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3 years ago

Given w= 4+,- 0.02 A=2.0+,-0.2 and 3.0+,-0.6 what is the value of w\ A y square

Physics

1 answer:

In-s [12.5K]3 years ago

8 0

Explanation:

w = (4.52 ± 0.02) cm, x = ( 2.0 ± 0.2) cm, y = (3.0 ± 0.6) cm. Find z = x + y - w and its uncertainty.

z = x + y - w = 2.0 + 3.0 - 4.5 = 0.5 cm

Dz = Dx + Dy + Dw = 0.2 + 0.6 + 0.02 = 0.82 rounding to 0.8 cm

So z = (0.5 ± 0.8) cm

Solution with standard deviations, Eq. 1b, Dz = 0.633 cm

z = (0.5 ± 0.6) cm

Notice that we round the uncertainty to one significant figure and round the answer to match.

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To apply the principle of superposition to overlapping waves, you should _____ of the individual waves.

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The answer is d I believe

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4 years ago

What is the energy Q released when 131 53Idecays and 131 54Xe is formed? The atomic mass of 131 53I is 130.906118 u and the atom

DanielleElmas [232]

Answer:0.967meV

Explanation:

-Find the difference in u, so 130.906118-130.90508= 0.001038u

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1 u = 931.494meV

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4 years ago

A light source shines light consisting of two wavelengths, λ1 = 540 nm (green) and λ2 = 450 nm (blue), on two slits separated by

densk [106]

Answer:

The maximun distance is $z_1 = z_2 = 0.0138m$

Explanation:

From the question we are told that

The wavelength are $\lambda _ 1 = 540nm (green) = 540 *10^{-9}m$

$\lambda_2 = 450nm(blue) = 450 *10^{-9}m$

The distance of seperation of the two slit is $d = 0.180mm = 0.180 *10^{-3}m$

The distance from the screen is $D = 1.53m$

Generally the distance of the bright fringe to the center of the screen is mathematically represented as

$z = \frac{m \lambda D}{d}$

Where m is the order of the fringe

For the first wavelength we have

$z_1 = \frac{m_1 (549 *10^{-9} * (1.53))}{0.180*10^{-3}}$

$z_1=0.00459m_1 m$

$z_1= 4.6*10^{-3}m_1 m ----(1)$

For the second wavelength we have

$z_2 = m_2 \frac{450*10^{-9} * 1.53 }{0.180*10^{-3}}$

$z_2 = 0.003825m_2$

$z_2 = 3.825 *10^{-3} m_2 m$ ----(2)

From the question we are told that the two sides coincides with one another so

$zy_1 =z_2$

$4.6*10^{-3}m_1 m = 3.825 *10^{-3} m_2 m$

$\frac{m_1}{m_2} = \frac{3.825 *10^{-3}}{4.6*10^{-3}}$

Hence for this equation to be solved

$m_1 = 3$

and $m_2 = 4$

Substituting this into the equation

$z_1 = z_2 = 3 * 4.6*10^{-3} = 4* 3.825*10^{-3}$

Hence $z_1 = z_2 = 0.0138m$

7 0

3 years ago

An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 2060 × 103 seconds (about

Fittoniya [83]

Answer:

Acceleration, $a=2.22\times 10^{-3}\ m/s^2$

Explanation:

It is given that,

Time period of revolution of the moon, $T=2060\times 10^3\ s$

If the distance from the center of the moon to the surface of the planet is, $h=235\times 10^6\ m$

The radius of the planet, $r=3.9\times 10^6\ m$

Let a is the moon's radial acceleration. Mathematically, it is given by :

$a=R\times \omega^2$ , R is the radius of orbit

Since, $\omega=\dfrac{2\pi}{T}$

The radius of orbit is,

$R=r+h$

$R=3.9\times 10^6\ m+235\times 10^6\ m=238900000\ m$

So, $a=\dfrac{4\pi^2 R}{T^2}$

$a=\dfrac{4\pi^2 \times 238900000}{(2060\times 10^3)^2}$

$a=2.22\times 10^{-3}\ m/s^2$

Hence, this is the required solution for the radial acceleration of the moon.

5 0

3 years ago

Is the wavelength comparable to the size of atoms?

Helen [10]

It totally depends on what kind of wave you're talking about.

-- a sound wave from a trumpet or clarinet playing a concert-A pitch is about 78 centimeters long ... about 2 and 1/2 feet. This is bigger than atoms.

-- a radio wave from an AM station broadcasting on 550 KHz, at the bottom of your radio dial, is about 166 feet long ... maybe comparable to the height of a 10-to-15-story building. This is bigger than atoms.

-- a radio wave heating the leftover meatloaf inside your "microwave" oven is about 4.8 inches long ... maybe comparable to the length of your middle finger. this is bigger than atoms.

-- a deep rich cherry red light wave ... the longest one your eye can see ... is around 750 nanometers long. About 34,000 of them all lined up will cover an inch. These are pretty small, but still bigger than atoms.

-- the shortest wave that would be called an "X-ray" is 0.01 nanometer long. You'd have to line up 2.5 billion of <u>those</u> babies to cover an inch. Hold on to these for a second ... there's one more kind of wave to mention.

-- This brings us to "gamma rays" ... our name for the shortest of all electromagnetic waves. To be a gamma ray, it has to be shorter than 0.01 nanometer.

Talking very very very very roughly, atoms range in size from about 0.025 nanometers to about 0.26 nanometers.

The short end of the X-rays, and on down through the gamma rays, are in this neighborhood.

5 0

4 years ago