The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas -Maths 9th

The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas -Maths 9th
by

1 Answer

Answer :

If diameter of earth is said d, then the diameter of moon will be d/4 (as per given statement) Radius of earth = d/2 Radius of moon = ½×d/4 = d/8 Surface area of moon = 4π(d/8)2 Surface area of earth = 4π(d/2)2 Ncert solutions class 9 chapter 13-6
by

Related questions

The diameter of the moon is approximately -Maths 9th

Description : The diameter of the moon is approximately -Maths 9th

Last Answer : Dmoon = 1/4.Dearth ⇒ 2. Rmoon = 1/4. 2 Rearth ⇒ Rearth = 4 Rmoon Surface area of moon (Smoon)/Surface area earth (Searth) = 4πR2moon/4πR2earth ⇒ Smoon/Searth = R2moon/(4Rmoon)2 = 1/16.R2moon/R2moon = 1/16 ∴ Smoon : Searth = 1:16

The diameter of the moon is approximately -Maths 9th

Description : The diameter of the moon is approximately -Maths 9th

Last Answer : Let, the diameter of the moon = 2RM Diameter of the earth = 2RE According to statement = 2RM = 1/4(2RE) RE = 4RM Volume of Moon (VM)/Volume of Earth VE) = 4/3πRM3/4/3πRE3 = 4/3πRM3/4/3π (4 RM)3 = 1RM3/64RM3 VM/VE = 1/64 ⇒ VM = 1/64VE i.e., volume of the moon is 1/64 of the volume of earth.

The outer curved surface areas of the hemisphere and sphere are in ratio 2:9. find their ratio of their raddii -Maths 9th

Description : The outer curved surface areas of the hemisphere and sphere are in ratio 2:9. find their ratio of their raddii -Maths 9th

Last Answer : This answer was deleted by our moderators...

The outer curved surface areas of the hemisphere and sphere are in ratio 2:9. find their ratio of their raddii -Maths 9th

Description : The outer curved surface areas of the hemisphere and sphere are in ratio 2:9. find their ratio of their raddii -Maths 9th

Last Answer : This answer was deleted by our moderators...

The volumes of two spheres are in the ratio 64 : 27. Find the difference of their surface areas, if the sum of their radii is 7 cm. -Maths 9th

Description : The volumes of two spheres are in the ratio 64 : 27. Find the difference of their surface areas, if the sum of their radii is 7 cm. -Maths 9th

Last Answer : answer:

Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes. -Maths 9th

Description : Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes. -Maths 9th

Last Answer : Let each side of a cube = a cm Then surface area = 6a² cm² and surface area of 3 such cubes = 3 x 6a² = 18a² cm² By placing three cubes side by side we get a cuboid whose ... + 3a²] = 14 a² ∴ Ratio between their surface areas = 14a² : 18a² = 7 : 9

The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases. -Maths 9th

Description : The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases. -Maths 9th

Last Answer : Let r1 and r2 be the radii of spherical balloon and spherical balloon when air is pumped into it respectively. So r1 = 7cm r2 = 14 cm Now, Required ratio = (initial surface area)/(Surface area after pumping air into ... = (7/14)2 = (1/2)2 = ¼ Therefore, the ratio between the surface areas is 1:4.

Surface Areas and Volumes Class 9th Formulas -Maths 9th

Description : Surface Areas and Volumes Class 9th Formulas -Maths 9th

Last Answer : Circle is the locus of all such points which are equidistant from a fixed point, this point is known as centre while distance of any point from centre defined as radius of circle. Here O is fixed ... the exterior angle is equal to the interior opposite angle. An isosceles trapezium is cyclic.

NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.1 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.1 -Maths 9th

Last Answer : 1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine: (i) The area of the sheet required for making the box ... of a cylinder = 2πrh + 2πr2 = 2πr(h + r) Note: Unless it is mentioned assume π = (22/7)

NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.2 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.2 -Maths 9th

Last Answer : 1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. 2. It is required to make a closed cylindrical tank of height 1 m ... open from the top. ∴ Surface area of a penholder (cylinder) = [Lateral surface area] + [Base area]

NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.3 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.3 -Maths 9th

Last Answer : 1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area. 2.Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 ... m2, what will be the cost of painting all these cones? (Use π = 3.14 and take √1.04= 1.02)

NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.4 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.4 -Maths 9th

Last Answer : 1. Find the surface area of a sphere of radius: (i) 10.5 cm (ii) 5.6 cm (iii) 14 cm 2. Find the surface area of a sphere of diameter: (i) 14 cm (ii) 21 cm (iii) 3.5 m 3. Find the ... area of the sphere, (ii) curved surface area of the cylinder, (iii) ratio of the areas obtained in (i) and (ii).

NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.5 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.5 -Maths 9th

Last Answer : 1. A matchbox measures 4 cm x 2.5 cm. x 1.5 cm. What will be the volume of a packet containing 12 such boxes ? 2. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of ... 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute ?

NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.6 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.6 -Maths 9th

Last Answer : 1. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold ? (1000 cm3 = 1 l) . 2. The inner diameter of a cylindrical ... with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients ?

Cbqs (case base study ) of chapter 13 surface areas and volume of maths class 9th -Maths 9th

Description : Cbqs (case base study ) of chapter 13 surface areas and volume of maths class 9th -Maths 9th

Last Answer : CBQs Ch- 13 Surface area and volume - Maths Class 9th 1 . Dev was doing an experiment to find the radius r of a sphere.For this he took a cylindrical container with radius R = 7 cm and height 10 cm. He filled the container ... .14)*6²*8 = 301.44 m³ area of the floor = πR² = 3.14 (6)² = 113.04 m²

The median of a trapezoid cuts the trapezoid into two regions whose areas are in the ratio 1 : 2. Compute the ratio of the smaller base of the -Maths 9th

Description : The median of a trapezoid cuts the trapezoid into two regions whose areas are in the ratio 1 : 2. Compute the ratio of the smaller base of the -Maths 9th

Last Answer : answer:

A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m2 . -Maths 9th

Description : A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m2 . -Maths 9th

Last Answer : Diameter of the pillar = 50 cm ∴ Radius (r) = 502m = 25 m = 14m and height (h) = 3.5m Curved surface area of a pillar = 2πrh ∴ Curved surface area to be painted = 112m2 ∴ Cost of painting of 1 m2 pillar = Rs. 12.50 ∴ Cost of painting of 112 m2 pillar = Rs. ( 112 x 12.50 ) = Rs. 68.75.

The slant height and base diameter of conical tomb are 25m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m2 -Maths 9th

Description : The slant height and base diameter of conical tomb are 25m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m2 -Maths 9th

Last Answer : Slant height of conical tomb, l = 25m Base radius, r = diameter/2 = 14/2 m = 7m CSA of conical tomb = πrl = (22/7)×7×25 = 550 CSA of conical tomb= 550m2 Cost of white-washing 550 m2 area, which is Rs (210×550)/100 = Rs. 1155 Therefore, cost will be Rs. 1155 while white-washing tomb.

Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m -Maths 9th

Description : Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m -Maths 9th

Last Answer : Radius of cone, r = 24/2 m = 12m Slant height, l = 21 m Formula: Total Surface area of the cone = πr(l+r) Total Surface area of the cone = (22/7)×12×(21+12) m2 = 1244.57m2

Find (i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5m high. -Maths 9th

Description : Find (i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5m high. -Maths 9th

Last Answer : Height of cylindrical tank, h = 4.5m Radius of the circular end , r = (4.2/2)m = 2.1m (i) the lateral or curved surface area of cylindrical tank is 2πrh = 2 (22/7) 2.1 4.5 m2 = (44 0.3 ... ) = 87.12 m2 This implies, S = 95.04 m2 Therefore, 95.04m2 steel was used in actual while making such a tank.

Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area -Maths 9th

Description : Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area -Maths 9th

Last Answer : Radius of the base of cone = diameter/ 2 = (10.5/2)cm = 5.25cm Slant height of cone, say l = 10 cm CSA of cone is = πrl = (22/7)×5.25×10 = 165

The inner diameter of a circular well is 3.5m. It is 10m deep. Find (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2. -Maths 9th

Description : The inner diameter of a circular well is 3.5m. It is 10m deep. Find (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2. -Maths 9th

Last Answer : Inner radius of circular well, r = 3.5/2m = 1.75m Depth of circular well, say h = 10m (i) Inner curved surface area = 2πrh = (2 (22/7 ) 1.75 10) = 110 Therefore, the inner curved surface ... area = Rs (110 40) = Rs.4400 Therefore, the cost of plastering the curved surface of the well is Rs. 4400.

