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yanalaym [24]
4 years ago
14

What is the work done by a 20 newton force applied at an angle of 45.0° to move a box a horizontal distance of 40 meters?

Physics
2 answers:
bearhunter [10]4 years ago
8 0
Your answer would be B.

<span>votre réponse serait B</span>
Artist 52 [7]4 years ago
3 0

Explanation :

Given that,

Force, F = 20 N

The force is applied at an angle of 45 degrees to move a box a horizontal distance of 40 meters. We have to find the work done by the force. It is given by :

W = F. d

W=Fd\ cos\theta

W=20\ N\times 40/ m cos(45)

W = 565.6 Joules

or

W=5.6\times 10^2\ Joules

Hence, this is the required solution.

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The mainly property of the conservative fields is \vec{\nabla} \times \vec E=\vec 0

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and the curl expression is:

\nabla\times \vec E=\frac{1}{r^2{\sin}\,\theta}\left|\begin{matrix}\hat{r} & r\,\hat{\theta} & r\,{\sin}\,\theta\,\hat{\varphi}  \\& & \\\frac{\partial}{\partial r} & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial \varphi}\\ & & \\E_r & rE_\theta & r{\sin}\,\theta\, E_\varphi\end{matrix}\right|=(0, 0, 0)

to find the expression for the potential function associated:

\vec E=\vec \nabla . V, \Delta V= V_b-V_a=-\int _c \vec E.d\vec l=-\int _c E\^r.dr\^r=-\int _c Edr=\int \limits^a_b \frac{Q}{4\pi \epsilon _0 r^2} dr= \frac{Q}{4\pi \epsilon _0}.(\frac{1}{r}|^b_a)= \frac{Q}{4\pi \epsilon _0}.(\frac{1}{b}-\frac{1}{a})

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A ray of light bends upon entering a new medium. What is this phenomenon called?
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An object has a mass of 6kg. calculate it's gpe​
m_a_m_a [10]

Explanation:

When m=<em>mass</em>

G=<em>a</em><em>c</em><em>c</em><em>e</em><em>l</em><em>e</em><em>r</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em> </em><em>d</em><em>u</em><em>e</em><em> </em><em>t</em><em>o</em><em> </em><em>gravity</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>H</em><em>=</em><em>h</em><em>e</em><em>i</em><em>g</em><em>h</em><em>t</em>

<em>U</em><em>s</em><em>i</em><em>n</em><em>g</em><em> </em><em>f</em><em>o</em><em>r</em><em>m</em><em>u</em><em>l</em><em>a</em>

<em>M</em><em>g</em><em>h</em>

<em>(</em><em>M</em><em>=</em><em>6</em><em>, </em><em>g</em><em>=</em><em>10</em><em>,</em><em>h</em><em>=</em><em>?</em><em>) </em>

6×10×h

=60joules

7 0
3 years ago
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