Answer:
W = 0.562 J
Explanation:
Initial potential energy of two point charges is given as
![U_i = \frac{kq_1q_2}{r_1}](https://tex.z-dn.net/?f=U_i%20%3D%20%5Cfrac%7Bkq_1q_2%7D%7Br_1%7D)
here we have
![q_1 = 3.10 \mu C](https://tex.z-dn.net/?f=q_1%20%3D%203.10%20%5Cmu%20C)
![q_2 = -4.50 \mu C](https://tex.z-dn.net/?f=q_2%20%3D%20-4.50%20%5Cmu%20C)
![r_1 = 0.140 m](https://tex.z-dn.net/?f=r_1%20%3D%200.140%20m)
now we have
![U_i = \frac{(9\times 10^9)(3.10 \mu C)(-4.50 \mu C)}{0.140}](https://tex.z-dn.net/?f=U_i%20%3D%20%5Cfrac%7B%289%5Ctimes%2010%5E9%29%283.10%20%5Cmu%20C%29%28-4.50%20%5Cmu%20C%29%7D%7B0.140%7D)
![U_i = -0.897 J](https://tex.z-dn.net/?f=U_i%20%3D%20-0.897%20J)
now at final position the distance between two charges is given as
![r_2 = \sqrt{0.265^2 + 0.265^2} = 0.375 m](https://tex.z-dn.net/?f=r_2%20%3D%20%5Csqrt%7B0.265%5E2%20%2B%200.265%5E2%7D%20%3D%200.375%20m)
Now final energy is given as
![U_f =\frac{kq_1q_2}{r_2}](https://tex.z-dn.net/?f=U_f%20%3D%5Cfrac%7Bkq_1q_2%7D%7Br_2%7D)
![U_f = \frac{(9\times 10^9)(3.10 \mu C)(-4.50 \mu C)}{0.375}](https://tex.z-dn.net/?f=U_f%20%3D%20%5Cfrac%7B%289%5Ctimes%2010%5E9%29%283.10%20%5Cmu%20C%29%28-4.50%20%5Cmu%20C%29%7D%7B0.375%7D)
![U_f = -0.335 J](https://tex.z-dn.net/?f=U_f%20%3D%20-0.335%20J)
Now the work done to change the position of charge is given as
![W = U_f - U_i](https://tex.z-dn.net/?f=W%20%3D%20U_f%20-%20U_i)
![W = (-0.335) - (-0.897) = 0.562 J](https://tex.z-dn.net/?f=W%20%3D%20%28-0.335%29%20-%20%28-0.897%29%20%3D%200.562%20J)
We know, work done = Force * displacement
Here, w = 12,000 J
f = 300 N
Substitute their values into the expression:
12,000 = 300 * d
d = 12,000 / 300
d = 120/3
d = 40 m
In short, Your Answer would be 40 meters
Hope this helps!
Answer:
The refractive index of fluid 2 is 1.78
Explanation:
Refractive index , n = real depth/apparent depth
For the first fluid, n = 1.37 and apparent depth = 9.00 cm.
The real depth of the container is thus
real depth = n × apparent depth = 1.37 × 9.00 cm = 12.33 cm
To find the refractive index of fluid index of fluid 2, we use the relation
Refractive index , n = real depth/apparent depth.
Now,the real depth = 12.33 cm and the apparent depth = 6.86 cm.
So, n = 12.33 cm/6.86 cm = 1.78
So the refractive index of fluid 2 is 1.78
<span>The Balmer series or Balmer lines in atomic physics, is the designation of one of a set of six named series describing the spectral line emissions of the hydrogen atom. The Balmer series is calculated using the Balmer formula, an empirical equation discovered by Johann Balmer in 1885. this is all I know sorry</span>