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3 years ago

A standard 1 kilogram weight is a cylinder 40.0 mm in height and 53.5 mm in diameter. What is the density of the material?

Physics

1 answer:

jek_recluse [69]3 years ago

8 0

Answer:

1.15*10^-5 kg/m^3

Explanation:

Given data

mass= 1kg

hieght= 40 mm

diameter= 52.5mm

radius= 53.5/2= 26.25mm

The volume of the cylinder

V=πr^2h

V=3.142*26.25^2*40

V=3.142*689.0625*40

V=86601.375 mm^3

Density= mass/volume

Density= 1/86601.375

Density=0.00001154716

Density= 1.15*10^-5 kg/m^3

Hence the density is 1.15*10^-5 kg/m^3

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Ksenya-84 [330]

Answer:

A.2.95 m

B.7

Explanation:

We are given that

Diffraction grating=600 lines/mm

d= $\frac{1 mm}{600}=\frac{1\times 10^{-3} m}{600}=1.67\times 10^{-6} m$

Wavelength of light, $\lambda=510 nm=510\times 10^{-9} m$

l=4.6 m

A.We have to find the distance between the two m=1 bright fringes

$sin\theta=\frac{m\lambda}{d}$

For first bright fringe, =1

$sin\theta=\frac{1\times 510\times 10^{-9}}{1.67\times 10^{-6}}=0.305$

$\theta=sin^{-1}(0.305)=17.76^{\circ}$

The distance between two m=1 fringes

$x=2ltan\theta=2\times 4.6 tan(17.76^{\circ})=2.95 m$

Hence, the distance between two m=1 fringes=2.95 m

B.For maximum number of fringes,

$sin\theta=1$

$sin\theta=\frac{m\lambda}{d}$

Substitute the values

$1=\frac{m\times 510\times 10^{-9}}{1.67\times 10^{-6}}$

$m=\frac{1.67\times 10^{-6}}{510\times 10^{-9}}=3.3\approx 3$

Maximum number of bright fringes on the scree= $2m+1=2(3)+1=7$

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4 years ago