Water pumps use electrical energy to create mechanical energy needed to move water.

What is/are the difference between wavelength and spectral lines?

Gre4nikov [31]

Answer:

Definition of spectral line: one of a series of linear images formed by a spectrograph or similar instrument and corresponding to a narrow portion of the spectrum of the radiation emitted or absorbed by a particular source.

Definition of Wavelength: can be defined as the distance between two successive crests or troughs of a wave. It is measured in the direction of the wave. ... Wavelength is inversely proportional to frequency. This means the longer the wavelength, lower the frequency.

So, the spectrum is the range of wavelength in visible light. While, wavelength is the length of a wave.

Explanation:

I hope this helps!

8 0

2 years ago

A damped mass/spring system takes 14.0 s for its amplitude of the oscillator to decrease by a factor of 9. By what factor does t

fiasKO [112]

Answer:

The correct answer is "0.246".

Explanation:

Given that the amplitude is decreased by a factor of 9, then

$A \rightarrow (A-\frac{A}{9} )$

$A \rightarrow \frac{8A}{9}$

As we know,

Energy will be:

⇒ $E_{1}=\frac{1}{2}KA^2$

and,

⇒ $E_{2}=\frac{1}{2}K(\frac{8A}{9} )^2$

$=\frac{64KA^2}{162}$

⇒ $\Delta E=E_1-E_2$

On putting the estimated values, we get

$=\frac{1}{2}KA^2-\frac{64KA^2}{162}$

⇒ $\frac{\Delta E}{E}=\frac{\frac{20}{162}KA^2}{\frac{1}{2}KA^2}$

$=\frac{40}{162}$

$=0.246$

3 0

3 years ago

Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ

bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; $H_0 = 51km/s/Mly$

Age of the universe; $t = \ ?$

We know that, the reciprocal of the Hubble's constant ( $H_0$ ) gives an estimate of the age of the universe ( $t$ ). It is expressed as:

$Age\ of\ Universe; t = \frac{1}{H_0}$

Now,

Hubble's constant; $H_0 = 51km/s/Mly$

We know that;

$1\ light\ years = 9.46*10^{15}m$

so

$1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m$

Therefore;

$H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly} \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 = 5.39 *10^{-18}s^{-1}\\$

Now, we input this Hubble's constant value into our equation;

$Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t = \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years$