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Setler [38]
3 years ago
15

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

Physics
1 answer:
Ber [7]3 years ago
7 0
What is the difference between<span> a</span>size declarator<span> and a </span>subscript<span>? The </span>size declarator<span> is ... When writing a function that accepts a two-dimensional </span>array<span> as an argument, which </span>size declarator<span> must you provide in the parameter </span>for<span> the</span>array<span>? The second size ...</span>
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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
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