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# Tutor profile: Rana K.

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Rana K.
Friendly, Energetic, and Highly Skilled Physics and Mechanical Engineering subjects Tutor
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## Questions

### Subject:Basic Math

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Question:

A rectangular building lot has a width of 32.0 m and a length of 45.0 m. Determine the area of this lot in square meters.

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Rana K.

Given in the question, Width of the rectangular building lot, w = 22.0 m Length of the rectangular building plot, l = 35.0 m We will use the area of a rectangle to find the area of the given rectangular lot, it is given as $$A = w\times l$$ Where, A is the area Substituting given values in the above formula $$A = 22.0\times 35.0$$ $$A = 770.0\ m^2$$ Thus, the area of the given building lot will be $$A = 770.0\ m^2$$ (Answer)

### Subject:Mechanical Engineering

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Question:

Calculate the natural frequency in rad/s of a hanging spring-mass system, given m = 8 kg and k = 1 kN/m. Write your answer in seconds to 2 decimal places.

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Rana K.

We have the following known information, Mass m = 8 kg Spring constant $$k = 1 \ kN/m = 1\times1000 \ N/m = 1000 \ N/m$$ (The symbol "k" (kilo) is a decimal unit prefix in the metric system denoting multiplication by 1000) The formula for the natural frequency of a simple spring-mass system is given by $$\omega_n = \sqrt{\frac{k}{m}}$$ Where, $$\omega_n$$ is the natural frequency Substituting known values in the above formula $$\omega_n = \sqrt{\frac{1000}{8}}$$ $$\omega_n = \sqrt{125}$$ $$\omega_n = 11.1803\ rad/s$$ Rounding to 2 decimal places $$\omega_n = 11.18\ rad/s$$ Thus, the natural frequency of the given spring-mass sytem will be $$\omega_n = 11.18\ rad/s$$ (Answer)

### Subject:Physics

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Question:

A metal wire 750 mm long and 1.30 mm in diameter stretches 0.35 mm when an external force of 80 N is applied on its ends. Find the (a) stress, (b) strain, and (c) Young's modulus for the material of the wire.

Inactive
Rana K.

We have the following known information, Length of the metal wire $$L = 750\ mm$$ Diameter of the metal wire $$d = 1.30\ mm$$ Change in length of the metal wire $$\Delta L = 0.35 \ mm$$ Magnitude of the applied force $$F = 80\ N$$ a) The formula for the stress is given as $$\sigma = \frac{F}{A}$$ ------(1) Where, A is the cross-sectional are of the wire given by the formula $$A =\frac{\pi}{4}d^2$$ -------(2) Using equation (2) in equation (1) $$\sigma = \frac{F}{ \frac{\pi}{4}d^2}$$ $$\sigma = \frac{4F}{ \pi d^2}$$ Substituting known values in the above derived formula $$\sigma = \frac{4\times80}{ \pi\times 1.30^2}$$ $$\sigma = 60.3 \ N/mm^2$$ We know 1 MPa (Mega Pascal) = 1 $$N/mm^2$$ Thus, stress developed in the metal wire will be $$\sigma$$ = 60.3 MPa (Answer) b) Now, the formula for the strain is given as $$\epsilon = \frac{\Delta L}{L}$$ Substituting known values in the above formula $$\epsilon = \frac{0.35}{750}$$ $$\epsilon = 4.67\times10^{-4}$$ Thus, the strain in the metal wire will be $$\epsilon = 4.67\times10^{-4}$$ (Answer) c) Now, the formula for the Young's modulus is given as $$E = \frac{\sigma}{\epsilon}$$ Substituting calculated values in the above formula $$E = \frac{60.3}{4.67\times10^{-4}}$$ E = 129122.0 Mpa Thus, the Young's modulus of the metal wire will be E = 129122.0 Mpa (Answer)

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