Answer:
The charge on the dust particle is 
Explanation:
From the question we are told that
The length is 
The width is 
The charge is 
The mass suspended in mid-air is 
Generally the electric field on the carpet is mathematically represented as

Where
is the permittivity of free space with value 
substituting values


Generally the electric force keeping the dust particle on the air equal to the force of gravity acting on the particles

=> 
=> 
=> 
=> 
Answer:
a. 8.33 x 10 ⁻⁶ Pa
b. 8.19 x 10 ⁻¹¹ atm
c. 1.65 x 10 ⁻¹⁰ atm
d. 2.778 x 10 ⁻¹⁴ kg / m²
Explanation:
Given:
a.
I = 2500 W / m² , us = 3.0 x 10 ⁸ m /s
P rad = I / us
P rad = 2500 W / m² / 3.0 x 10 ⁸ m/s
P rad = 8.33 x 10 ⁻⁶ Pa
b.
P rad = 8.33 x 10 ⁻⁶ Pa *[ 9.8 x 10 ⁻⁶ atm / 1 Pa ]
P rad = 8.19 x 10 ⁻¹¹ atm
c.
P rad = 2 * I / us = ( 2 * 2500 w / m²) / [ 3.0 x 10 ⁸ m /s ]
P rad = 1.67 x 10 ⁻⁵ Pa
P₁ = 1.013 x 10 ⁵ Pa /atm
P rad = 1.67 x 10 ⁻⁵ Pa / 1.013 x 10 ⁵ Pa /atm = 1.65 x 10 ⁻¹⁰ atm
d.
P rad = I / us
ΔP / Δt = I / C² = [ 2500 w / m² ] / ( 3.0 x 10 ⁸ m/s)²
ΔP / Δt = 2.778 x 10 ⁻¹⁴ kg / m²
Answer:
The angle is 
Explanation:
From the question we are told that
The distance of separation is 
The wavelength of light is 
Generally the condition for destructive interference is mathematically represented as
![dsin(\theta ) =[m + \frac{1}{2} ]\lambda](https://tex.z-dn.net/?f=dsin%28%5Ctheta%20%29%20%20%3D%5Bm%20%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20%5D%5Clambda)
Here m is the order of maxima, first minimum (dark space) m = 0
So
![100 *10^{-6 } * sin(\theta ) =[0 + \frac{1}{2} ]600 *10^{-9}](https://tex.z-dn.net/?f=100%20%2A10%5E%7B-6%20%7D%20%2A%20%20sin%28%5Ctheta%20%29%20%20%3D%5B0%20%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20%5D600%20%2A10%5E%7B-9%7D)
=> ![\theta = sin^{-1} [0.003]](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%20sin%5E%7B-1%7D%20%5B0.003%5D)
=> 
Answer:
James is correct here as the force of hand pushing upwards is always more than the force of hand pushing down
Explanation:
Here we know that one hand is pushing up at some distance midway while other hand is balancing the weight by applying a force downwards
so here we can say
Upwards force = downwards Force + weight of snow
while if we find the other force which is acting downwards
then for that force we can say that net torque must be balanced
so here we have

so here we have

so here we can say that upward force by which we push up is always more than the downwards force
DE which is the differential equation represents the LRC series circuit where
L d²q/dt² + Rdq/dt +I/Cq = E(t) = 150V.
Initial condition is q(t) = 0 and i(0) =0.
To find the charge q(t) by using Laplace transformation by
Substituting known values for DE
L×d²q/dt² +20 ×dq/dt + 1/0.005× q = 150
d²q/dt² +20dq/dt + 200q =150