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Artyom0805 [142]
4 years ago
5

nciado: Frequentemente, em ciências exatas, utilizamos potências de base 10 para expressar números muito grandes ou muito pequen

os. Para as potências de base 10 mais utilizadas foram desenvolvidos os prefixos e símbolos. Por exemplo, o prefixo centi (simbolizado pela letra minúscula c) se refere à potência 10-2. Então, se fizermos uma medida de 0,01 metro, ela poderá ser escrita como 1.10-2m ou, então, como 1 cm (um centímetro). Existem vários outros prefixos, é importante saber usá-los corretamente e adotar a simbologia adequada. Dentre as opções abaixo, qual seria a opção correta para escrevermos a medida de 30 000 gramas (g) em potência de base 10? Escolha uma:
Physics
1 answer:
Mariulka [41]4 years ago
4 0
Hi. The language here looks as though it's Spanish/Portugese ??? It would help to answer the q if the q were posted in english. I speak a little spanish and french, but it's mostly guesswork.
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Answer:

Explanation:

Given

altitude of the Plane h=6\ miles

When Airplane is s=10\ miles away

Distance is changing at the rate of \frac{\mathrm{d} s}{\mathrm{d} t}=290\ mph

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h^2+x^2=s^2

differentiate above equation w.r.t time

2h\frac{\mathrm{d} h}{\mathrm{d} t}+2x\frac{\mathrm{d} x}{\mathrm{d} t}=2s\frac{\mathrm{d} s}{\mathrm{d} t}

as altitude is not changing therefore \frac{\mathrm{d} h}{\mathrm{d} t}=0

0+x\frac{\mathrm{d} x}{\mathrm{d} t}=s\frac{\mathrm{d} s}{\mathrm{d} t}

at s=10\ miles\ and\ h=6\ miles

substitute the value we get x=\sqrt{10^2-6^2}=8\ miles

8\times \frac{\mathrm{d} x}{\mathrm{d} t}=10\times 290

\frac{\mathrm{d} x}{\mathrm{d} t}=362.5\ mph

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Two charges, X and Y, are placed along the x-axis. Charge X is +18 nC and is placed at x = 0. Charge Y is placed at a location o
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Answer:

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m.

Explanation:

The Coulomb force between two charges, Q_1 and Q_2, separated by a distance, d, is given

F = k\dfrac{Q_1Q_2}{r^2}

<em>k</em> is a constant.

For the charge Z to be at equilibrium, the force exerted on it by charge X must be equal and opposite to the force exerted on it by charge Y.

It is to be placed along the <em>x</em>-axis. Hence, it is on the same line as charges X and Y.

Let the charge on Z be <em>Q</em>. It is positive.

Let the distance from charge X be <em>x m.</em> Then the distance from charge Y will be (0.60 - <em>x</em>) m.

Force due to charge X

F_X = k\dfrac{18Q}{x^2}

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x^2+2.40x-0.72 = 0

Applying the quadratic formula,

x = \dfrac{-2.40\pm\sqrt{2.40^2 - (4)(1)(-0.72)}}{2} = \dfrac{-2.40\pm\sqrt{8.64}}{2}

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