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3 years ago

4 what is the difference between an array's size declarator and a subscript?

1 answer:

Ber [7]3 years ago

7 0

What is the difference between asize declarator and a subscript? The size declarator is ... When writing a function that accepts a two-dimensional array as an argument, which size declarator must you provide in the parameter for thearray? The second size ...

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A spaceship from a friendly, extragalactic planet flies toward Earth at 0.201 times the speed of light and shines a powerful las

bagirrra123 [75]

Answer:

The wavelength of observed light on earth is 568.5 nm

Explanation:

Given that,

Velocity of spaceship $v= 0.201c$

Wavelength of laser $\lambda= 697\ nm$

We need to calculate the wavelength of observed light on earth

Using formula of wavelength

$\lambda_{0}=\lambda_{e}\times\sqrt{\dfrac{1-\dfrac{v}{c}}{1+\dfrac{v}{c}}}$

$\lambda_{0}=697\times10^{-9}\times\sqrt{\dfrac{1-\dfrac{0.201 c}{c}}{1+\dfrac{0.201c}{c}}}$

$\lambda_{0}=697\times10^{-9}\times\sqrt{\dfrac{1-0.201}{1+0.201}}$

$\lambda=5.685\times10^{-7}\ m$

$\lambda=568.5\times10^{-9}\ m$

$\lambda=568.5\ nm$

Hence, The wavelength of observed light on earth is 568.5 nm

8 0

4 years ago

A person hangs from a nylon rope (Young's modulus of 5 x 109 N/m2) as seen in the picture below. The rope stretches by 2 % and h

disa [49]

Answer:

959183.7 kg

Explanation:

from the question we have :

young modulus = 5 x 10^{9} N/m^{2}

strain = 2% = 2÷100 = 0.02

diameter = 0.03 m

radius = 0.015 m

acceleration due to gravity (g) = 9.8 m/s^{2}

we can get the mass from the formula below

young modulus = stress ÷ strain

where

stress = \frac[force}{area} = \frac {mass x g}{area}

area = 2πr = 2π x 0.015 = 0.094

therefore

young modulus = \frac{\frac {mass x g}{area}}{strain}

5 x 10^{9} = \frac{\frac {mass x 9.8}{0.094}}{0.02}

mass = \frac{5 x 10^{9} x 0.02 x 0.094}{9.8}

mass = 959183.7 kg

8 0

3 years ago

In measuring the width of a hair sample, a light of wavelength 500 nm is used. The hair sample is 40 um in radius. With the scre

sergejj [24]

Answer:

The distance of the second dark band away from the central bright spot be located is $5.625\times10^{-2}\ m$

Explanation:

Given that,

Wave length = 500 nm

Radius $d= 40\ \mu m$

Distance from the hair sample D= 6 m

We need to calculate the distance of the second dark band away from the central bright spot be located

$\sin\theta=\dfrac{y}{D}$

$\sin\theta=\dfrac{y}{6}$

Using formula for dark fringe

$(n-\dfrac{1}{2})\lambda=2d\sin\theta$

Put the value into the formula

$(2-\dfrac{1}{2})\times500\times10^{-9}=2\times40\times10^{-6}\times\dfrac{y}{6}$

$y=\dfrac{(2-\dfrac{1}{2})\times500\times10^{-9}\times6}{2\times40\times10^{-6}}$

$y=0.05625\ m$

$y=5.625\times10^{-2}\ m$

Hence, The distance of the second dark band away from the central bright spot be located is $5.625\times10^{-2}\ m$

6 0

3 years ago

Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's m

ira [324]

Answer:

$\Delta L=15\,mm$

Explanation:

Given:

length of a steel-string, $L=1m$

area of the string, $A=0.5\,mm^2$

Young's modulus of the steel, $Y=2\times 10^{11} Pa$

force of tension on the string, $F=1500\,N$

We have the relation for change in length:

$\Delta L=\frac{F.L}{A.Y}$

$\Delta L=\frac{1500\times 1000}{0.5\times 10^{-6}\times 2\times 10^{11}}$

$\Delta L=0.015m$

$\Delta L=15\,mm$

6 0

3 years ago

3. As the mass of an object increases, the force of gravity

kondaur [170]

Answer:

As the mass of an object increases, the force of gravity increases as well.

Explanation:

Objects with more mass have more gravity. They work together.

4 0

3 years ago