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Margaret [11]
3 years ago
14

joe spend 8 on lunch and 6.50 on dry cleaning . He also buys 2 shirts that each cost the same smount. joe spend a total of $52 w

hat is the cost of each shirt
Physics
1 answer:
Gnesinka [82]3 years ago
6 0
8+6.5+2x=52
52/2=8+6.5+x
26=8+6.5+x
26-8-6.5=x
11.5=x
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A force Ě = F, î + Fy h acts on a particle
AVprozaik [17]

Answer:

Pt 1: W=39J

Pt 2: θ = 19.7

Explanation:

Part 1:

W=FS

W=F_{_xs_x}+F{_ys_y}

imagine the "i"and the "j" with the hat

W=(10i-1j)(4i+3j)

W=(10*4)-(1*1)

W=40-1

W=39J

Part 2:

|F|=\sqrt{10^2+(-1)^2}

|F|=\sqrt{101}

|S|=\sqrt{4^2+1^2}

|S|=\sqrt{17}

W=|F||S|cosθ

39=(\sqrt{101})(\sqrt{17} )cosθ

θ = 19.7

Hope it helps

8 0
3 years ago
A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i
Arturiano [62]

Answer:

q / q_{1} = 2, assuming that q_{1} and (q - q_{1}) are point charges.

Explanation:

Let k denote the coulomb constant. Let r denote the distance between the two point charges. In this question, neither k and r depend on the value of q_{1}.

By Coulomb's Law, the magnitude of electrostatic force between q_{1} and (q - q_{1}) would be:

\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}.

Find the first and second derivative of F with respect to q_{1}. (Note that 0 < q_{1} < q.)

First derivative:

\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}.

Second derivative:

\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}.

The value of the coulomb constant k is greater than 0. Thus, the value of the second derivative of F with respect to q_{1} would be negative for all real r. F\! would be convex over all q_{1}.

By the convexity of \! F with respect to \! q_{1} \!, there would be a unique q_{1} that globally maximizes F. The first derivative of F\! with respect to q_{1}\! should be 0 for that particular \! q_{1}. In other words:

\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0<em>.</em>

2\, q_{1} = q.

q_{1} = q / 2.

In other words, the force between the two point charges would be maximized when the charge is evenly split:

\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}.

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Reflecting telescopes are popular because they are more durable than a refracting telescopes.
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Reflecting telescopes also provides a good focus.
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3 years ago
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NNADVOKAT [17]

Answer: Joule

Explanation:

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