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3 years ago

What are continuous, emission, and absorption spectra? How are they produced?

Physics

1 answer:

Svetradugi [14.3K]3 years ago

3 0

Answer: An emission line occurs when an electron drops down to a lower orbit around the nucleus of an atom and looses energy.

An absorption line also occurs when any electron move to a higher orbit by absorbing energy.

Each atom has a unique way of using some space of the orbits and can absorb only certain energies or wavelengths.

Explanation:

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The opening to a cave is a tall, 30.0-cm-wide crack. A bat is preparing to leave the cave emits a 30.0 kHz ultrasonic chirp. How

Vlada [557]

Answer:

The value is $w = 7.54 \ m$

Explanation:

From the question we are told that

The length of the crack is $a = 0.3 \ m$

The frequency is $f = 30.0 \ kHz = 30 *10^{3} \ Hz$

The distance outside the cave that is being consider is $D = 100 \ m$

The speed of sound is $v_s = 340 \ m/s$

Generally the wavelength of the wave is mathematically represented as

$\lambda = \frac{v}f}$

=> $\lambda = \frac{340 }{30*10^{3}}$

=> $\lambda = 0.0113 \ m/s$

Generally for a single slit the path difference between the interference patterns of the sound wave and the center is mathematically represented as

$y = \frac{ n * \lambda * D}{a}$

=> $y = \frac{ 1 * 0.0113 * 100}{0.3}$

=> $y = 3.77 \ m$

Generally the width of the sound beam is mathematically represented as

$w = 2 * y$

=> $w = 2 * 3.77$

=> $w = 7.54 \ m$

4 0

3 years ago

g A loop circuit has a resistance of R1 and a current of 2 A. The current is reduced to 1.4 A when an additional 2.2 Ω resistor

Mrrafil [7]

Answer:

R1 = 5.13 Ω

Explanation:

From Ohm's law,

V = IR............... Equation 1

Where V = Voltage, I = current, R = resistance.

From the question,

I = 2 A, R = R1

Substitute into equation 1

V = 2R1................ Equation 2

When a resistance of 2.2Ω is added in series with R1,

assuming the voltage source remain constant

R = 2.2+R1, and I = 1.4 A

V = 1.4(2.2+R1)................. Equation 3

Substitute the value of V into equation 3

2R1 = 1.4(2.2+R1)

2R1 = 3.08+1.4R1

2R1-1.4R1 = 3.08

0.6R1 = 3.08

R1 = 3.08/0.6

R1 = 5.13 Ω

6 0

3 years ago

Write a hypothesis about the use of an object’s physical characteristics to determine its density. Use the format "if . . . then

mash [69]

For this case you must first know the definition of density.
D = m / v
where,
m: mass
v: volume.
You can then write the following hypothesis:
IF you know two physical characteristics of an object then you can determine the density. First weigh the object, THEN measure its volume BECAUSE the density is the quotient between the mass and the volume of an object.

5 0

3 years ago

Read 2 more answers

An electron is a negatively charged particle that has a charge of magnitude, e - 1.60 x 10-19 C. Which one of the following stat

Ivahew [28]

Answer:

The electric field is directed toward the electron and has a magnitude of $E=\frac{ke}{r^{2} }$ .

Explanation:

An electric field is define as the surrounding of charges which exert a force on each other and this force can be attractive or repulsive depends on the charge.

In the given case electron is given and the magnitude of charge on electron is $e=1.6\times 10^{-19}C$

Electric field can be represented as,

$E=\frac{kQ}{r^{2} }$

Here, r is the distance between the point ande charge, k is the electric field constant and Q is the charge.

In the given question an electron is given so electric field will be,

$E=\frac{ke}{r^{2} }$

As we know that electric field start from the positive charge and vanish in the negative charge.

So, here the electric field will be $E=\frac{ke}{r^{2} }$ and it is directed toward the electron because of negative charge on the electron.

6 0

2 years ago

What is the approximate uncertainty in the area of a circle of radius 4.5×10^4 cm? Express your answer using one significant fig

Igoryamba

Answer:

$A=6.36\times 10^5\ m^2$

Explanation:

It is given that,

Radius of the circle, $r=4.5\times 10^4\ cm=4.5\times 10^2\ m$

The area of the circle is given by :

$A=\pi r^2$

$A=\pi (4.5\times 10^2\ m)^2$

$A=636172.51\ m^2$

$A=6.36\times 10^5\ m^2$

As there is no uncertainty given in the radius of the circle. So, the area of the circle is $6.36\times 10^5\ m^2$ . Hence, this is the required solution.

5 0

3 years ago