The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².
The given parameters;
- <em>Current flowing in the wire, I = 4.00 mA</em>
- <em>Initial diameter of the wire, d₁ = 4 mm = 0.004 m</em>
- <em>Final diameter of the wire, d₂ = 1 mm = 0.001 m</em>
- <em>Length of wire, L = 2.00 m</em>
- <em>Density of electron in the copper, n = 8.5 x 10²⁸ /m³</em>
<em />
The initial area of the copper wire;
![A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2](https://tex.z-dn.net/?f=A_1%20%3D%20%5Cfrac%7B%5Cpi%20d%5E2%7D%7B4%7D%20%3D%20%5Cfrac%7B%5Cpi%20%5Ctimes%20%280.004%29%5E2%7D%7B4%7D%20%3D1.257%5Ctimes%2010%5E%7B-5%7D%20%5C%20m%5E2)
The final area of the copper wire;
![A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2](https://tex.z-dn.net/?f=A_2%20%3D%20%5Cfrac%7B%5Cpi%20d%5E2%7D%7B4%7D%20%3D%20%5Cfrac%7B%5Cpi%20%280.001%29%5E2%7D%7B4%7D%20%3D%207.86%5Ctimes%2010%5E%7B-7%7D%20%5C%20m%5E2)
The initial drift velocity of the electrons is calculated as;
![v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s](https://tex.z-dn.net/?f=v_d_1%20%3D%20%5Cfrac%7BI%7D%7BnqA_1%7D%20%5C%5C%5C%5Cv_d_1%20%3D%20%5Cfrac%7B4%5Ctimes%2010%5E%7B-3%7D%20%7D%7B8.5%5Ctimes%2010%5E%7B28%7D%20%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%20%5Ctimes%201.257%5Ctimes%2010%5E%7B-5%7D%7D%20%5C%5C%5C%5Cv_d_1%20%3D%202.34%20%5Ctimes%2010%5E%7B-8%7D%20%5C%20m%2Fs)
The final drift velocity of the electrons is calculated as;
![v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7} \ m/s](https://tex.z-dn.net/?f=v_d_2%20%3D%20%5Cfrac%7BI%7D%7BnqA_2%7D%20%5C%5C%5C%5Cv_d_2%20%3D%20%5Cfrac%7B4%5Ctimes%2010%5E%7B-3%7D%20%7D%7B8.5%5Ctimes%2010%5E%7B28%7D%20%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%20%5Ctimes%207.86%5Ctimes%2010%5E%7B-7%7D%7D%20%5C%5C%5C%5Cv_d_2%20%3D%203.74%5Ctimes%2010%5E%7B-7%7D%20%20%5C%20m%2Fs)
The change in the mean drift velocity is calculated as;
![\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s](https://tex.z-dn.net/?f=%5CDelta%20v%20%3D%20v_d_2%20-v_d_1%5C%5C%5C%5C%5CDelta%20v%20%3D%203.74%5Ctimes%2010%5E%7B-7%7D%20%5C%20m%2Fs%20%5C%20-%5C%202.34%20%5Ctimes%2010%5E%7B-8%7D%20%5C%20m%2Fs%20%3D%203.506%5Ctimes%2010%5E%7B-7%7D%20%5C%20m%2Fs)
The time of motion of electrons for the initial wire diameter is calculated as;
![t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s](https://tex.z-dn.net/?f=t_1%20%3D%20%5Cfrac%7BL%7D%7Bv_d_1%7D%20%5C%5C%5C%5Ct_1%20%3D%20%5Cfrac%7B2%7D%7B2.34%5Ctimes%2010%5E%7B-8%7D%7D%20%5C%5C%5C%5Ct_1%20%3D%208.547%5Ctimes%2010%5E%7B7%7D%20%5C%20s)
The time of motion of electrons for the final wire diameter is calculated as;
![t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s](https://tex.z-dn.net/?f=t_2%20%3D%20%5Cfrac%7BL%7D%7Bv_d_1%7D%20%5C%5C%5C%5Ct_2%3D%20%5Cfrac%7B2%7D%7B3.74%20%5Ctimes%2010%5E%7B-7%7D%7D%20%5C%5C%5C%5Ct_2%20%3D%205.348%20%5Ctimes%2010%5E%7B6%7D%20%5C%20s)
The average acceleration of the electrons is calculated as;
![a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B%5CDelta%20v%7D%7B%5CDelta%20t%7D%20%5C%5C%5C%5Ca%20%3D%20%5Cfrac%7B3.506%20%5Ctimes%2010%5E%7B-7%7D%20%7D%7B%288.547%5Ctimes%2010%5E7%29-%20%285.348%5Ctimes%2010%5E6%29%7D%20%5C%5C%5C%5Ca%20%3D%204.38%5Ctimes%2010%5E%7B-15%7D%20%5C%20m%2Fs%5E2)
Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².
Learn more here: brainly.com/question/22406248
Answer. Second Option: .85p_o=p_o e^-.00012h
Solution:
P(h)=Po e^(-0.00012h)
Air pressure: P(h)
Height above the surface of the Earth (in meters): h
Air pressure at the sea level: Po
Height at which air pressure is 85% of the air pressure at sea level:
h=?, P(h)=85% Po
P(h)=(85/100) Po
P(h)=0.85 Po
Replacing P(h) by 0.85 Po in the formula above:
P(h)=Po e^(-0.00012h)
0.85 Po = Po e^(-0.00012h)
Answer:
![v_f=-0.29\frac{m}{s}](https://tex.z-dn.net/?f=v_f%3D-0.29%5Cfrac%7Bm%7D%7Bs%7D)
Explanation:
The principle of conservation of momentum, states that if the sum of the forces acting on a system is null, the initial total momentum of the system before a collision equals the final total momentum of the system after the collision. The collision is completely inelastic, which means that the players remain stick to each other after the collision:
![p_i=p_f\\m_1v_1+m_2v_2=(m_1+m_2)v_f\\v_f=\frac{m_1v_1+m_2v_2}{(m_1+m_2)}\\\\v_f=\frac{(110kg)4.5\frac{m}{s}+150kg(-3.8\frac{m}{s})}{(110kg+150kg)}\\v_f=-0.29\frac{m}{s}](https://tex.z-dn.net/?f=p_i%3Dp_f%5C%5Cm_1v_1%2Bm_2v_2%3D%28m_1%2Bm_2%29v_f%5C%5Cv_f%3D%5Cfrac%7Bm_1v_1%2Bm_2v_2%7D%7B%28m_1%2Bm_2%29%7D%5C%5C%5C%5Cv_f%3D%5Cfrac%7B%28110kg%294.5%5Cfrac%7Bm%7D%7Bs%7D%2B150kg%28-3.8%5Cfrac%7Bm%7D%7Bs%7D%29%7D%7B%28110kg%2B150kg%29%7D%5C%5Cv_f%3D-0.29%5Cfrac%7Bm%7D%7Bs%7D)