- Question #5b97a + Example - Socratic
Here's what I got I don't think that you have enough information to provide a numerical solution, but you can find the mass of the gas that was released from the container in terms of P_(rho), the pressure at which the density of the gas is equal to "0 78 g L"^(-1) The first thing that you need to do here is to use the density of the gas, rho, to find its molar mass, let's say M_M, in terms
- Vivek R. on Socratic
High school student, law and history enthusiast, and aspiring SCOTUS justice
- Question #17b7c - Socratic
Using Gauss's Law for Gravity: int int_S mathbf g cdot d mathbfS = - 4 pi GM Where: S is any closed surface (the boundary of an arbitrary volume V), d mathbf A is the usual outward pointing surface area vector, mathbf g is the gravitational field, G is the universal gravitational constant, and M is the total mass enclosed within the surface S Say we are at radius R_o, inside the earth's
- Question #83f13 - Socratic
(1) Using Newton's Second Law of motion Force vecF=mveca=m (dvecv) (dt) where veca represents the acceleration of the particle and vecv its velocity In terms of acceleration equation (1) becomes (dvec v) dt = q m (vecE+vecv×vecB)
- Question #4b856 - Socratic
I think we have to model these as charged conducting spheres so the charges are at the surface as shown Gauss' Law tells us that: int int_S mathbf E cdot d mathbf S = (sum Q_(enc)) epsilon_o And by using a concentric Gaussian sphere, we can say of a general sphere of radius r that: mathbf E = (sum Q_(enc)) ( 4 pi epsilon_o r^2) For these conducting spheres, we have 2 different situations
- Question #43f8e - Socratic
LeChatelier's Principle => If a stress is applied to a reaction, the reaction will shift away from the applied stress and establish a new equilibrium Three 'stress' factors affect stability of an equilibrium 1 Changes in concentration of reactants or products, 2 Temperature changes, and 3 Pressure-Volume effects (Boyle's Law) While considering concentration effects, it is useful to
- Question #19cdd - Socratic
→ F = q→ v × → B Newton's second law of motion states that the force on the particle is equal to the rate of change of its momentum; ∴ → F = → p So we get → p = q→ v × → B As momentum → p = m→ v, if we take the dot product of both sides with → p
- Question #bded0 - Socratic
Explanation: We use Boyle's law, #P_1V_1=P_2V_2#, and solve for #P_1#
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