#### • Class 11 Physics Demo

Explore Related Concepts

#### • elliptic geometry

Question:can nyone help with derivation of formula of following 3d objects: cube cuboid cylinder cone sphere

Answers:i denote pi with P. volume area cube a^3 6a^2 cuboid abc 2(ab+bc+ca) cylinder Pr^2h 2Pr^2+h(2Pr) cone (1/3)bh Pr^2+Prs sphere (4/3)Pr^3 4Pr^2

Question:Calculate the force on an end when the tank is half full of oil that weighs 60 pounds per cubic feet?

Answers:Total force is the integral of pressure with respect to area. http://en.wikipedia.org/wiki/Pressure Pressure is a function of of depth so the integral should look like: from z = 0 to z = h integral of P(z) dA You are given that the maximum depth 3 feet. The pressure is a linear function of the depth and is a function of the density of the fluid. It corresponds to the weight of the fluid above it. You are given that this oil weighs to pounds per cubic foot, so at a depth of z feet, the pressure is 60 z per square foot. At depth z the trough has a width w(z) so: dA = w(z) dz The tricky part of this question as opposed to the trough with a triangular cross-section is that w(z) is not a linear function of z. The ellipse has the formula (z/a)^2 + (y/b)^2 = 1 for some a and b which you can determine. Note that at depth = z = 0, the width is 12 feet so y (the distance to the axis of symmetry) and b must be 6. Similarly for z = 3 So, w(z) will have a term involving a square root. Check your integral table for how to integrate this if you don't already know. http://www.sosmath.com/tables/integral/integ12/integ12.html http://www.sosmath.com/tables/integral/integ13/integ13.html etc. Or look at: http://integrals.wolfram.com/index.jsp

Question:Please............I need it right away. Please make it simple and easy to understand. Thanks :) volume-sphere and square even if you just answer one of the above it will really help me a lot.

Answers:You haven't specified in what grade you are, so I will just be as simple as possible: Formula for acceleration: This is according to Newton's second law: F= m*a Where a is equal to the acceleration, F is equal to the force in Newton applied on the object which you are moving, and m = the mass of the object in kilograms. This is the second law of Newton. Newton's unit is the measurement for force. 1 Newton = 1 kg*1m/second square. This is how force is measured. I will give you a simple example: A car has a mass of 1000 kilograms; a force of 1000 Newtons is applied upon the car. What is the acceleration of the car as a result of this applied force neglecting the resistance coming from friction of the wheels or from the wind if it were windy that day! 1000 = 1000 *a a = 1000/1000 = 1 Newton/kilogram. This is equivalent to: 1 meter per second square acceleration unit is = 1 meter/second square. One Newton =1 kg.1m/second square. This is the unit of Force. Now, the equation of velocity which is the same as speed accept that velocity denotes that an object is moving with a certain speed and towards a certain direction, which we call it in Physics a "vector". To calculate the velocity to any given object, the equation of motion would be: V final = v initial + 1/2 a*t^2 where t is time and a is acceleration and v final is the final velocity and v initial is the initial velocity; the unit is meter per second. This can also be written as: V2^2 = v1^2 +2as, where s is the distance in meter attained, a is the acceleration. Now what else you need: Volume of course is the space that any object occupies. I will give you the volume of few objects: The volume of a sphere is 4/3*pi*r^3, where pi is 3.14, and r is the radius of the sphere. For a cube, the volume is = the length of its size to the third power, or a^3 For a cylinder, the volume is pi*r^2*h, where h is the height of the cylinder. For a pyramid, V= 1/3( area of the base * height) For a cone, the volume is 1/3*pi*r^3*h

Question:I've got to try and finish this calculus homework but I can't remember the formula.

Answers:Volume of a hemi-sphee is 2*pi*r^3/3 If it is thesurface area of a solid hemisphere that you want to know,it is 3*pi*r^2 Perimeter of a semi-circle is pi*r+2r