Answer:
![(1)p(x)\geq 0\\(2)\int_{0}^{\infty} p(x) dx=0](https://tex.z-dn.net/?f=%281%29p%28x%29%5Cgeq%200%5C%5C%282%29%5Cint_%7B0%7D%5E%7B%5Cinfty%7D%20p%28x%29%20dx%3D0)
Explanation:
A function f(x) is a Probability Density Function if it satisfies the following conditions:
![(1)f(x)\geq 0\\(2)\int_{0}^{\infty} f(x) dx=0](https://tex.z-dn.net/?f=%281%29f%28x%29%5Cgeq%200%5C%5C%282%29%5Cint_%7B0%7D%5E%7B%5Cinfty%7D%20f%28x%29%20dx%3D0)
Given the function:
![p(x)=\dfrac{1}{r}e^{-x/r} \: on\: [0,\infty), where\:r=\dfrac{20}{ln(2)}](https://tex.z-dn.net/?f=p%28x%29%3D%5Cdfrac%7B1%7D%7Br%7De%5E%7B-x%2Fr%7D%20%5C%3A%20on%5C%3A%20%5B0%2C%5Cinfty%29%2C%20where%5C%3Ar%3D%5Cdfrac%7B20%7D%7Bln%282%29%7D)
(1)p(x) is greater than zero since the range of exponents of the Euler's number will lie in ![[0,\infty).](https://tex.z-dn.net/?f=%5B0%2C%5Cinfty%29.)
(2)
![\int_{0}^{\infty} p(x)=\int_{0}^{\infty} \dfrac{1}{r}e^{-x/r}\\=\dfrac{1}{r} \int_{0}^{\infty} e^{-x/r}\\=-\dfrac{r}{r}\left[e^{-x/r}\right]_{0}^{\infty}\\=-\left[e^{-\infty/r}-e^{-0/r}\right]\\=-e^{-\infty}+e^{-0}\\SInce \: e^{-\infty} \rightarrow 0\\e^{-0}=1\\\int_{0}^{\infty} p(x)=1](https://tex.z-dn.net/?f=%5Cint_%7B0%7D%5E%7B%5Cinfty%7D%20p%28x%29%3D%5Cint_%7B0%7D%5E%7B%5Cinfty%7D%20%5Cdfrac%7B1%7D%7Br%7De%5E%7B-x%2Fr%7D%5C%5C%3D%5Cdfrac%7B1%7D%7Br%7D%20%5Cint_%7B0%7D%5E%7B%5Cinfty%7D%20e%5E%7B-x%2Fr%7D%5C%5C%3D-%5Cdfrac%7Br%7D%7Br%7D%5Cleft%5Be%5E%7B-x%2Fr%7D%5Cright%5D_%7B0%7D%5E%7B%5Cinfty%7D%5C%5C%3D-%5Cleft%5Be%5E%7B-%5Cinfty%2Fr%7D-e%5E%7B-0%2Fr%7D%5Cright%5D%5C%5C%3D-e%5E%7B-%5Cinfty%7D%2Be%5E%7B-0%7D%5C%5CSInce%20%5C%3A%20e%5E%7B-%5Cinfty%7D%20%5Crightarrow%200%5C%5Ce%5E%7B-0%7D%3D1%5C%5C%5Cint_%7B0%7D%5E%7B%5Cinfty%7D%20p%28x%29%3D1)
The function p(x) satisfies the conditions for a probability density function.
We Know, F = m*a
F = 2200 * 3.4
F = 7480 Kg m/s²
So, your final answer is 7480
Answer:
Electric field due to two charges is given as
![E = 1.44 \times 10^7 N/C](https://tex.z-dn.net/?f=E%20%3D%201.44%20%5Ctimes%2010%5E7%20N%2FC)
Explanation:
As we know that two charges are opposite in nature
So the electric field at the mid point of two charges will add together
so the net field is given as
![E = 2\frac{kq}{r^2}](https://tex.z-dn.net/?f=E%20%3D%202%5Cfrac%7Bkq%7D%7Br%5E2%7D)
now we have
![q = 2\times 10^{-6} C](https://tex.z-dn.net/?f=q%20%3D%202%5Ctimes%2010%5E%7B-6%7D%20C)
![r = 5 cm = 0.05 m](https://tex.z-dn.net/?f=r%20%3D%205%20cm%20%3D%200.05%20m)
now we have
![E = 2\frac{(9\times 10^9)(2\times 10^{-6})}{0.05^2}](https://tex.z-dn.net/?f=E%20%3D%202%5Cfrac%7B%289%5Ctimes%2010%5E9%29%282%5Ctimes%2010%5E%7B-6%7D%29%7D%7B0.05%5E2%7D)
![E = 1.44 \times 10^7 N/C](https://tex.z-dn.net/?f=E%20%3D%201.44%20%5Ctimes%2010%5E7%20N%2FC)
A region within a magnetic material in which magnetization is in a uniform direction this means the individual magnetic moments of the atoms are aligned with one another and they point the same direction. when cooled bwlow a temperature called the curie temperature the magnetization of a piece of ferromagnetic material.<span />
Answer:
123 J transfer into the gas
Explanation:
Here we know that 123 J work is done by the gas on its surrounding
So here gas is doing work against external forces
Now for cyclic process we know that
![\Delta U = 0](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%200)
so from 1st law of thermodynamics we have
![dQ = W + \Delta U](https://tex.z-dn.net/?f=dQ%20%3D%20W%20%2B%20%5CDelta%20U)
![dQ = W](https://tex.z-dn.net/?f=dQ%20%3D%20W)
so work done is same as the heat supplied to the system
So correct answer is
123 J transfer into the gas