is some matter above the ball . Solution. Find the volume of water in the bowl. 90 ± 0. stone comes to rest at the surface D. [Bonus) A hole of radius r is bored through the center of a sphere of radius R (r<R). Correct answers: 2 question: A hole of radius 2 inches is bored through the center of a right cylinder with radius 6 inches and height 10 inches. Solution: ±1 0 1 ±0. R2… 0 Ra R Rp a2¡r2 ¡ p a2¡r2 rdzdrdµ 3. youtube. Find the ratio V2/V1 . I arrange them all on a table with the holes Also, recall we are using \(r$$ to represent the radius of the cylinder. Q. 8 z ±1 ±0. V [remainder] = (4pi/3)L*L*L = 36pi. (b) Show that the volume enclosed by the barrel is V = 13 πh(2R 2 + r 2 − 25 d 2 ) 85. Solution: The line y= rintersects the semicircle y= p R2 x2 when r= p R2 x2 implies that r2 = R 2 2x, so x = R2 r2 and x= p R2 r2. Long distance air travel sucks. Assume 0 < b < a. show that if a body were dropped into the hole , it would execute a simple hormonic motion. 4 ±0. 4 0 0. I used the formula for and drew a semi circle. You discover that both napkin rings have the same height 5h. Cylinders do not change their radius in the middle of a problem and so as we move along the center of the cylinder (i. In the generic case of a non-rotating and electrically neutral black hole [i. com/question-answer-physics/a-sphere-of-radius-r-has-its-centre-at-the-origin-it-has-a-uniform-mass-density-rho0-except-that-the-12230257 Please Subscribe here, thank you!!! https://goo. Gravity is a function of mass, and mass is a property of matter. 2 . if the earth were a homogenous sphere and a straight hole bored in it through its center . (b)Express the volume in part (a) in terms of the height h of the ring. 01 cm, and the diameter of the piston is 19. Once we’ve built the sphere up to a radius r, Gauss’ law tells us that the potential at the surface is just that of a point charge of radius r: V(r) = k eq(r) r Where q(r) is the charge built up so . 036 Review Problem - Circular hole bored through a sphere. Find the volume of the hole. 5 meters above the cg of the single block, the acceleration at belt height would be closer to 0. The radius of a sphere is r inches at time t seconds. Putting a hole in a sphere, when the eccentricity is zero (the hole is centered) is easy to calculate. 8 1 y Figure 9: Q5: Left: The solid E; Right: The image of E on xy-plane Sphere of radius a . Find the volume of the resulting spherical ring. Boost your resume with certification as an expert in up to 15 unique STEM subjects this summer. - Slader Answer: Πr²(4r/3 - h) Explanation: Volume of a sphere is 4/3Πr³. After time t, the depth it reached (inside the earth) = d A sphere has a definite center-point and the sphere’s axes can freely rotate around that point to become parallel to the line, and the line will then be some distance from the center point (marked eccentricity below). View Answer. A hole of radius r is bored through the center of a sphere of radius R > r . The biggest part of the material is just a cylinder. 0μC and radius 5. Consider a cross section through the sphere, which we have centred on the origin of a Cartesian coordinate system, and assuming R gt r: The red circle has radius R, hence its equation is: x^2+y^2=R^2 Method 1 - Calculate core and subtract from . Ignore the change in density of the fluid on combining the drops. doubtnut. Since this is uniform, this is why the resulting electric ﬁeld is radially symmetric. - 14565712 A hole of radius $r$ is bored through the center of a sphere of radius \$ R… 02:05 Find a the volume and b the surface area of the sphere with the given radius… Use the shell method to find the volume of a sphere of radius r r r with a vertical hole of radius a < r a < r a < r bored through the center of the sphere. A cylindrical drill with radius 5 is used to bore a hole through the center of a sphere of radius 8. Only at the surface, it appears to me, will the constant of acceleration be 9. In geometry, the napkin-ring problem involves finding the volume of a "band" of specified height around a sphere, i. 82. 2 m that is free to rotate about an axis that passes through its center. Therefore the answer couldn't depend on the radius of the sphere. e . Find the magnitude and . A hole of radius r is bored through the middle of a . A cylindrical drill with radius r1 is used to bore a hole throught the center of a sphere of radius r2. Find the volume of the bead. Note that the ordinary sphere is a 2 . 5b). Suppose that a region R has area A and lies above the x-axis. 4 y 0. They are then combined to form a bigger drop. It would cause the trans-Earth traveler to oscillate back and forth through the center of the Earth like a mass bobbing up and down on a spring. Find a formula for the volume of the remaining portion of the sphere. Example. 52) If the sphere has radius 4 and the cylinder has radius 2 find the volume of the spherical ring. Let h denote the height of the remaining solid. My friend wouldn't have posed a math problem [boring] and she wouldn't have left out critical information.  