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vovangra [49]
2 years ago
14

❤️❤️❤️❤️WILL MARK BRAINLIEST PLEASE ANSWER BOTH PARTS!❤️❤️❤️❤️

Physics
1 answer:
alekssr [168]2 years ago
8 0

Answer:

the answer to the first one is B and the second one is c I hope I got them right

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A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou
elena55 [62]

Answer:

3.2\times 10^{-7}\ m or 0.32 μm.

Explanation:

Given:

The radiations are UV radiation.

The frequency of the radiations absorbed (f) = 9.38\times 10^{14}\ Hz

The wavelength of the radiations absorbed (λ) = ?

We know that, the speed of ultraviolet radiations is same as speed of light.

So, speed of UV radiation (v) = 3\times 10^8\ m/s

Now, we also know that, the speed of the electromagnetic radiation is related to its frequency and wavelength and is given as:

v=f\lambda

Now, expressing the above equation in terms of wavelength 'λ', we have:

\lambda=\frac{v}{f}

Now, plug in the given values and solve for 'λ'. This gives,

\lambda=\frac{3\times 10^8\ m/s}{9.38\times 10^{14}\ Hz}\\\\\lambda=3.2\times 10^{-7}\ m\\\\\lambda=3.2\times 10^{-7}\times 10^{6}\ \mu m\ [1\ m=10^6\ \mu m]\\\\\lambda=3.2\times 10^{-1}=0.32\ \mu m

Therefore, the wavelength of the radiations absorbed by the ozone is nearly 3.2\times 10^{-7}\ m or 0.32 μm.

7 0
3 years ago
At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction
lbvjy [14]

Answer:

F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]

|F_T|=2\sqrt{34}k\frac{Q^2}{L}

\theta=tan^{-1}(\frac{5}{3})=59.03\°

Explanation:

I attached an image below with the scheme of the system:

The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:

F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]

the distances R1, R2 and R3, for a square arrangement is:

R1 = L

R2 = L

R3 = (√2)L

θ = 45°

F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]

and the magnitude is:

|F_T|=2k\frac{Q^2}{L}\sqrt{3^2+5^2}=2\sqrt{34}k\frac{Q^2}{L}

the direction is:

\theta=tan^{-1}(\frac{5}{3})=59.03\°

4 0
2 years ago
2. Neutrons have a ____<br> charge.<br> a. positive<br> b. negative<br> c. neutral
AnnyKZ [126]

Answer:

b. negative

Explanation:

neutrons have a negative charge and protons have a proton has a positive charge

3 0
2 years ago
Read 2 more answers
X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr
mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m

b) The change in photon energy is given by:

\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\

E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV

5 0
3 years ago
Give an example of a Claim.
Olin [163]

Answer:

Yeah

Explanation:

Look at the pic!!

5 0
3 years ago
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