In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. -Maths 9th

Description : In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. -Maths 9th

Last Answer : Height of cylindrical pipe = Length of cylindrical pipe = 28m Radius of circular end of pipe = diameter/ 2 = 5/2 cm = 2.5cm = 0.025m Now, CSA of cylindrical pipe = 2πrh, where r = radius and h = height of ... = 2 (22/7) 0.025 28 m2 = 4.4m2 The area of the radiating surface of the system is 4.4m2.

The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. (Assume π =22/7 ) -Maths 9th

Description : The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. (Assume π =22/7 ) -Maths 9th

Last Answer : Height of cylinder, h = 14cm Let the diameter of the cylinder be d Curved surface area of cylinder = 88 cm2 We know that, formula to find Curved surface area of cylinder is 2πrh. So 2πrh =88 cm2 (r is the ... 88 cm2 2r = 2 cm d =2 cm Therefore, the diameter of the base of the cylinder is 2 cm.

If the volume of a sphere is numerically equal to its surface area, then find the diameter of the sphere. -Maths 9th

Description : If the volume of a sphere is numerically equal to its surface area, then find the diameter of the sphere. -Maths 9th

Last Answer : Let r be the radius of the sphere. and Volume of a sphere = surface area of the sphere ⇒ 4 / 3πr3 = 4πr2 ⇒ r = 3 cm ∴ Diameter of the sphere = 2r = 2 × 3 = 6 cm

If the volume of a sphere is numerically equal to its surface area, then find the diameter of the sphere. -Maths 9th

Description : If the volume of a sphere is numerically equal to its surface area, then find the diameter of the sphere. -Maths 9th

Last Answer : Let r be the radius of the sphere. and Volume of a sphere = surface area of the sphere ⇒ 4 / 3πr3 = 4πr2 ⇒ r = 3 cm ∴ Diameter of the sphere = 2r = 2 × 3 = 6 cm

A hemispherical bowl has its external diameter equal to 10 cm and its thickness is 1 cm. What is the whole surface area of the bowl ? -Maths 9th

Description : A hemispherical bowl has its external diameter equal to 10 cm and its thickness is 1 cm. What is the whole surface area of the bowl ? -Maths 9th

Last Answer : External radius of hemispherical bowl = 5 cm Internal radius of the bowl = (5 – 1) cm = 4 cm Surface area of external portion = 2π(5)2 = 50 p sq. cm Surface area of internal portion = 2π(4)2 = ... = 91π sq. cm = (91×227)(91×227) sq. cm = 13 × 22 sq. cm = 286 cm2

A copper wire 4 mm in diameter is evenly wound about a cylinder whose length is 24 cm and diameter 20 cm so as to cover the surface. -Maths 9th

Description : A copper wire 4 mm in diameter is evenly wound about a cylinder whose length is 24 cm and diameter 20 cm so as to cover the surface. -Maths 9th

Last Answer : answer:

Find the ratio of the diameter of the circles inscribed in and circumscribing an equilateral triangle to its height? -Maths 9th

Description : Find the ratio of the diameter of the circles inscribed in and circumscribing an equilateral triangle to its height? -Maths 9th

Last Answer : (b) 2 : 4 : 3.For an equilateral triangle of side a units,In-radius = \(rac{a}{2\sqrt3}\) units⇒ Diameter of inscribed circle = \(rac{a}{\sqrt3}\) unitsCircumradius = \(rac{a}{\sqrt3}\)⇒ Diameter of circumscrible circle = \( ... \(rac{2a}{\sqrt3}\): \(rac{\sqrt3}{2}a\) = 2a : 4a : 3a = 2 : 4 : 3.

