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ss7ja [257]
3 years ago
15

A robot arm that controls the position of a video camera in an automated surveillance system is manipulated by a servo motor tha

t exerts a force on a push-rod. The force is given by F(x) = (12.9 N/m2)x2 ,
where x is the position of the end of the pushrod. If the push-rod moves from x1 = 1 m to x2 = 4 m, how much work did the servo motor do? Answer in units of J.
Physics
1 answer:
lesya [120]3 years ago
6 0

Answer:

W = 270.9 J

Explanation:

given,  

F(x) = (12.9 N/m²) x²  

work = Force x displacement  

dW = F  .dx  

the push-rod moves from x₁= 1 m to x₂ = 4 m

integrating the above  

\int dW = \int_{x_1}^{x_2}F. dx

W = \int_{x_1}^{x_2}F. dx

W = \int_{1}^{4} (12.9 x^2) dx

W = 12.9\times [\dfrac{x^3}{3}]_1^4 dx

`W = 12.9\times [\dfrac{4^3}{3}-\dfrac{1^3}{3}] dx

W = 270.9 J

work done by the motor is W = 270.9 J

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Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

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Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

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Therefore

R=\rho\frac{l}{A}

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The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

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R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }

\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }......(1)

Again for tungsten:

R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }

\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }........(2)

Given that R_1=R_2   and    l_1=l_2

Dividing the equation (1) and (2)

\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}

\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}   [since R_1=R_2   and    l_1=l_2]

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}

\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

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