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Wilburn Stoltenberg

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โˆ™ 2021-02-26 23:17:29
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Algebra

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A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

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โˆ™ 2012-02-23 18:50:13

lol do it yoursel'

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Theorem that states that an impulse produces a change in momentum?

It is called the momentum-impulse theorem and states that an impulse will change the momentum of an object. For example, if you drop an object when it hits the ground an impulse occurs. The momentum of the object also changes. Jnet = deltap, where deltap is the change in momentum.


What does it mean if the price elasticity of demand is 2?

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Can an exothermic reaction have positive enthalpy?

Typically, exothermic reactions have negative enthalpy, But this may not always necessarily be the case. To prove this we say that H=U+PV and U=Q+W (H=enthalpy, U=internal energy, P=pressure, V=volume, Q=heat, W=work). H=U+PV Substitute for U (see above) H=Q+W+PV Take the differential of dH: dH=dQ+dW+d(PV) Plug in for dW with dW=-P(deltaV)) (this is true for a ideal gas expanding against constant external pressure) dH=dQ-P(deltaV)+d(PV) Take the derivative of d(PV) using the chain rule dH=dQ-P(deltaV)+P(deltaV)+V(deltaP) Cancel like terms dH=dQ+V(deltaP) At constant pressure (which is what is normally observed in almost all naturally occurring environments the pressure is constant so the V(deltaP) term is 0 (change in volume is zero) and therefore dH=dQ. So in this case, if Q is negative so is enthalpy. However, if are at constant volume conditions (example: a bomb calormiter) where the pressure can change it may be possible to have a V(deltaP) term that is greater than the negative Q (indicating the reaction is exothermic while still giving you a positive enthalpy. Example: dQ=-1000J (Exothermic process) V=(5L) deltaP=2atm dH=dQ+V(deltaP) dH=-1000J+10L*atm=-1000J+1013.25J=13.25... (10Latm=1013.25J) dH is positive while the process is exothermic. I understand these conditions are extreme and unlikely, but they are being used to illustrate a point - mathematically, and potentially IRL, it is plausible to have an exothermic process with a positive change in enthalpy.


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