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2 years ago

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A 4.5 kg box is being pulled by a girl using a string that makes an angle of 20° horizontal. Find the force of friction if the f

orce exerted on the string is 15 N and the coefficient of friction is 0.5.

1 answer:

Mashutka [201]2 years ago

5 0

Answer:

Fcoso-f =ma. F= ma 15 = 4.5 a then get acceleration. Fcoso- umg = ma then get friction

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Solution :

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$E_P=\int \frac{KdQ}{(\sqrt{r^2+x^2})^2}\times \frac{x}{\sqrt{r^2+x^2}}$

$\vec{E_P}=\frac{Kx}{r^2+x^2} \int dQ$

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If we put an electron on point P, then force on point e is :

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Then, $\frac{-eKQ}{r^3}x$

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Compare, a = -ω²x

We get,

$\omega^2 = \frac{eKQ}{R^3m}$

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3 years ago

Football player A has a mass of 210 pounds and is running at a rate of 5.0 mi/hr. He collides with player B. Player B has a mass

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Answer:

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3 years ago

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3 years ago

Delvig [45]

Answer:

Approximately $73\; {\rm N}$ , assuming that the acceleration of this ball is constant during the descent.

Explanation:

Assume that the acceleration of this ball, $a$ , is constant during the entire descent.

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$\begin{aligned} a &= \frac{2\, x}{t^{2}}\end{aligned}$ .

Note that $x = 11\; {\rm m}$ and $t = 1.5\; {\rm s}$ in this question. Thus:

$\begin{aligned} a &= \frac{2\, x}{t^{2}} \\ &= \frac{2 \times 11\; {\rm m}}{(1.5\; {\rm s})^{2}} \\ &\approx 9.78\; {\rm m \cdot s^{-2}}\end{aligned}$ .

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Since $m = 7.5\; {\rm kg}$ and $a \approx 9.78\; {\rm m\cdot s^{-2}}$ , the net external force on this ball would be:

$\begin{aligned} (\text{net force}) &= m\, a \\ &\approx 7.5\; {\rm kg} \times 9.78\; {\rm m\cdot s^{-2}} \\ &\approx 73\; {\rm kg \cdot m \cdot s^{-2} \\ &= 73\; {\rm N} && (1\; {\rm N} = 1\; {\rm kg \cdot m\cdot s^{-2}}) \end{aligned}$ .

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2 years ago