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2 years ago

3. Calculate the wavelength of wave that has a frequency of 4.75 x 1012Hz.

Physics

1 answer:

Klio2033 [76]2 years ago

4 0

Frequency=v=4.75×10^12Hz
Wavelength=?

We know

$\boxed{\sf \lambda=\dfrac{C}{V}}$

$\\ \sf\longmapsto \lambda=\dfrac{3\times 10^8ms^{-1}}{4.75\times 10^{12}s^{-1}}$

$\\ \sf\longmapsto \lambda=0.631\times 10^{-4}m$

$\\ \sf\longmapsto \lambda=6.31\times 10^{-5}m$

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Nikolas had an idea that he could use the compressed carbon dioxide in a fire extinguisher to propel him on his skateboard.

Vikentia [17]

The Newton’s law Nikolas would use to come up with this idea is the Third law that states:

When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.

So, in this case, let's name the first Body A which is the skateboard and the second body B which is the compressed carbon dioxide in a fire extinguisher. Then, as shown in the figure below, according to the Third law:

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2 years ago

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A natural atom can be negatively charged by

svet-max [94.6K]

Answer:

By Gaining Electrons

Explanation:

A nuetral atom is negative when it gains electrons, and it can be positive when it loses electrons.

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2 years ago

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ome metal oxides can be decomposed to the metal and oxygen under reasonable conditions. 2 Ag2O(s) → 4 Ag(s) + O2(g) Thermodynami

yuradex [85]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

Explanation :

The given balanced chemical reaction is,

$2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{product}$

$\Delta H^o=[n_{Ag}\times \Delta H_f^0_{(Ag)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta H_f^0_{(Ag_2O)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)]$

$\Delta H^o=62.2kJ=62200J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{product}$

$\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)]$

$\Delta S^o=132.67J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 298 K.

$\Delta G^o=(62200J)-(298K\times 132.67J/K)$

$\Delta G^o=22664.34J=22.66kJ$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

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3 years ago

Why are we not crushed by the weight of the atmosphere on our shoulders?

antiseptic1488 [7]

Answer:

Due to equal pressure in all the direction at a particular level in a fluid medium (Pascal's Law)

Explanation:

We are not crushed by the weight of the atmosphere because atmosphere is a fluid and we are immersed into it. So, according to the Pascal's law the the pressure a each point in a horizontal level is equal in all the direction irrespective of the orientation of a body.

Variation of pressure in term of the height of a fluid medium is given as:

$P=\rho.g.h$

$\rho=$ density of fluid

g = acceleration due to gravity

h = height of the free surface of the fluid from the immersed object.

And atmosphere has very less variation of pressure with change in height as it is a rare medium fluid and so for a human height there is very negligible variation of pressure at the heat of a human with respect to his toe.

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3 years ago

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Wewaii [24]

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