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Step2247 [10]
4 years ago
9

A man pulls on a (massless) rope tied to a tree with a force of 500 N. Later, two men pull on opposite ends of the same rope wit

h opposing forces of 500 N each. The tension in the rope is __________.
Physics
2 answers:
qwelly [4]4 years ago
7 0

Answer:

  • <em>In both cases the tension in the rope is </em><u>equal to 500N</u>

Explanation:

It may be that in the case of the <em>tree</em>, the result is more intuitive, because you can think that there is only one force. But this is misleading.

To find the <em>tension in the rope</em>, you should draw a free body diagram. By doing so, you would find that the rope is static because there are two opposite forces. Assuming, for simplicity, that the rope is horizontal,  a force of 500N is pulling to one direction (let's say to the right) and a force of 500N is pulling to the opposite direction (to the left). Else, the rope would not be static.

That analysys is the same for the<em> rope tied to the tree</em> ( the tree is pulling with 500N, such as the man, but in opposite direction) and when the rope is pulled  by <em>two men</em> on opposite ends, each with<em> forces of 500N.</em>

Hence, the tension is the same and equal to 500N.

garri49 [273]4 years ago
3 0

Answer:

Explanation:

In first case, the tension is equal to the applied force that means it is 500 N.

Now two persons pull the rope with a force of 500 N each, so the tension in each rope is also 500 N.

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Answer:

\begin{array}{lll}&\underline{Object \ at \ more \ than \ 2\times Focus} & \underline{Object   \ at \ less \ than \  Focus}\\\\1.  \ Location \ of \ image &Same \ side \ as \ object&Opposite \ side \ of \ lens\\\\2. \ Orientation \ of \ image &Inverted&Upright\\\\3. \ Type \ of \ image&Real&Virtual\\\\4. \ Size\ of \ image&Smaller \ than \ object& Larger \ than \ object\end{array}

Explanation:

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Answer:

charge      Qint = 7.17 10⁻⁴ C

Explanation:

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let's form a Gaussian surface that is parallel to the surface, for example, a Cube. As the field is vertical and perpendicular to the surface, the field lines and the area vector are parallel whereby the scalar product is reduced to an ordinary product.

    Φ = E A = Qint / ε₀

   

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We can calculate the charge

    Qint = E A ε₀

    Qint = 81 1 10⁶ 8.85 10⁻¹²

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The answer to this question is:

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Your Welcome :)
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