Physics, 11.11.2019 05:31 aons

Kinetic
which has more a 0.0014 bullet travel @ 396m/'s or 7.1 10^7kg ocean liner travel @ 13m/'s (25 knots)?

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Acetic acid /əˈsiːtɪk/, systematically named ethanoic acid /ˌɛθəˈnoʊɪk/, is a colourless liquid organic compound with the chemical formula ch3cooh (also written as ch3co2h or c2h4o2). when undiluted, it is sometimes called glacial acetic acid. vinegar is no less than 4% acetic acid by volume, making acetic acid the main component of vinegar apart from water. acetic acid has a distinctive sour taste and pungent smell. in addition to household vinegar, it is mainly produced as a precursor to polyvinyl acetate and cellulose acetate. it is classified as a weak acid since it only partially dissociates in solution, but concentrated acetic acid is corrosive and can attack the skin.acetic acidnamespreferred iupac nameacetic acidsystematic iupac nameethanoic acidother namesvinegar (when dilute); hydrogen acetate; methanecarboxylic acid[1][2]identifierscas number64-19-7 3d model (jsmol)interactive image3dmetb00009 reference506007chebichebi: 15366 chemblchembl539 chemspider171 drugbankdb03166 echa infocard100.000.528ec number200-580-7e numbere260 (preservatives)gmelin reference1380iuphar/bps1058keggd00010 meshacetic+acidpubchem cid176rtecs numberaf1225000uniiq40q9n063p un number2789inchiinchi=1s/c2h4o2/c1-2(3)4/h1h3,(h,3,4) key: qtbsbxvteameqo-uhfffaoysa-n smilescc(o)=o

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Are a.larger

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Potato’s are the right way to go promise not apples i’m 5

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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

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