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lana [24]
3 years ago
13

Myoncedyret

Physics
1 answer:
konstantin123 [22]3 years ago
8 0

Answer:

i need help too

Explanation:

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Be sure to answer all parts. Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 k
Harrizon [31]

Explanation:

Given that,

(a) Speed, v=6.66\times 10^6\ m/s

Mass of the electron, m_e=9.11\times 10^{-31}\ kg

Mass of the proton, m_p=1.67\times 10^{-27}\ kg

The wavelength of the electron is given by :

\lambda_e=\dfrac{h}{m_ev}

\lambda_e=\dfrac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times 6.66\times 10^6}

\lambda_e=1.09\times 10^{-10}\ m

The wavelength of the proton is given by :

\lambda_p=\dfrac{h}{m_p v}

\lambda_p=\dfrac{6.63\times 10^{-34}}{1.67\times 10^{-27}\times 6.66\times 10^6}

\lambda_p=5.96\times 10^{-14}\ m

(b) Kinetic energy, K=1.71\times 10^{-15}\ J

The relation between the kinetic energy and the wavelength is given by :

\lambda_e=\dfrac{h}{\sqrt{2m_eK}}

\lambda_e=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.11\times 10^{-31}\times 1.71\times 10^{-15}}}

\lambda_e=1.18\times 10^{-11}\ m

\lambda_p=\dfrac{h}{\sqrt{2m_pK}}

\lambda_p=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 1.67\times 10^{-27}\times 1.71\times 10^{-15}}}

\lambda_p=2.77\times 10^{-13}\ m

Hence, this is the required solution.

6 0
3 years ago
PLEASE HLEP I WILL GIVE BRAINLIEST TO CORRECT ANSWER!!!!!!
Hoochie [10]

Answer:

75 km/h

Explanation:

Speed = Distance divided by Time.

The train went a distance of 300 km in 4 hours. Therefore, the speed of the train is: 300 divided by 4 = 75 km/h

8 0
3 years ago
Read 2 more answers
Calculate the electric field at the center of a square
pantera1 [17]

Answer:

E_y=1175510.2\ N.C^{-1}

The Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

Explanation:

Given:

  • side of a square, a=52.5\ cm
  • charge on one corner of the square, q_1=+45\times 10^{-6}\ C
  • charge on the remaining 3 corners of the square,q_2=q_3=q_4=-27\times 10^{-6}\ C

<u>Distance of the center from each corners</u>=\frac{1}{2} \times diagonals

diagonal=\sqrt{52.5^2+52.5^2}

diagonal=74.25\cm=0.7425\ m

∴Distance of center from corners, b=0.3712\ m

Now, electric field due to charges is given as:

E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}

<u>For charge q_1 we have the field lines emerging out of the charge since it is positively charged:</u>

E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}

  • E_1=2938775.5\ N.C^{-1}

<u>Force by each of the charges at the remaining corners:</u>

E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}

  • E_2=E_3=E_4=1763265.3\ N.C^{-1}

<u> Now, net electric field in the vertical direction:</u>

E_y=E_1-E_4

E_y=1175510.2\ N.C^{-1}

<u>Now, net electric field in the horizontal direction:</u>

E_y=E_2-E_3

E_y=0\ N.C^{-1}

So the Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

8 0
3 years ago
Dr. james van allen and his group of united states scientists discovered a set of ___________. magnetic equators planets magneti
Fofino [41]

RADIATION BELTS....... I think but it should be radiation belt

3 0
3 years ago
Read 2 more answers
Heat is transferred at a rate of 2 kW from a hot reservoir at 800 K to a cold reservoir at 300 K. Calculate the rate at which th
vekshin1

Answer:

0.00417 kW/K or 4.17 W/K

Second law is satisfied.

Explanation:

Parameters given:

Rate of heat transfer, Q = 2kW

Temperature of hot reservoir, Th = 800K

Temperature of cold reservoir, Tc = 300K

The rate of entropy change is given as:

ΔS = Q * [(1/Tc) - (1/Th)]

ΔS = 2 * (1/300 - 1/800)

ΔS = 2 * 0.002085

ΔS = 0.00417 kW/K or 4.17 W/K

Since ΔS is greater than 0, te the second law of thermodynamics is satisfied.

6 0
3 years ago
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