Explanation:
it is given that, the linear charge density of a charge, ![\lambda=\dfrac{\lambda_ox_o}{x}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7B%5Clambda_ox_o%7D%7Bx%7D)
Firstly, we can define the electric field for a small element and then integrate for the whole. The very small electric field is given by :
..........(1)
The linear charge density is given by :
![\lambda=\dfrac{dq}{dx}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7Bdq%7D%7Bdx%7D)
![dq=\lambda.dx=\dfrac{\lambda_ox_o}{x}dx](https://tex.z-dn.net/?f=dq%3D%5Clambda.dx%3D%5Cdfrac%7B%5Clambda_ox_o%7D%7Bx%7Ddx)
Integrating equation (1) from x = x₀ to x = infinity
![E=\int\limits^\infty_{x_o} {\dfrac{k\lambda_ox_o}{x^3}}.dx](https://tex.z-dn.net/?f=E%3D%5Cint%5Climits%5E%5Cinfty_%7Bx_o%7D%20%7B%5Cdfrac%7Bk%5Clambda_ox_o%7D%7Bx%5E3%7D%7D.dx)
![E=-\dfrac{k\lambda_ox_o}{2}\dfrac{1}{x^2}|_{x_o}^\infty}](https://tex.z-dn.net/?f=E%3D-%5Cdfrac%7Bk%5Clambda_ox_o%7D%7B2%7D%5Cdfrac%7B1%7D%7Bx%5E2%7D%7C_%7Bx_o%7D%5E%5Cinfty%7D)
![E=\dfrac{k\lambda_o}{2x_o}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7Bk%5Clambda_o%7D%7B2x_o%7D)
Hence, this is the required solution.
To solve this problem we will apply the principles of energy conservation. On the one hand we have that the work done by the non-conservative force is equivalent to -30J while the work done by the conservative force is 50J.
This leads to the direct conclusion that the resulting energy is 20J.
The conservative force is linked to the movement caused by the sum of the two energies, therefore there is an increase in kinetic energy. The decrease in the mechanical energy of the system is directly due to the loss given by the non-conservative force, therefore there is a decrease in mechanical energy.
Therefore the correct answer is A. Kintetic energy increases and mechanical energy decreases.
The helium may be treated as an ideal gas, so that
(p*V)/T =constant
where
p = pressure
V = volume
T = temperature.
Note that
7.5006 x 10⁻³ mm Hg = 1 Pa
1 L = 10⁻³ m³
Given:
At ground level,
p₁ = 752 mm Hg
= (752 mm Hg)/(7.5006 x 10⁻³ mm Hg/Pa)
= 1.0026 x 10⁵ Pa
V₁ = 9.47 x 10⁴ L = (9.47 x 10⁴ L)*(10⁻³ m³/L)
= 94.7 m³
T₁ = 27.8 °C = 27.8 + 273 K
= 300.8 K
At 36 km height,
P₂ = 73 mm Hg = 73/7.5006 x 10⁻³ Pa
= 9.7326 x 10³ Pa
T₂ = 235 K
If the volume at 36 km height is V₂, then
V₂ = (T₂/p₂)*(p₁/T₁)*V₁
= (235/9.7326 x 10³)*(1.0026 x 10⁵/300.8)*94.7
= 762.15 m³
Answer: 762.2 m³
Answer:
Explanation:
The equivalent resistance for three resistors connected in parallel is given as
(1/R)=(1/R₁)+(1/R₂)+(1/R₃)
now we.need to.insert the value of 3 resistances but only 2 are given in the question.