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inna [77]
3 years ago
10

Calculate the density 1.33*10⁻⁷ gcm⁻³ into kgm⁻³.

Physics
1 answer:
Alex Ar [27]3 years ago
5 0

1 Kg = 1,000g

Therefore 1.33 × 10^-7 g = 1.33 × 10^-10 kg

1m³ = 1,000,000cm³

Therefore 1 cm³ = 1 × 10^-6 m³

Dividing the mass per unit volume you get:

(1.33 × 10^-10 kg) ÷ (1 × 10^-6 m³)

= 1.33 × 10^(-10--6) = 1.33 × 10^(-10 + 6) = 1.33 × 10^-4 kg/m³

Density = 1.33 × 10^-4 kg/m³

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An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and
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Answer:

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Explanation:

Given that:

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The relation used in  calculating the contraction on the bar is:

\delta L = \dfrac{P *L }{A*E}

The relation used in  calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Total contraction on the Bar  = \dfrac{180 *10^3 \N  }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]

Total contraction on the Bar  = 2.117( 0.398 + 0.182)

Total contraction on the Bar  = 2.117*(0.58)

Total contraction on the Bar  = 1.22786 mm

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