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4 years ago

Some examples of solar ejections

1 answer:

8 0

Coronal Mass ejections or Solar ejections are activities on the surface of the sun termed to the reaction of the gas composition of the sun's surface attributing to explosion of these gases. One example most commonly known as Solar Flares, and another example is termed as erupting prominence.

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Amy put some salt in a weighing dish. Then she put the weighing dish on a sensitive balance to measure the mass of the salt. She

tatyana61 [14]

Answer:

Explanation:

i. Accuracy in measurement is a term that is used to describe the closeness of a measured value to its actual value. It relates the actual value to the measured values.

Therefore in the given question, Amy's measurement of the salt's mass is accurate because her final value is close to all the measured values.

ii. Precision in measurement is a term that is used to describe the closeness of different measured values to each other. It relates the keenness in the measured values.

Thus, Amy's measurements of the salt's mass is precise because the values of the masses are close to one another.

8 0

3 years ago

1. (I) If the magnetic field in a traveling EM wave has a peak magnitude of 17.5 nT at a given point, what is the peak magnitude

Elan Coil [88]

Answer:

The electric field is $E = 5.25 V/m$

Explanation:

From the question we are told that

The peak magnitude of the magnetic field is $B = 17.5 nT = 17.5 *10^{-9}\ T$

Generally the peak magnitude of the electric field is mathematically represented as

$E = c * B$

Where c is the speed of light with value $c = 3.0 *10^{8} \ m/s$

$E = 3.0 *10^{8} * 17.5 *10^{-9}$

$E = 5.25 V/m$

8 0

4 years ago

Find the values of the root mean square translational speed v, molecules in gaseous diatomic oxygen (O2), gaseous carbon dioxide

tiny-mole [99]

Answer:

For diatomic oxygen:V=539.06 m/s

For carbon dia oxide:V=459.71 m/s

For dia atomic hydrogen:V=2156.25 m/s

Explanation:

As we know that

Root mean square velocity V

$V=\sqrt{\dfrac{3RT}{M}}$

Where

R is the gas constant

$R=8.31\ \frac{kg.m^2}{s^2.mol.K}$

T is the temperature (K).

M is the molecular weight.

For diatomic oxygen:

M=32 g/mol

T=273+100 = 373 K

$R=8.31\ \frac{kg.m^2}{s^2.mol.K}$

$V=\sqrt{\dfrac{3RT}{M}}$

$V=\sqrt{\dfrac{3\times 8.31\times 373}{32\times 10^{-3}}}$

V=539.06 m/s

For carbon dia oxide

M=44 g/mol

T=273+100 = 373 K

$R=8.31\ \frac{kg.m^2}{s^2.mol.K}$

$V=\sqrt{\dfrac{3RT}{M}}$

$V=\sqrt{\dfrac{3\times 8.31\times 373}{44\times 10^{-3}}}$

V=459.71 m/s

For dia atomic hydrogen:

M= 2 g/mol

T=273+100 = 373 K

$R=8.31\ \frac{kg.m^2}{s^2.mol.K}$

$V=\sqrt{\dfrac{3RT}{M}}$

$V=\sqrt{\dfrac{3\times 8.31\times 373}{2\times 10^{-3}}}$

V=2156.25 m/s

3 0

4 years ago

Why is the work output always less than the work input?

Elena L [17]

Because energy is wasted either in the form of heat energy or something else (like friction)

8 0

3 years ago

Read 2 more answers

Doubly ionized lithium Li2+ (Z = 3) and triply ionized beryllium Be3+ (Z = 4) each emit a line spectrum. For a certain series of

EleoNora [17]

Answer:

tex]\lambda_{Be}[/tex] = 22.78 nm

Explanation:

Bohr's model for the hydrogen atom has been used by other atoms with a single electric charge by changing the number of charges by the charge of the new atom (atomic number)

$E_{n}$ = k e² / 2a₀ (1 /n²)

ao = h'² / k m e² h' = h/2πi

For another atom with a single electron in the last layer

a₀ ’= h’² / k m (Ze)²

a₀ ’= a₀ / Z²

Therefore, when replacing in the equation

$E_{n}$ = - Z² Eo/n²

E₀ = 13,606 eV

The transition occurs when the electron stops from one level to another

$E_{n}$ - $E_{m}$ = Z² E₀ (1 / n² - 1 / m²) = Z² ΔE

Let's relate this expression to the wavelength

c = λ f

E = h f

E = h c /λ

h c / λ = Z² ΔE

λ = 1 / Z² (hc / ΔE)

λ = 1 / Z² λ_hydrogen

Let's apply this last equation to our case

Lithium Z = 3

$E_{n}$ = - 9 Eo / n²

40.5 10-9 = 1/9 λ_hydrogen

Beryllium Z = 4

λ = 1/16 λ_hydrogen

Let's write our two equations is and solve

40.5 10-9 = 1/9 λ_hydrogen

tex]\lambda_{Be}[/tex] = 1/ 16 λ_hydrogen

40.5 10⁻⁹ = 1/9 (16 $\lambda_{Be}$ )

tex]\lambda_{Be}[/tex] = 40.5 9/16

tex]\lambda_{Be}[/tex] = 22.78 nm

6 0

3 years ago