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ss7ja [257]
4 years ago
5

Choose two forces and compare and contrast these forces. These must be different forces than used in the prior question. Provide

two ways that they are similar and two ways that they are different. You may make a list, write it out, or make a chart.
Physics
1 answer:
igor_vitrenko [27]4 years ago
6 0
There are 4 forces. These are 1) Gravity, 2) Weak Nuclear Force, 3) Electromagnetism, and 4) Strong Nuclear Force. 

Order of strength from weakest to strongest: Gravity, Weak Nuclear Force, Electromagnetism, Strong Nuclear Force

Type of Range:
Gravity - Unlimited range
Weak Nuclear Force - Limited range
Electromagnetism - Infinite range
Strong Nuclear Force - Limited Range

Found in:
Gravity - Exists between all objects with mass
Weak Nuclear Force - Governs over beta decays like the emission of electron or positron
Electromagnetism - the attraction found between particles that are electrically charged
Strong Nuclear Force - Found in atoms and subatomic particles. It is responsible for holding the atoms' nucleus together. 

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10. A triply ionized beryllium atom is in the ground state. It absorbs energy and makes a transition to the n = 5 excited state.
Xelga [282]

Answer:

\lambda=1282\ nm

Explanation:

E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules

For transitions:

Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

\Delta E=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

Also, \Delta E=\frac {h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

So,  

\frac {h\times c}{\lambda}=2.179\times 10^{-18}(|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)\ J

\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m

So,  

\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m

Given, n_i=5\ and\ n_f=3

\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (\frac{1}{5^2} - \dfrac{1}{3^2})}\ m

\lambda=\frac{10^{-26}\times \:19.878}{10^{-18}\times \:2.179\left(|\frac{1}{25}-\frac{1}{9}\right)|}\ m

\lambda=\frac{19.878}{10^8\times \:2.179\left(|-\frac{16}{225}\right|)}\ m

\lambda= 1.2828\times10^{-6}

1 m = 10⁻⁹ nm

\lambda=1282\ nm

6 0
3 years ago
A circuit has an AC voltage source in series with a 50 ohm resistor and a 113 mH inductor. The frequency is 100 cycles/sec, and
user100 [1]

Answer:

Explanation:

The rms voltage = 140/√2 = 140/1.414 = 99 V.

Reactance of inductor  = wL = 2 X 3.14 X 100 X 113 X 10⁻³ =70.96 ohm.

Total resistance in terms of vector = 50+70.96j

j is imaginary unit  number

Magnitude of this resistance = √ 50² + 70.96² = 86.80 ohm

current in resistance (rms) ( I ) = 99/86.80 = 1.14 A.

Power dissipated in resistor = I² R = 1.14 X 1.14 X 50 = 65 W( approx)

7 0
4 years ago
What was Neptune's distance from the Sun in 1989?
Studentka2010 [4]

Answer:

4.5 Billion km :)

Explanation:

thats how it is

7 0
4 years ago
Read 2 more answers
Suppose you have two chains available to suspend an object in the air. Let’s also say that you can arrange the suspension howeve
Liono4ka [1.6K]
<h2>your answer is going to be image 1 that one looks the most decent </h2>
5 0
3 years ago
Read 2 more answers
A purse at radius 2.30 m and a wallet at radius 3.45 m travel in uniform circular motion on the floor of a merry-go-round as the
ivolga24 [154]

Answer:

The acceleration of the wallet is 3\hat{i}+6\hat{j}

Explanation:

Given that,

Radius of purse r= 2.30 m

Radius of wallet r'= 3.45 m

Acceleration of the purse a=2\hat{i}+4.00\hat{j}

We need to calculate the acceleration of the wallet

Using formula of acceleration

a=r\omega^2

Both the purse and wallet have same angular velocity

\omega=\omega'

\sqrt{\dfrac{a}{r}}=\sqrt{\dfrac{a'}{r'}}

\dfrac{a}{r}=\dfrac{a'}{r'}

\dfrac{a'}{a}=\dfrac{r'}{r}

\dfrac{a'}{a}=\dfrac{3.45}{2.30}

\dfrac{a'}{a}=\dfrac{3}{2}

a'=\dfrac{3}{2}\times(2\hat{i}+4.00\hat{j})

a'=3\hat{i}+6\hat{j}

Hence, The acceleration of the wallet is 3\hat{i}+6\hat{j}

4 0
3 years ago
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