Answer:
Explanation:
Given that,
Force is downward I.e negative y-axis
F = -2 × 10^-14 •j N
Magnetic field is westward, +x direction
B = 8.3 × 10^-2 •i T
Charge of an electron
q = 1.6 × 10^-19C
Velocity and it direction?
Force in a magnetic field is given as
F = q(V×B)
Angle between V and B is 270, check attachment
The cross product of velocity and magnetic field
F =qVB•Sin270
2 × 10^-14 = 1.6 × 10^-19 × V × 8.3 × 10^-2
Then,
v = 2 × 10^-14 / (1.6 × 10^-19 × 8.3 × 10^-2)
v = 1.51 × 10^6 m/s
Direction of the force
Let x be the direction of v
-F•j = v•x × B•i
From cross product
We know that
i×j = k, j×i = -k
j×k =i, k×j = -i
k×i = j, i×k = -j OR -k×i = -j
Comparing -k×i = -j to given problem
We notice that
-F•j = q ( -V•k × B×i)
So, the direction of V is negative z- direction
V = -1.51 × 10^6 •k m/s
Answer:
d = 380 feet
Explanation:
Height of man = perpendicular= 130 feet
Angle of depression = ∅ = 70 °
distance to bus stop from man = hypotenuse = d = 130 sec∅
As sec ∅ = 1 / cos∅
so d = 130 sec∅ or d = 130 / cos∅
d = 130 / cos(70°)
d = 380 feet
Refraction is the bending of a wave disturbance as it passes at an angle from one medium into another
Explanation:
Area of ring 
Charge of on ring 
Charge on disk
![\begin{aligned}d v &=\frac{k d q}{\sqrt{x^{2}+a^{2}}} \\&=2 \pi-k \frac{a d a}{\sqrt{x^{2}+a^{2}}} \\v(1) &=2 \pi c k \int_{0}^{R} \frac{a d a}{\sqrt{x^{2}+a^{2}}} \cdot_{2 \varepsilon_{0}}^{2} R \\&=2 \pi \sigma k[\sqrt{x^{2}+a^{2}}]_{0}^{2} \\&=\frac{2 \pi \sigma}{4 \pi \varepsilon_{0}}[\sqrt{z^{2}+R^{2}}-(21)] \\&=\frac{\sigma}{2}(\sqrt{2^{2}+R^{2}}-2)\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Dd%20v%20%26%3D%5Cfrac%7Bk%20d%20q%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%7D%20%5C%5C%26%3D2%20%5Cpi-k%20%5Cfrac%7Ba%20d%20a%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%7D%20%5C%5Cv%281%29%20%26%3D2%20%5Cpi%20c%20k%20%5Cint_%7B0%7D%5E%7BR%7D%20%5Cfrac%7Ba%20d%20a%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%7D%20%5Ccdot_%7B2%20%5Cvarepsilon_%7B0%7D%7D%5E%7B2%7D%20R%20%5C%5C%26%3D2%20%5Cpi%20%5Csigma%20k%5B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%5D_%7B0%7D%5E%7B2%7D%20%5C%5C%26%3D%5Cfrac%7B2%20%5Cpi%20%5Csigma%7D%7B4%20%5Cpi%20%5Cvarepsilon_%7B0%7D%7D%5B%5Csqrt%7Bz%5E%7B2%7D%2BR%5E%7B2%7D%7D-%2821%29%5D%20%5C%5C%26%3D%5Cfrac%7B%5Csigma%7D%7B2%7D%28%5Csqrt%7B2%5E%7B2%7D%2BR%5E%7B2%7D%7D-2%29%5Cend%7Baligned%7D)
Note: Refer the image attached
Edison's education is most unique and relevant.
1. The first teacher he had was his mother
2. He found vital lessons and was influenced greatly by the book of R.G. Parker called School of Natural Philosophy
3. Another educating piece he had was a book entitled The Cooper Union for the Advancement of Science and Art
His style of learning was though reading books on a variety of subjects, a self-educating environment that fosters independent learning which can be useful through his life.
