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timama [110]
2 years ago
10

Please I need help with this :(

Physics
1 answer:
Ann [662]2 years ago
3 0

three charged particals are located at the corners of an equil triangle shown in the figure showing let (q 2.20 Uc) and L 0.650

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Which characeristic makes omasis different from diffusion.
Ksju [112]
 <span>haha I used to think biology was so hard, i find it quite easy now. 
Okay, so basically Osmosis is the movement of water molecules from a higher concentration to a lower concentration. Diffusion is generally the movement of a gradient from higher concentration to an area of lower concentration. Osmosis applies to water only, whereas diffusion, you have many types such as Passive transport [ movement of molecules from high- low, NO CELLULAR ENERGY needed! ] then you have faciliated diffusion ( basically uses a channel protein to allow big substances to go through the membrane : NO ENERGY needed] 
OSMOSIS, the important thing to remember is that water ALWAYS flow towards the region with the higher concentration of the solute (ex: Salt is solute, water is solvent) solute is the thing that is being dissolved. Solvent is the one doing the dissolving. Hope this helped!</span>
5 0
4 years ago
Help me please! i will give brainliest i need it done i know its easy but i have a lot of work to do
cricket20 [7]

Answer:

Down below :)

Explanation:

1. Solid - A

2. Liquid - C

3. Gas - E

4. Melting - G

5. Evaporation - F

6. Condensation - D

7. Freezing - B

Hope it is right!

5 0
2 years ago
Read 2 more answers
What do we mean when we say that energy levels are quantized in atoms?
jek_recluse [69]

Answer:

Electrons are located in specific orbit corresponding to discrete energy levels

Explanation:

In Bohr's model of the atom, electron orbit the nucleus in specific levels, each of them corresponding to a specific energy. The electrons cannot be located in the space between two levels: this means that only some values of energy are possible for the electrons, so the energy levels are quantized.

A confirmation of Bohr's model is found in the spectrum of emission of gases. In fact, when an electron jumps from a higher energy level to a lower energy level, it emits a photon whose energy is exactly equal to the difference in energy between the two levels: since the energy levels are discrete, this means that the emitted photons cannot have any value of wavelength, but also their wavelength will appear as a discrete spectrum. This is exactly what it is observed in the spectrum of emission of gases.

3 0
3 years ago
What is the formula of the compound between calcium and sulfur that has the percent composition 55.6?
Yanka [14]

CaS is the empirical formula of the compound between calcium and sulphur that has the percent composition 55.6.

When percentages are given, we take a total mass of 100 grams.

Therefore the mass of each element is equal to the percentage given.

Mass of Ca = 55.6 g (given) of

S Mass = 44.4 g (100 - 55.6 = 44.4)

Step 1: Convert the given mass to moles.

moles Ca = given mass Ca / molar mass Ca

moles = 55.6 / 40 = 1.39 moles

mol S = specific mass S / molar mass S

mol = 44.4 / 32 = 1.39 mol

Step 2: Divide the molar ratio of each molar value by the smallest number of moles calculated.

For Ca = 1.39 / 1.39 = 1

For S = 1.39 / 1.39 = 1

The ratio of Ca : S = 1:1

Hence the empirical formula of the given compound will be CaS.

Learn more about empirical formula here : brainly.com/question/1496676

#SPJ4

8 0
1 year ago
A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v
atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

y(x, t) = Acos(kx -wt)

The relationship between velocity and propagation constant is

v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\

v = 15.87m/s

Time taken, t = \frac{\lambda}{v}

= \frac{1.3}{15.87}\\\\=0.0819 sec

t = 0.0819s

2)

The velocity of transverse wave is given by

v = \sqrt{\frac{T}{\mu}}

v = \sqrt{\frac{W}{\frac{m}{\lambda}}}

mass of string is calculated thus

mg = 0.0125N

m = \frac{0.0125N}{9.8N/s}

m = 0.00128kg

\omega = \frac{v^2m}{\lambda}

\omega = \frac{(15.87^2)(0.00128)}{1.30}

\omega = 0.25N

3)

The propagation constant k is

k=\frac{2\pi}{\lambda}

hence

\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}

\lambda = 0.036 m

No of wavelengths, n is

n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\

n = 36

4)

The equation of wave travelling down the string is

y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]

without, unit\\\\y(x , t)= Acos[172x + 2730t]

7 0
3 years ago
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