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1 year ago

A homogeneous 800kg bar AB is supported at either end by a cable asshown in the figure

1 answer:

4 0

The smallest area of each cable if the stress is not to exceed 90MPa in bronze is 43.6 mm² and 120MPa in steel is 32.7 mm².

<h3>What is normal stress?</h3>

If the direction of deformation force is perpendicular to the cross-sectional area of the body, the stress is called normal stress. Changes in wire length and body volume will be normal.

σ = P/A

Where, σ = Normal stress

P = Pressure

A = Area

1 Kg = 9.81 N

800 kg = 7848 N

Since the rod is half bronze and half steel

800 kg = 7848/2

= 3924 N

Pₙ = Fₙ = 3924 N [n = Bronze]

Pₓ = 3924 N [x = steel]

Given,

σₙ = 90MPa

σₓ = 120MPa

Aₙ = ?

Aₓ = ?

Aₙ = Pₙ/σₙ

Aₙ = 3924/90

Aₙ = 43.6 mm²

Aₓ = Pₓ/σₓ

Aₓ = 3924/120

Aₓ = 32.7 mm²

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A 15.00 mL sample of a solution of H2SO4 of unknown concentration was titrated with 0.3200M NaOH. the titration required 21.30 m

natima [27]

Answer:

a. 0.4544 N

b. $5.112 \times 10^{-5 M}$

Explanation:

For computing the normality and molarity of the acid solution first we need to do the following calculations

The balanced reaction

$H_2SO_4 + 2NaOH = Na_2SO_4 + 2H_2O$

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= 0.006816 mol

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$= \frac{0.006816}{2}$

= 0.003408 mol

$Mass\ of\ H_2SO_4 = moles \times molecular\ weight$

$= 0.003408\ mol \times 98g/mol$

= 0.333984 g

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$= \frac{0.333984 g}{49 \times 0.015}$

= 0.4544 N

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= 0.00005112

= $5.112 \times 10^{-5 M}$

We simply applied the above formulas

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