Explanation:
1. Mass of the proton, ![m_p=1.67\times 10^{-27}\ kg](https://tex.z-dn.net/?f=m_p%3D1.67%5Ctimes%2010%5E%7B-27%7D%5C%20kg)
Wavelength, ![\lambda_p=4.23\times 10^{-12}\ m](https://tex.z-dn.net/?f=%5Clambda_p%3D4.23%5Ctimes%2010%5E%7B-12%7D%5C%20m)
We need to find the potential difference. The relationship between potential difference and wavelength is given by :
![\lambda=\dfrac{h}{\sqrt{2m_pq_pV}}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7Bh%7D%7B%5Csqrt%7B2m_pq_pV%7D%7D)
![V=\dfrac{h^2}{2q_pm_p\lambda^2}](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7Bh%5E2%7D%7B2q_pm_p%5Clambda%5E2%7D)
![V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 1.67\times 10^{-27}\times (4.23\times 10^{-12})^2}](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7B%286.62%5Ctimes%2010%5E%7B-34%7D%29%5E2%7D%7B2%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%5Ctimes%201.67%5Ctimes%2010%5E%7B-27%7D%5Ctimes%20%284.23%5Ctimes%2010%5E%7B-12%7D%29%5E2%7D)
V = 45.83 volts
2. Mass of the electron, ![m_p=9.1\times 10^{-31}\ kg](https://tex.z-dn.net/?f=m_p%3D9.1%5Ctimes%2010%5E%7B-31%7D%5C%20kg)
Wavelength, ![\lambda_p=4.23\times 10^{-12}\ m](https://tex.z-dn.net/?f=%5Clambda_p%3D4.23%5Ctimes%2010%5E%7B-12%7D%5C%20m)
We need to find the potential difference. The relationship between potential difference and wavelength is given by :
![\lambda=\dfrac{h}{\sqrt{2m_eq_eV}}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7Bh%7D%7B%5Csqrt%7B2m_eq_eV%7D%7D)
![V=\dfrac{h^2}{2q_em_e\lambda^2}](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7Bh%5E2%7D%7B2q_em_e%5Clambda%5E2%7D)
![V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 9.1\times 10^{-31}\times (4.23\times 10^{-12})^2}](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7B%286.62%5Ctimes%2010%5E%7B-34%7D%29%5E2%7D%7B2%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%5Ctimes%209.1%5Ctimes%2010%5E%7B-31%7D%5Ctimes%20%284.23%5Ctimes%2010%5E%7B-12%7D%29%5E2%7D)
![V=6.92\times 10^{34}\ V](https://tex.z-dn.net/?f=V%3D6.92%5Ctimes%2010%5E%7B34%7D%5C%20V)
V = 84109.27 volt
Hence, this is the required solution.
Answer:
Please make image clearer
Explanation:
You are too far back please move closer to the paper
You would need to use the equation a= (v-u)÷t
You need to substitute in the correct numbers.
a= (10-20)÷1
Your answer is -10m/s^2
Answer:
O The particles of the medium move more slowly and there are fewer chances to transfer energy.
Explanation:
Various media are made up of particles. These particles are in constant motion according to the kinetic theory of matter. Recall that temperature has been defined as the average kinetic energy of the particles in a medium. Hence, for any given medium, the velocity of particle motion increases or decreases linearly with temperature.
The speed of particles in any medium increases or decreases as the temperature of the medium increases or decreases as emphasised above. Hence, at low temperature, the velocity of waves set up by the motion of particles in a medium decreases and transfer the wave energy to neighbouring particles occurs more slowly than at high temperatures.
Explanation:
Given that,
Mass of the car, m₁ = 1250 kg
Initial speed of the car, u₁ = 7.39 m/s
Mass of the truck, m₂ = 5380 kg
It is stationary, u₂ = 0
Final speed of the truck, v₂ = 2.3 m/s
Let v₁ is the final velocity of the car. Using the conservation of momentum as :
![m_1u_1+m_2u_2=m_1v_1+m_2v_2](https://tex.z-dn.net/?f=m_1u_1%2Bm_2u_2%3Dm_1v_1%2Bm_2v_2)
![1250\times 7.39+5380\times 0=1250\times v_1+5380\times 2.3](https://tex.z-dn.net/?f=1250%5Ctimes%207.39%2B5380%5Ctimes%200%3D1250%5Ctimes%20v_1%2B5380%5Ctimes%202.3)
![v_1=-2.5\ m/s](https://tex.z-dn.net/?f=v_1%3D-2.5%5C%20m%2Fs)
So, the final velocity of the car is 2.5 m/s but in opposite direction. Hence, this is the required solution.