A hole is cut through the center of a sphere of radius r. Homework Equations The Attempt at a Solution Wouldn't is be (4/3)∏(R^3-r^3)? Answer: Πr²(4r/3 - h) Explanation: Volume of a sphere is 4/3Πr³. 31. If the earth were a homogeneous sphere of radius R and a straight hole bored in it through its centre, show that a body dropped into the hole will execute SHM and find its time period. A hole of radius 2 centimeters is bored completely through a solid metal sphere of radius 5 cm. A hole of radius r is bored through the middle of a cylinder of radius R > r at right angles to the axis of the cylinder. circumference of a solid disk of radius R = 2. It is designed to connect two or more sections of pipe or tubing to each other. A hole of radius is bored through the center of a sphere of radius > r. A. 34 m/sec^2. We have a hollow metallic sphere with charge -5. Round to the nearest tenth. 2 0. A sphere of radius r is kept on a concave mirror of radius of curvature R. A hole of radius {eq}r {/eq} is bored through the center of a sphere of radius {eq}\displaystyle R \gt r {/eq}. The acceleration due to gravity at the surface of the earth = g. A stone is dropped into the hole. The height of the remaining spherical ring is h. 4 0. Pythagoras shows that it has height 2·1, hence its volume is V C = π · √ 3 2 ·2 = 6π. R g xt =x This result shows that the mass m is undergoing simple harmonic motion with the same angular frequency ω=(2π/τ)=sqrt(mg/R) as a mass falling through a shaft passing through the center of the earth. Find the volume of material removed from the sphere. Find the volume of the solid that results when the region enclosed by y = 9 − x 2 and y = 0 is revolved about the x -axis. 111 . Write. 02 cm. Bore a hole of radius a a down the axis of a right cone and through the base of radius b, b, as seen here. 1 . Find the volume that remains after a hole of radius 1 is bored through the center of a solid sphere of radius 2. The volume of the remaining portion of the sphere is equal to the difference between the outer sphere and innner sphere . Volume = ? The hole has radius 1. Spherical coordinates are somewhat more difficult to understand. R and r have the relationship R^2 = K^2 + r^2. The main character is tossed down the shaft through the planet and calculates how long the journey will take him, which is given as this equation in the 1989 paperback edition: time = " ˇ 2 tan 1 v R q 4ˇ 3 Gˆ # 3 4ˇGˆ; (1) Use Pythagoras theorem twice to derive a distance formula, hence the equation: x^2+y^2+z^2 = 2^2 The distance of a point (x, y, z) from (0, 0, 0) is sqrt(x^2+y^2+z^2) To see this you can use Pythagoras twice: The points (0, 0, 0), (x, 0, 0) and (x, y, 0) form the vertices of a right-angled triangle with sides of length x, y and sqrt(x^2+y^2). 48. Find the volume remaining in a sphere of radius a after a hole of radius b is drilled through the centre. Consider a hole bored diametrically through the earth. 6 1 x ±1 ±0. stone stops at the center of the earth E. Sign In. A round hole of radius √ 3 ft is bored through the center of a solid sphere of radius 3 ft. The solid left over is like a napkin ring. Find the volume of the ring-shaped solid that remains. After time t, the depth it reached (inside the earth) = d cylindrical hole of radius R is bored through a sphere of radius a,where0<R<a, passing through the center of the sphere along the pole. Question From – Cengage BM Sharma ELECTROSTATICS AND CURRENT ELECTRICITY ELECTRIC FLUX AND GAUSS LAW JEE Main, JEE Advanced, NEET, KVPY, AIIMS, CBSE, RBSE, U. Radius of the earth is 6. the center of each sphere lies on the surface of the other sphere. i. The curve y = 4 − 5x 4 through the point (0, 4, 4), parallel to the xy-plane. The n-sphere of unit radius centered at the origin is denoted S n and is often referred to as "the" n-sphere. Most people would say that there’s not enough information to solve it: how big is the sphere? MA 113 220 1 Summer 2013 Assignment 2 Journey through the Center of the Earth Have you ever speculated as to what would happen to an object dropped into a tunnel bored through the center of the earth to the other side? In this exercise, we will explore this problem. Calculus is designed for the typical two- or three-semester general calculus course, incorporating innovative features to enhance student learning. surface of a cylinder of radius R;(R>a), as shown in the gure. 2y = 2 \sqrt{r^2 - x^2} . Use integrals to find the volume of the remaining portion of the sphere. stone falls through and out the back C. Determine the time period of its oscillation . 2 R 2π 0 [R 2a 0 (R√ 4a2−r2 0 dz)rdr]dθ = 4 √ 3πa3 (c) Find the volume of the region bounded above by the plane z = 2x and below by the paraboloid z = x 2+y . 8. 👎. 4/24) (a) A cylindrical drill with radius r 1 is used to bore a hole through the center of a sphere with radius r 2. also find its time period . The force applied to the rope has a magnitude of 35 N; and the disk has a mass M of 7. R. Expert Answer 100% (7 ratings) Volume of a sphere with a hole drilled through its centre. 68 . Find the volume V of the remaining portion of the sphere. Find the total volume and surface area. 5. If the sphere is displaced from its equilibrium position and left, then it executes S. Forgot your password? Sign in with Facebook Bore a hole from the earth surface through the center of the earth to a point directly opposite the entry hole. 4. H. 38Let D be the disk with center the origin and . Let R^2 = h^2 + r^2, ( R = 11, r = 8 ). Consider a cross section through the sphere, which we have centred on the origin of a Cartesian coordinate system: The red circle has radius #2a#, hence its equation is: # x^2+y^2=(2a)^2 => x^2+y^2=4a^2# Method 1 - Calculate core and subtract from Sphere. a hole of radius r is bored through the center of a sphere

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