The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Description : The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Last Answer : Let the length, breadth and height of the rectangular box be 4x, 3x and 2x, respectively. ∵ Total surface area = 1300 cm2 2(4x × 3x + 3x × 2x + 4x × 2x) = 1300 52x2 =1300x2 = 25x = 5 ∴ Volume of rectangular box = 4x × 3x × 2x = 24(5)2 = 3000 cm3

The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Description : The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Last Answer : Let the length, breadth and height of the rectangular box be 4x, 3x and 2x, respectively. ∵ Total surface area = 1300 cm2 2(4x × 3x + 3x × 2x + 4x × 2x) = 1300 52x2 =1300x2 = 25x = 5 ∴ Volume of rectangular box = 4x × 3x × 2x = 24(5)2 = 3000 cm3

A sphere and a cone have equal bases. If their heights are also equal, the ratio of their curved surface will be : -Maths 9th

Description : A sphere and a cone have equal bases. If their heights are also equal, the ratio of their curved surface will be : -Maths 9th

Last Answer : answer:

If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Description : If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Last Answer : (b) In point P (-1, 1), x-coordinate is -1 unit and y-coordinate is 1 unit, so it lies in llnd quadrant. Similarly, we can plot all the points Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4), It is clear from the graph that points R and Q lie in fourth quadrant.

The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Description : The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Last Answer : Solution of this question

If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Description : If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Last Answer : (b) In point P (-1, 1), x-coordinate is -1 unit and y-coordinate is 1 unit, so it lies in llnd quadrant. Similarly, we can plot all the points Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4), It is clear from the graph that points R and Q lie in fourth quadrant.

The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Description : The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Last Answer : Solution of this question

A fair dice is thrown twenty times. The probability that on the tenth throw the fourth six appears is -Maths 9th

Description : A fair dice is thrown twenty times. The probability that on the tenth throw the fourth six appears is -Maths 9th

Last Answer : (c) \(rac{84 imes5^6}{6^{10}}\)In the first nine throws we should have three sixes and six non-sixes and a six in the tenth throw and thereafter whatever face appears, it doesn't matter. ∴ Required probability = 9C3 \(\bigg(rac{1} ... x 1 x 1 ............x 1 {10 times} = \(rac{84 imes5^6}{6^{10}}\).

In how many ways can a pack of 52 cards be divided into 4 sets, three of them having 16 cards each and the fourth just 4 cards? -Maths 9th

Description : In how many ways can a pack of 52 cards be divided into 4 sets, three of them having 16 cards each and the fourth just 4 cards? -Maths 9th

Last Answer : First we divide 52 cards into two sets which contains 1 and 51 cards respectively is 1! 51! 52! Now 51 cards can be divided equally in three sets each contains 17 cards (Here order of sets is not important) in 3!(17!) ... ways Hence, the required number of ways = 1! 51! 52! 3! (17!) 3 51!

A cylindrical rod of iron whose radius is one-fourth of its height is melted and cast into spherical balls of the same radius as that of the cylinder. -Maths 9th

Description : A cylindrical rod of iron whose radius is one-fourth of its height is melted and cast into spherical balls of the same radius as that of the cylinder. -Maths 9th

Last Answer : Let radius of cylindrical rod =r ⇒ height =4r Volume of cylindrical rod =πr2h =πr2(4r) =4πr3 Volume of spherical balls of radius r=34​πr3 No. of balls =34​πr34πr3​=

If A(3, 5), B(– 5, – 4), C(7, 10) are the vertices of a parallelogram taken in order, then the co-ordinates of the fourth vertex are: -Maths 9th

Description : If A(3, 5), B(– 5, – 4), C(7, 10) are the vertices of a parallelogram taken in order, then the co-ordinates of the fourth vertex are: -Maths 9th

Last Answer : (c) RhombusCo-ordinates of P are \(\bigg(rac{-1-1}{2},rac{-1+4}{2}\bigg)\)i.e, \(\big(-1,rac{3}{2}\big)\)Co-ordinates of Q are \(\bigg(rac{-1+5}{2},rac{4+4}{2}\bigg)\)i.e, (2, 4)Co-ordinates of R ... \sqrt{(2-2)^2+(4+1)^2}\) = \(\sqrt{25}\) = 5⇒ PR ≠ SQ ⇒ Diagonals are not equal ⇒ PQRS is a rhombus.

