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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.
Answer:

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.
Answer:

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.
Answer:

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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