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Data: Youngs modulus of steel: 2.0 × 1011 Pa density of steel: 7850 kg m-3 basic SI units for pressure: kg m-1s-2 Assume Hookes law where applicable (a) A cylindrical wire of material with ultimate tensile strength (maximum tensile stress) 109 Pa, just breaks due to a tensile force of 50 N. Find the radius, in SI units, giving your answer in standard scientific (power of ten) notation, to 4 significant figures, with one significant figure to the left of the decimal point. (b) Find the spring constant for a stecl wire of cross-sectional area 10-4 m2 and length 0.5 m. (c) A spherical object has initial radius 0.50 m. Due to an increase in pressure of 1010 Pa, the final radius is 0.49 m. Find the bulk modulus of the material. (d) A cylindrical steel rod of cross-sectional area 10-4 m2 increases in lengtlh by 10-4 m when an external tensile force (load) of 104 N is applicd. Find the mass of the rod. (e) The bulk modulus B of a gas can be written 1 where ρ is density, M. is relative molecular mass, T is temperature and C is another quantity depending on the gas properties: Find the basic SI units of quantitity C.CAN I HAVE A DETAILED EXPLANATION FOR ALL OF THEM ( ESPECIALLY A AND D ) PLEASE ( CLEAR HANDWRITING)

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Answer #1

a)

P = tensile stress = 109 Pa

F = tensile force = 50 N

A = area of cross-section

r = radius

tensile stress is given as

P = F/A

inserting the values

109 = 50/A

A = 50 x 10-9 m2

Area of cross-section is given as

A = \pi r2

50 x 10-9 = (3.14) r2

r = 1.261 x 10-4 m

D)

Y = young's modulus = 2 x 1011 Pa

F = tensile force applied = 104 N

A = cross-sectional area = 10-4 m2

\DeltaL = change in length = 10-4 m

L = original length

using the equation

Y = FL/(A \Delta L)

2 x 1011 = (104) L /((10-4) (10-4))

L = 0.2 m

Volume is given as

V = A L

inserting the values

V = (10-4) (0.2)

V = 2 x 10-5 m3

\rho = density of wire = 7850 kgm-3

mass is given as

m = \rho V

m = (7850) (2 x 10-5)

m = 0.157 kg

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