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ivann1987 [24]
3 years ago
12

How much current is running through a circuit powered by a 1.5 volt battery wit 18

Physics
1 answer:
FromTheMoon [43]3 years ago
6 0
Um honestly I’m not sure but I need points but good luck god be with you mate
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The table provided shows data collected during an experiment:
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Explanation:

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True or False. Light is a form of energy.
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What is the moon phase a week before waning crescent?
Charra [1.4K]

Waning Crescent lasts a week.

One week before any point in that week, it's a Waning Gibbous.

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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
a) Two protons in a molecule are 3.40 10-10 m apart. Find the electric force exerted by one proton on the other. magnitude 1.99e
kap26 [50]

Answer: a) F_{e}= 1.99 x 10⁻⁹N

              b) Electric force is 1.24 x 10³⁶ times larger than gravitational force

Explanation: <u>Electric</u> <u>Force</u> is an interaction between two charges at a distance. It is calculated as

F_{e}=k\frac{|q||Q|}{r^{2}}

k is electric constant that values 9 x 10⁹N.m²/C²

q and Q are the quantity of charges

r is distance between the charges

a) For the two protons, charge = 1.602 x 10⁻¹⁹C:

F_{e}=9.10^{9}\frac{(1.602.10^{-19})(1.602.10^{-19})}{(3.4.10^{-10})^{2}}

F_{e}= 1.99 x 10⁻⁹ N

Magnitude of electric force between 2 protons is 1.99 x 10⁻⁹ N and it is repulsive because they are both positive

<u>Gravitational</u> <u>Force</u> is the force of attraction that acts between all materials.

It is proportional to the object's masses and inversely proportional to their squared distance:

F_{g}=G\frac{mM}{r^{2}}

G is gravitational constant that values G = 6.67 x 10⁻¹¹ N.m²/kg²

b) For the two protons, m = 1.67 x 10⁻²⁷kg

F_{g}=6.67.10^{-11}\frac{(1.67.10^{-27})(1.67.10^{-27})}{(3.4.10^{-10})^{2}}

F_{g}= 1.61 x 10⁻⁴⁵N

Comparing electrical and gravitational forces:

\frac{F_{e}}{F_{g}} =\frac{1.99.10^{-9}}{1.61.10^{-45}}

\frac{F_{e}}{F_{g}}= 1.24 x 10³⁶

This shows that electrical force is <u>1.24 x 10³⁶</u> times greater than gravitational force.

4 0
3 years ago
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