Use the concept or conservation of energy and conservation of momentum to explain how you would find the final height of a wooden block pendulum that is shot with a bullet. You know all physical properties of the block and the bullet and the initial speed of the bullet.
Finding the final height of the bullet/block combination is a two part process. The first part uses conservation of momentum to find the speed of the bullet/block combination, and the second uses conservation of energy to find the final height of the bullet/block combination. When the bullet collides with the wooden block, it is a perfectly inelastic collision because the two colliding objects stick together. At this point, all of the energy in the system is in the form of kinetic energy (KE = .5mv^2). The bullet has an initial momentum that is found by multiplying its speed by its mass (given information), and if you add the mass of the bullet and the block, the speed after the collision can be found because momentum is conserved. After the collision, conservation of energy can be used to determine the final height of the block/bullet combination. At the top of the path, the pendulum will stop, so all of the kinetic energy has been converted into potential energy at this point. The kinetic energy found earlier can be set equal to the potential energy (PE = mgh), and since mass and gravity are known values, the final height can be found.
What determines the grain size of an igneous rock?
The grain size of an igneous rock is determined by the rate at which the magma or lava cools. Typically, when an igneous rock solidifies underground in an intrusion, the grains are large because it will take longer for the magma to cool. When the rock is extrusive and solidifies above ground, the grains will be small because the rock has cooled rapidly. Rock grains grow around points called nucleation sites, and the number of nucleation sites is determined by the number of stable grains that are formed. A grain becomes stable enough to continue growing once its radius reaches a certain size (the critical radius), and the grains will grow until they are stopped by another grain. In extrusive rocks, the grains solidify before they are able to reach the critical radius, so a lot of nucleation sites are present and are close together. An intrusive rock cools slowly enough to allow for some grains to reach the critical radius before melting, but because not all grains are able to reach the critical radius, the nucleation sites have a greater distance between them.
Integrate 7xe^x dx.
When presented with a problem that has a cyclical term--something that can be integrated over and over without disappearing--and some sort of polynomial, integration by parts is a useful tool. For this problem, you will need to use the method of integration by parts because e^x is a cyclical term and 7x is a polynomial term. In order to integrate by parts, the following equation is used: the integral of udv = uv - the integral of vdu. First, you will need to determine which part of the problem statement will be used as "u" and what will be used as "dv". The mnemonic "LIPET" can help you decide which term should be used as "u". LIPET stands for "ln (natural log), inverse trigonometric, polynomial, exponential, and trigonometric". In this problem, we will select 7x as the "u" and e^x as "dv". 7x is chosen because "exponential" is the first letter in the mnemonic to be satisfied by the terms in the problem. It will help to write this down on your paper: u= 7x v= du= dv= e^x dx (dx has to be put here because it is the derivative of something) At this point, you need to take the derivative of "u" to get "du", and take the integral of "dv" to get "v" u= 7x v= e^x du= 7dx dv= e^x dx Next, you plug all of the terms into the integration by parts equation: the integral of udv = 7xe^x -integral of 7e^x dx Since the polynomial term has been removed, you can now integrate 7e^x dx without using integration by parts a second time. The answer becomes 7xe^x- 7e^x, and you can check this answer by taking the derivative.