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4 years ago

If you were looking for sodium in the solar spectrum, it would be a[n] _______ line. blue

2 answers:

8 0

<span>We are tasked to fill in the right term which describes the blank space provided in the question if you were looking for sodium in the solar spectrum, it would be an absorption line. The line was originally seen as a dark light which in turn realized by Joseph Fraunhofer that it is an absorption line which he called it Fraunhofer lines.</span>

Crank4 years ago

5 0

In solar spectrum when looking for sodium, it would be a (n) absorption line. A gas of hydrogen will produce absorption spectrum it is between teloscope and spectrograph and a continuum light source. An emission line is spectrum is viewed from a different angle.

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Write a short summary paragraph that could go at the end of the text

UNO [17]

Answer:

The electro magnesium of the atoms cause the magnetic force to charge thus creating electricity. just kidding

Explanation:

If you read on the second paragraph it states that magnesium charged created electricity.

3 0

3 years ago

Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect

Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a) 0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b) r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 $\pi$ $r^{2}$ = $\frac{Q1}{E_{0} }$

So,

Rearranging the above equation to get Electric field, we will get:

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$

Multiply and divide by $r1^{2}$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ x $\frac{r1^{2} }{r1^{2} }$

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

c) r > r2 :

Electric Field = ?

E x 4 $\pi$ $r^{2}$ = $\frac{Q1 + Q2}{E_{0} }$

Rearranging the above equation for E:

E = $\frac{Q1+Q2}{E_{0} . 4 \pi. r^{2} }$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

As we know from above, that:

$\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Then, Similarly,

$\frac{Q2}{E_{0} . 4 \pi. r^{2} }$ = (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

So,

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

Replacing the above equations to get E:

E = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ ) + (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Now, for

d) Under what conditions, E = 0, for r > r2?

For r > r2, E =0 if

σ1 x $r1^{2}$ = - σ2 x $r2^{2}$

4 0

3 years ago

Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.8 x 105 kg on springs that hav

Vikki [24]

Answer:

The force constant is $k =1.316 *10^{7} \ N/m$

The energy stored in the spring is $E = 1.68 *10^{7} \ J$

Explanation:

From the question we are told that

The mass of the object is $M = 4.8*10^{5} \ kg$

The period is $T = 1.2 \ s$

The period of the spring oscillation is mathematically represented as

$T =2 \pi \sqrt{ \frac{M}{k}}$

where k is the force constant

So making k the subject

$k = \frac{4 \pi ^2 M }{T^2}$

substituting values

$k = \frac{4 (3.142) ^2 (4.8 *10^{5}) }{(1.2)^2}$

$k =1.316 *10^{7} \ N/m$

The energy stored in the spring is mathematically represented as

$E = \frac{1}{2} k x^2$

Where x is the spring displacement which is given as

$x = 1.6 \ m$

substituting values

$E = \frac{1}{2} (1.316 *10^{7}) (1.6)^2$

$E = 1.68 *10^{7} \ J$

7 0

3 years ago

Please help

vladimir2022 [97]

Answer:

true true false true false

5 0

2 years ago

Read 2 more answers

To develop muscle tone, a woman lifts a 2.50 kg weight held in her hand. She uses her biceps muscle to flex the lower arm throug

Romashka [77]

To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

$\tau = Fr$