To answer the problem we would be using this formula which isE = hc/L where E is the energy, h is Planck's constant, c is the speed of light and L is the wavelength
L = hc/E = 4.136×10−15 eV·s (2.998x10^8 m/s)/10^4 eV
= 1.240x10^-10 m
= 1.240x10^-1 nm
Answer:
yes
Explanation:
because it has the potential to move
Answer:
a) ![n=5.98*10^{26}/m^3](https://tex.z-dn.net/?f=n%3D5.98%2A10%5E%7B26%7D%2Fm%5E3)
b) ![i=2010000A/m^2](https://tex.z-dn.net/?f=i%3D2010000A%2Fm%5E2)
c) ![I_w=14.207A](https://tex.z-dn.net/?f=I_w%3D14.207A)
d) ![V_e=1.92*10^{-4}m/s](https://tex.z-dn.net/?f=V_e%3D1.92%2A10%5E%7B-4%7Dm%2Fs)
Explanation:
From the question we are told that:
Diameter ![d=3mm=>3*10^{-3}](https://tex.z-dn.net/?f=d%3D3mm%3D%3E3%2A10%5E%7B-3%7D)
Conductivity ![\sigma= 6.7 10^7 (0.M),](https://tex.z-dn.net/?f=%5Csigma%3D%206.7%2010%5E7%20%280.M%29%2C)
Electron mobility ![\phi= 0.0064 m2 /V sec](https://tex.z-dn.net/?f=%5Cphi%3D%200.0064%20m2%20%2FV%20sec)
Electric field ![E= 30 mV/m](https://tex.z-dn.net/?f=E%3D%2030%20mV%2Fm)
a)
Generally the equation for Charge Density is mathematically given by
![\phi=\frac{\sigma}{n e}](https://tex.z-dn.net/?f=%5Cphi%3D%5Cfrac%7B%5Csigma%7D%7Bn%20e%7D)
Therefore
![n=\frac{\sigma}{\phi e}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B%5Csigma%7D%7B%5Cphi%20e%7D)
![n=\frac{6.7 10^7}{1.6*10^{-19} *0.0064}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B6.7%2010%5E7%7D%7B1.6%2A10%5E%7B-19%7D%20%2A0.0064%7D)
![n=5.98*10^{26}/m^3](https://tex.z-dn.net/?f=n%3D5.98%2A10%5E%7B26%7D%2Fm%5E3)
b)
Generally the equation for current density is mathematically given by
![i=\sigma*E](https://tex.z-dn.net/?f=i%3D%5Csigma%2AE)
![i= 30*10^{-3] *6.7 10^7](https://tex.z-dn.net/?f=i%3D%2030%2A10%5E%7B-3%5D%20%2A6.7%2010%5E7)
![i=2010000A/m^2](https://tex.z-dn.net/?f=i%3D2010000A%2Fm%5E2)
c)
Generally the equation for current in wire is mathematically given by
![I_w=iA](https://tex.z-dn.net/?f=I_w%3DiA)
![I_w=i*\pi r^2](https://tex.z-dn.net/?f=I_w%3Di%2A%5Cpi%20r%5E2)
![I_w=(2010000)*\pi( 1.5*10^{-3})^2](https://tex.z-dn.net/?f=I_w%3D%282010000%29%2A%5Cpi%28%201.5%2A10%5E%7B-3%7D%29%5E2)
![I_w=14.207A](https://tex.z-dn.net/?f=I_w%3D14.207A)
d)
Generally the equation for electron draft velocity. is mathematically given by
![V_e=\phi E](https://tex.z-dn.net/?f=V_e%3D%5Cphi%20E)
![V_e=(0.0064)*(30*10^{-3})](https://tex.z-dn.net/?f=V_e%3D%280.0064%29%2A%2830%2A10%5E%7B-3%7D%29)
![V_e=1.92*10^{-4}m/s](https://tex.z-dn.net/?f=V_e%3D1.92%2A10%5E%7B-4%7Dm%2Fs)
Answer:
tension in rope = 25.0 N
Explanation:
- Two forces act on the suspended weight. The force coming down is the gravitational force and the upward force by the tension in the rope.
- Since the suspended weight is not accelerating so that the net force will be zero. Therefore the tension in the rope should be 25 N.
∑F = F - W = 0
so
F = W
so tension in rope = F = T = 25 N
Resistance of our body is given as
![R = 30,000 ohm](https://tex.z-dn.net/?f=R%20%3D%2030%2C000%20ohm)
voltage applied across the body is
![V = 120 V](https://tex.z-dn.net/?f=V%20%3D%20120%20V)
now by ohm's law current pass through our body is given by
![i = \frac{V}{R}](https://tex.z-dn.net/?f=i%20%3D%20%5Cfrac%7BV%7D%7BR%7D)
![i = \frac{120}{30,000}[\tex][tex]i = 4 * 10^{-3} A](https://tex.z-dn.net/?f=i%20%3D%20%5Cfrac%7B120%7D%7B30%2C000%7D%5B%5Ctex%5D%3C%2Fp%3E%3Cp%3E%5Btex%5Di%20%3D%204%20%2A%2010%5E%7B-3%7D%20A)
So current from our body will be 4 * 10^-3 A