Physics, 18.01.2020 03:31 cjrogers4401

Two security professionals - jo and chris - are discussing the contracting process.

jo says that the federal acquisition regulation governs the process the federal government uses to acquire or purchase goods and services.

chris says that although the federal acquisition regulation's intent is to provide uniform and government-wide policies and procedures for acquisition, the department of
defense has issued a supplemental acquisition regulation called the dfar.

who is correct?
a. jo is correct
b. chris is correct
c. jo and chris are both correct
d. jo and chris are both incorrect

Answers: 2

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Answer from: Quest

the components of a vector are the parts of the vectors that are to each other.

b perpendicular

Answer from: Quest

An example of environmentally learned behavior is aggressive behavior, i grew up seeing many aggressive people in my life and a slowly started picking up those traits however i was able to call myself out and not give in to those traits.

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

Answer from: Quest

answer: the potential energy decreases from 379 j to 243 j

explanation:

for the law of conservation of energy, the total energy in a system must be conserved. the total energy of the system is given by the sum of kinetic (k) and potential (u) energy:

e = k + u

the total energy is e=675 j, and this does not change.

initially, the kinetic energy is k=296 j, so the potential energy is

$u_i = e-k_i = 675 j-296 j=379 j$

at the end, the kinetic energy is k=432 j, so the potential energy is

$u_f = e-k_f = 675 j-432 j=243 j$

so, we see that the potential energy has decreased from 379 j to 243 j.

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Two security professionals - jo and chris - are discussing the contracting process.

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