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grigory [225]
2 years ago
11

Why did the author most likely write "Meeting Juli a

Physics
2 answers:
svet-max [94.6K]2 years ago
8 0

Answer:

To share a positive experience she had with a pen pal

Explanation:

Just read the story :)

Jobisdone [24]2 years ago
5 0

Answer:

c

Explanation:

c7uh grvv

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An electromagnet is a device in which moving electric charges (current) in a coil of wire create a magnet. What’s one advantage
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Electromagnets have certain advantages over permanent magnets. One of the major advantage is, the amount of electric current can easily be controlled, in turn, magnetic field can be controlled. Out of all the options, option D would be the best answer.

In short, Your Answer would be option D) Electromagnets can easily be turned on and off.

Hope this helps!
5 0
3 years ago
Can someone please help me it would mean alot​
Lunna [17]
It’s the wavelength I believe
7 0
3 years ago
What type of force act on a man at room temperature and in a ground?
lakkis [162]
Isaac Newton’s second law of motion
6 0
3 years ago
what is the energy (in j) of a photon required to excite an electron from n = 2 to n = 8 in a he⁺ ion? submit an answer to three
grin007 [14]

Answer:

Approximately 5.11 \times 10^{-19}\; {\rm J}.

Explanation:

Since the result needs to be accurate to three significant figures, keep at least four significant figures in the calculations.

Look up the Rydberg constant for hydrogen: R_{\text{H}} \approx 1.0968\times 10^{7}\; {\rm m^{-1}.

Look up the speed of light in vacuum: c \approx 2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}.

Look up Planck's constant: h \approx 6.6261 \times 10^{-34}\; {\rm J \cdot s}.

Apply the Rydberg formula to find the wavelength \lambda (in vacuum) of the photon in question:

\begin{aligned}\frac{1}{\lambda} &= R_{\text{H}} \, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\end{aligned}.

The frequency of that photon would be:

\begin{aligned}f &= \frac{c}{\lambda}\end{aligned}.

Combine this expression with the Rydberg formula to find the frequency of this photon:

\begin{aligned}f &= \frac{c}{\lambda} \\ &= c\, \left(\frac{1}{\lambda}\right) \\ &= c\, \left(R_{\text{H}}\, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\right) \\ &\approx (2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}) \\ &\quad \times (1.0968 \times 10^{7}\; {\rm m^{-1}}) \times \left(\frac{1}{2^{2}} - \frac{1}{8^{2}}\right)\\ &\approx 7.7065 \times 10^{14}\; {\rm s^{-1}} \end{aligned}.

Apply the Einstein-Planck equation to find the energy of this photon:

\begin{aligned}E &= h\, f \\ &\approx (6.6261 \times 10^{-34}\; {\rm J \cdot s}) \times (7.7065 \times 10^{14}\; {\rm s^{-1}) \\ &\approx 5.11 \times 10^{-19}\; {\rm J}\end{aligned}.

(Rounded to three significant figures.)

6 0
2 years ago
Which of the following describes the relationship between the weight of fluid
ioda
The answer is archimedes principle
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