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lorasvet [3.4K]
3 years ago
6

An example of a Natural electric field is

Physics
2 answers:
liberstina [14]3 years ago
6 0

Answer:

Lightning

Explanation

user100 [1]3 years ago
5 0
Possibly an eel
I’m not too sure ... but definitely lightning
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The owners want the speed at the bottom of the first hill doubled. How much higher must the first hill be built?
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Neglecting friction and air resistance, the first hill must be built 4 times higher than it is now.
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What does it mean when it says a scientific question must be supportable
Shkiper50 [21]

A good scientific question has certain characteristics. It should have some answers (real answers), should be testable (can be tested by someone through an experiment or measurements), leads to a hypothesis that is falsifiable (means it should generate a hypothesis that can be shown to fail), etc.

3 0
3 years ago
A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou
Setler [38]

Answer:

a. wavelength of the sound, \vartheta = 1.315\vartheta_{o}

b. observed frequecy, \lambda = 0.7604\lambda_{o}

Given:

speed of sound source, v_{s} = 80 m/s

speed of sound in air or vacuum, v_{a} = 343 m/s

speed of sound observed, v_{o} = 0 m/s

Solution:

From the relation:

v = \vartheta \lambda        (1)

where

v = velocity of sound

\vartheta = observed frequency of sound

\lambda = wavelength

(a) The wavelength of the sound between source and the listener is given by:

\lambda = \frac{v_{a}}{\vartheta }         (2)

(b) The observed frequency is given by:

\vartheta = \frac{v_{a}}{v_{a} - v_{s}}\vartheta_{o}

\vartheta = \frac{334}{334 - 80}\vartheta_{o}

\vartheta = 1.315\vartheta_{o}                (3)

Using eqn (2) and (3):

\lambda = \frac{334}{1.315} = \frac{1}{1.315}\frac{v_{a}}{\vartheta_{o}}

\lambda = 0.7604\lambda_{o}

4 0
3 years ago
Which is a freshwater region characterized as a large standing body of water? A. lake B. river C. ocean D. stream
Alexus [3.1K]
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4 0
3 years ago
Read 2 more answers
Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed o
kobusy [5.1K]

Answer:

\Delta \lambda=14.3\ nm

Explanation:

It is given that,

The number of lines per unit length, N = 900 slits per cm

Distance between the formed pattern and the grating, l = 2.3 m

n the first-order spectrum, maxima for two different wavelengths are separated on the screen by 2.98 mm, \Delta Y=2.98\ mm = 0.00298\ m

Let d is the slit width of the grating,

d=\dfrac{1}{N}

d=\dfrac{1}{900\ cm}

d=1.11\times 10^{-5}\ m

For the first wavelength, the position of maxima is given by :

y_1=\dfrac{L\lambda_1}{d}

For the other wavelength, the position of maxima is given by :

y_2=\dfrac{L\lambda_2}{d}

So,

\Delta \lambda=\dfrac{\Delta y d}{l}

\Delta \lambda=\dfrac{0.00298\times 1.11\times 10^{-5}}{2.3}

\Delta \lambda=1.43\times 10^{-8}\ m

or

\Delta \lambda=14.3\ nm

So, the difference between these wavelengths is 14.3 nm. Hence, this is the required solution.

3 0
2 years ago
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