Answer:
Refractive index of a medium = Speed of light in vacuum/Speed of light in a medium
1.5 = 3 x 108 / Speed of light in medium
Speed of light in the medium = 3 x 108 /1.5
= 2 x 108 m/s.
Explanation:
Answer:
![F_{net} = 31.88 N](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%2031.88%20N)
Explanation:
When top block is just or about to slide on the lower block then we can say that the frictional force on it will be maximum static friction
So we will have
![F_{net} = ma](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20ma)
![F_s = ma](https://tex.z-dn.net/?f=F_s%20%3D%20ma)
![\mu_s mg = ma](https://tex.z-dn.net/?f=%5Cmu_s%20mg%20%3D%20ma)
![a = \mu_s g](https://tex.z-dn.net/?f=a%20%3D%20%5Cmu_s%20g)
![a = (0.50)(9.81)](https://tex.z-dn.net/?f=a%20%3D%20%280.50%29%289.81%29)
![a = 4.905 m/s^2](https://tex.z-dn.net/?f=a%20%3D%204.905%20m%2Fs%5E2)
now for the Net force on two blocks to move together
![F_{net} = (m_1 + m_2) a](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20%28m_1%20%2B%20m_2%29%20a)
![F_{net} = (2.3 + 4.2)(4.905)](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20%282.3%20%2B%204.2%29%284.905%29)
![F_{net} = 31.88 N](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%2031.88%20N)
If it is just a reseach investigatory project, you could monitor it by using different variables (controlled), acurate measuring devices and camera for documentations
Answer:
![400000\ \text{N/C}](https://tex.z-dn.net/?f=400000%5C%20%5Ctext%7BN%2FC%7D)
Explanation:
= Charge at 3000 m = 40 C
= Charge at 1000 m = -40 C
= 3000 m
= 1000 m
k = Coulomb constant = ![9\times10^9\ \text{Nm}^2/\text{C}^2](https://tex.z-dn.net/?f=9%5Ctimes10%5E9%5C%20%5Ctext%7BNm%7D%5E2%2F%5Ctext%7BC%7D%5E2)
Electric field due to the charge at 3000 m
![E_1=\dfrac{k|q_1|}{r_1^2}\\\Rightarrow E_1=\dfrac{9\times 10^9\times 40}{3000^2}\\\Rightarrow E_1=40000\ \text{N/C}](https://tex.z-dn.net/?f=E_1%3D%5Cdfrac%7Bk%7Cq_1%7C%7D%7Br_1%5E2%7D%5C%5C%5CRightarrow%20E_1%3D%5Cdfrac%7B9%5Ctimes%2010%5E9%5Ctimes%2040%7D%7B3000%5E2%7D%5C%5C%5CRightarrow%20E_1%3D40000%5C%20%5Ctext%7BN%2FC%7D)
Electric field due to the charge at 1000 m
![E_2=\dfrac{k|q_2|}{r_2^2}\\\Rightarrow E_2=\dfrac{9\times 10^9\times 40}{1000^2}\\\Rightarrow E_2=360000\ \text{N/C}](https://tex.z-dn.net/?f=E_2%3D%5Cdfrac%7Bk%7Cq_2%7C%7D%7Br_2%5E2%7D%5C%5C%5CRightarrow%20E_2%3D%5Cdfrac%7B9%5Ctimes%2010%5E9%5Ctimes%2040%7D%7B1000%5E2%7D%5C%5C%5CRightarrow%20E_2%3D360000%5C%20%5Ctext%7BN%2FC%7D)
Electric field at the aircraft is
.