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3 years ago

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A photon of ultraviolet (uv) light possesses enough energy to mutate a strand of human dna. what is the energy of a single uv ph

oton and a mole of uv photons having a wavelength of 25 nm?

1 answer:

sergey [27]3 years ago

4 0

Energy of single uv photon= 7.95 x 10⁻¹⁸ J

Energy of a mole of photons=4788213 J

Explanation:

the energy of a photon is given by $E= \frac{h c}{\lambda}$

h= planck's constant=6.626 x 10⁻³⁴ Js

λ= wavelength=25 nm= 25 x 10⁻⁹ m

c= velocity of light= 3 x 10⁸ m/s

so $E= \frac{6.626\times10^{-34}(3\times10^{8})}{25\times10^{-9}}$

E=7.95 x 10⁻¹⁸ J

so energy of one photon=7.95 x 10⁻¹⁸ J

Energy of one mole of photons= 6.022 x 10²³ (7.95 x 10⁻¹⁸ J)

Energy of one mole of photons=4788213 J

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Consider a system to be two train cars traveling toward each other. What is the total momentum of the system before the train ca

Brut [27]

Let say the two train cars are of masses $m_1$ and $m_2$

now if the speed of two cars are $v_1$ and $v_2$

then we can say that the momentum of two cars before they collide is given by

$P = m_1v_1 - m_2v_2$

here two cars are moving in opposite direction so we can say that the net momentum is subtraction of two cars momentum.

Now since in these two car motion there is no external force on them while they collide

So the momentum of two cars are always conserved.

hence we can say that the final momentum of two cars will be same after collision as it is before collision

$P = m_1v_1 - m_2v_2$

5 0

2 years ago

Read 2 more answers

At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81 m/s2 . A watermelon has a weight of 54.0 N at the s

butalik [34]

Answer:

$m=5.51Kg$

Explanation:

The weight of an object on Earth is given by $W=mg$ , so we can calculate its mass by doing $m=W/g$ , which for our values is:

$m=(54N)/(9.8m/s^2)=5.51Kg$

<em>Nothing is being asked</em> about Io but if one wanted to know the weight <em>W'</em> of the watermelon there one just have to do:

$W'=ma=(5.51Kg)(1.81m/s^2)=9.97N$

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2 years ago

A satellite of mass m is moving in a circular orbit around the earth at a constant speed v and at an altitude h above the earth'

qwelly [4]

The satellite executes a rotation motion around the earth, because Earth's force of attraction plays the role of centripetal force:

Fa=Fcp=>k*Mp*m/(Rp+r)²=mv²/(Rp+r)=>v=√(k*Mp/(Rp+r))=√(6.67*10⁻¹¹*5.98*10²⁴/(6371*10³+1000*10³))=√(39.88*10¹³/(7371*10³))=√(5.41*10⁷)=7355.53 m/s

Check the calculations again !

7 0

2 years ago

Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has

faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

Charge on first charged particle, $q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.$
Charge on the second charged particle, $q_2=3.80\ nC=3.80\times 10^{-9}\ C.$

Position of the first charge = $(x_1=0.00\ m,\ y_1=0.600\ m).$
Position of the second charge = $(x_2=1.50\ m,\ y_2=0.650\ m).$

The electric field at a point due to a charge $q$ at a point $r$ distance away is given by

$\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.$

where,

$k$ = Coulomb's constant, having value $\rm 8.99\times 10^9\ Nm^2/C^2.$

$\vec r$ = position vector of the point where the electric field is to be found with respect to the position of the charge $q$ .

$\hat r$ = unit vector along $\vec r$ .

The electric field at the origin due to first charge is given by

$\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.$

$\vec r_1$ is the position vector of the origin with respect to the position of the first charge.

Assuming, $\hat i,\ \hat j$ are the units vectors along x and y axes respectively.

$\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.$

Using these values,

$\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.$

The electric field at the origin due to the second charge is given by

$\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.$

$\vec r_2$ is the position vector of the origin with respect to the position of the second charge.

$\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.$

Using these values,

$\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\ N/C.$

The net electric field at the origin due to both the charges is given by

$\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.$

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

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2 years ago

When measuring espresso for a drink, which instrument would give the<br> greatest precision?

Gemiola [76]

How many mL is an espresso?

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2 years ago