Find the ratio of the total surface area and lateral surface area of a cube. -Maths 9th

Description : Find the ratio of the total surface area and lateral surface area of a cube. -Maths 9th

Last Answer : Let a be the edge of the cube, then Total surface area = 6a2² and lateral surface area = 4a² Now ratio between total surface area and lateral surface area = 6a² : 4a² = 3 : 2

If the ratio of curved surface area and total surface area of a cylinder is 1 : 3, then find the volume of cylinder when the height is 2 cm. -Maths 9th

Description : If the ratio of curved surface area and total surface area of a cylinder is 1 : 3, then find the volume of cylinder when the height is 2 cm. -Maths 9th

Last Answer : Let the radius and height of the cylinder be r and h, respectively . Given that, Curved surface area / Total surface area = 1/3 ⇒ 2πrh / 2πr(h + r) = 1/3 ⇒ 3h = h + r ⇒ r = 2h = 4cm ∴ volume of cylinder πr2h = π × (4)2 × 2 = 32π cm3

If the ratio of curved surface area and total surface area of a cylinder is 1 : 3, then find the volume of cylinder when the height is 2 cm. -Maths 9th

Description : If the ratio of curved surface area and total surface area of a cylinder is 1 : 3, then find the volume of cylinder when the height is 2 cm. -Maths 9th

Last Answer : Let the radius and height of the cylinder be r and h, respectively . Given that, Curved surface area / Total surface area = 1/3 ⇒ 2πrh / 2πr(h + r) = 1/3 ⇒ 3h = h + r ⇒ r = 2h = 4cm ∴ volume of cylinder πr2h = π × (4)2 × 2 = 32π cm3

Find the ratio of surface area and volume of the sphere of unit radius. -Maths 9th

Description : Find the ratio of surface area and volume of the sphere of unit radius. -Maths 9th

Last Answer : Required ratio = 4πr2 / 4/3.πr3 = 3 x 4 x π x (1)2 / 4 x π x (1)3 = 3/1 (Since, r = 1) i.e., 3 : 1

Find the ratio of the lateral surface -Maths 9th

Description : Find the ratio of the lateral surface -Maths 9th

Last Answer : Lateral surface area of cube : Total surface area of cube = 4a2 : 6a2 = 2 : 3.

A sphere and a cube have the same surface area. What is the ratio of the square of volume of the sphere to the square of volume of the cube ? -Maths 9th

Description : A sphere and a cube have the same surface area. What is the ratio of the square of volume of the sphere to the square of volume of the cube ? -Maths 9th

Last Answer : answer:

If the distance between the earth and moon were halved, the force of the attraction between them would be: w) one fourth as great x) one half as great y) twice as great z) four times as great

Description : If the distance between the earth and moon were halved, the force of the attraction between them would be: w) one fourth as great x) one half as great y) twice as great z) four times as great 

Last Answer : ANSWER: Z -- FOUR TIMES AS GREAT

A round viewing window with a 20 centimeter diameter is installed in a tank at the Shedd Aquarium. It is 5 meters below the water surface. The force on the window is approximately: w) 1,500 Newtons x) 10,000 Newtons y) 25,000 Newtons z) 50,000 Newtons

Description : A round viewing window with a 20 centimeter diameter is installed in a tank at the Shedd Aquarium. It is 5 meters below the water surface. The force on the window is approximately: w) 1,500 Newtons x) 10,000 Newtons y) 25,000 Newtons z) 50,000 Newtons

Last Answer : ANSWER: W -- 1,500 NEWTONS

The areas of three adjacent faces of a cuboid -Maths 9th

Description : The areas of three adjacent faces of a cuboid -Maths 9th

Last Answer : Let, length, breadth and height of the cuboid be a, b and c respectively. ∴ Volume = abc Also, A1 = ab, A2 = bc , A3 = ca ∴ A 1.A 2.A 3 = (ab)(bc)(ca) = a2 b2 c2 = (abc)2 = V2 ⇒ V2 = A 1.A 2.A 3

In a histogram, the areas of the rectangles are proportional -Maths 9th

Description : In a histogram, the areas of the rectangles are proportional -Maths 9th

Last Answer : No. It is true only when the class sizes are the same.