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bagirrra123 [75]
3 years ago
11

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

g), each having (a) a speed of 6.66 × 106 m/s and (b) a kinetic energy of 1.71 × 10−15 J.
Physics
1 answer:
Harrizon [31]3 years ago
6 0

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.

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A kettle is rated at 1 kW, 220 V. Calculate the working resistance of the kettle.
Anna [14]

Explanation:

Power of electric kettle, P = 1 kW

Voltage, V = 220 V

(a) Electric power is given by the formula as follows :

P=\dfrac{V^2}{R}

R is resistance

R=\dfrac{V^2}{P}\\\\R=\dfrac{(220)^2}{10^3}\\\\R=48.4\ \Omega

(b) When connected to a 220 V supply, it takes 3 minutes for the water in the kettle to reach boiling point.

Energy supplied is given by :

E=P\times t

P is power, P=\dfrac{V^2}{R}

E=\dfrac{V^2}{R}t\\\\E=\dfrac{(220)^2}{48.4}\times 180\\\\E=180000\ J\\\\E=180\ kJ

5 0
3 years ago
Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.
Leni [432]

Answer:

T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}

Explanation:

The given data :-

  • Diameter of rod AB ( d₁ ) = 200 mm.
  • Diameter of rod BC ( d₂ ) = 150 mm.
  • The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C
  • The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
  • Consider the required temperature increase to close the gap at C = T °C
  • Consider the change in length of the rod = бL

Solution :-

\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}

\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}

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4 years ago
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larisa86 [58]

\sf{Answer:}

A system that can be affected by the outside environment, by an exchange of matter or energy is an open physical system .

4 0
2 years ago
The equation shows neutralization of an acid and a base to produce a salt and water.
Lyrx [107]
This equation will be balanced if the x is a 2 because there are two sodiums on the reactants sides so there must be two sodiums on the products side

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7 0
3 years ago
Read 2 more answers
A 0.75 kg book is pushed across the table with an acceleration of 0.3 m/s2. What force is being applied to the
Brums [2.3K]

Answer:

\boxed {\boxed {\sf 0.225 \ Newtons}}

Explanation:

We are asked to find the force being applied to a book. According to Newton's Second Law of Motion, force is the product of mass and acceleration.

F= ma

The mass of the book is 0.75 kilograms and the acceleration is 0.3 meters per square second. Substitute these values into the formula.

  • m= 0.75 kg
  • a= 0.3 m/s²

F= 0.75 \ kg * 0.3 \ m/s^2

Multiply.

F =0.225 \ kg * m/s^2

1 kilogram meter per second squared is equal to 1 Newton. Therefore, our answer of 0.225 kilogram meters per second squared is equal to 0.225 Newtons.

F= 0.225  \ N

<u>0.225 Newtons of force</u> are applied to the book.

5 0
3 years ago
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