Answer:
The magnitude of the electric field is ![5.1 \times 10^{11} \frac{N}{C}](https://tex.z-dn.net/?f=5.1%20%5Ctimes%2010%5E%7B11%7D%20%20%5Cfrac%7BN%7D%7BC%7D)
Explanation:
Given:
Charge of electron
C
Separation between two charges
m
For finding the magnitude of the electric field,
![E= \frac{kq}{r^{2} }](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7Bkq%7D%7Br%5E%7B2%7D%20%7D)
Where ![k = 9 \times 10^{9}](https://tex.z-dn.net/?f=k%20%3D%209%20%5Ctimes%2010%5E%7B9%7D)
![E = \frac{9 \times 10^{9} \times 1.6 \times 10^{-19} }{(5.3 \times 10^{-11} )^{2} }](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B9%20%5Ctimes%2010%5E%7B9%7D%20%5Ctimes%201.6%20%5Ctimes%2010%5E%7B-19%7D%20%7D%7B%285.3%20%5Ctimes%2010%5E%7B-11%7D%20%29%5E%7B2%7D%20%7D)
![\frac{N}{C}](https://tex.z-dn.net/?f=%5Cfrac%7BN%7D%7BC%7D)
Therefore, the magnitude of the electric field is ![5.1 \times 10^{11} \frac{N}{C}](https://tex.z-dn.net/?f=5.1%20%5Ctimes%2010%5E%7B11%7D%20%20%5Cfrac%7BN%7D%7BC%7D)
I'm pretty sure that the only right answer is <span>B) a north-seeking pole of unit strength with a south-seeking pole of the same strength. Because there are no single magnetic poles.</span>
Answer:
The momentum principle tells us that impulse transfers momentum to an object. If an object has 2 kg m/s of momentum, a 1 kg m/s impulse delivered to the object increases its momentum to 3 kg m/s. That is, pfx = pix + Jx. Just as we did with energy, we can represent this “momentum accounting” with a momentum bar chart.
Answer:
![{\dfrac{\Delta V}{V}}=11\times 10^{-7}](https://tex.z-dn.net/?f=%7B%5Cdfrac%7B%5CDelta%20V%7D%7BV%7D%7D%3D11%5Ctimes%2010%5E%7B-7%7D)
Explanation:
Given that
Bulk modulus ,K = 7.1 x 10¹⁰ Pa
The pressure P₁ = 0.781 x 10⁵ Pa
The final pressure P ₂ = 0 Pa
We know that bulk modulus given as
![K=\dfrac{\Delta P}{\dfrac{\Delta V}{V}}](https://tex.z-dn.net/?f=K%3D%5Cdfrac%7B%5CDelta%20P%7D%7B%5Cdfrac%7B%5CDelta%20V%7D%7BV%7D%7D)
Now by putting the values
![K=\dfrac{\Delta P}{\dfrac{\Delta V}{V}}](https://tex.z-dn.net/?f=K%3D%5Cdfrac%7B%5CDelta%20P%7D%7B%5Cdfrac%7B%5CDelta%20V%7D%7BV%7D%7D)
![7.1\times 10^{10}=\dfrac{0.781\times 10^5}{\dfrac{\Delta V}{V}}](https://tex.z-dn.net/?f=7.1%5Ctimes%2010%5E%7B10%7D%3D%5Cdfrac%7B0.781%5Ctimes%2010%5E5%7D%7B%5Cdfrac%7B%5CDelta%20V%7D%7BV%7D%7D)
![{\dfrac{\Delta V}{V}}=\dfrac{0.781\times 10^5}{7.1\times 10^{10}}](https://tex.z-dn.net/?f=%7B%5Cdfrac%7B%5CDelta%20V%7D%7BV%7D%7D%3D%5Cdfrac%7B0.781%5Ctimes%2010%5E5%7D%7B7.1%5Ctimes%2010%5E%7B10%7D%7D)
![{\dfrac{\Delta V}{V}}=11\times 10^{-7}](https://tex.z-dn.net/?f=%7B%5Cdfrac%7B%5CDelta%20V%7D%7BV%7D%7D%3D11%5Ctimes%2010%5E%7B-7%7D)
Therefore fractional change will be 11 x 10⁻⁷.