Answer:
Option C. Its image falls on the periphery of the retina, denser in rods
Explanation:
This can be explained as the photo receptors present in the outer periphery are more sensitive to light and enable us to see clearly dimly lit objects.
Also the peripheral vision is the the one on the retinal periphery which is highly rich in concentration of rods than cones and thus a dimly lit object at night with peripheral vision gives a clearer image of the object.
The central vision makes use of the central part, i.e., fovea which is dense with the concentration of cones, photo-receptors which enable us to see different colors and functions well in bright light.
Answer:
Explanation:
Time period of satellite T = 6.43 X 60 X 60 s
acceleration = ω²R
4π²x R / T²
4 x 3.14²x 6.38 x 10⁶ / (6.43 x60 x 60)²
= 0.47 m /s²
Explanation:
(a) It is known that relation between charge and volume is as follows.
![q_{enclosed} = \rho V_{cube}](https://tex.z-dn.net/?f=q_%7Benclosed%7D%20%3D%20%5Crho%20V_%7Bcube%7D)
= ![(509 \times 10^{-9} C/m^{3}) \times (0.04 m)^{3}](https://tex.z-dn.net/?f=%28509%20%5Ctimes%2010%5E%7B-9%7D%20C%2Fm%5E%7B3%7D%29%20%5Ctimes%20%280.04%20m%29%5E%7B3%7D)
= ![509 \times 10^{-9} \times 6.4 \times 10^{-5}](https://tex.z-dn.net/?f=509%20%5Ctimes%2010%5E%7B-9%7D%20%5Ctimes%206.4%20%5Ctimes%2010%5E%7B-5%7D)
=
Now, according to Gauss's law
![\phi = \frac{q_{enclosed}}{\epsilon_{o}}](https://tex.z-dn.net/?f=%5Cphi%20%3D%20%5Cfrac%7Bq_%7Benclosed%7D%7D%7B%5Cepsilon_%7Bo%7D%7D)
= ![\frac{3.26 \times 10^{-11} C}{8.85 \times 10^{-12}C^{2}N^{-1}m^{-2}}](https://tex.z-dn.net/?f=%5Cfrac%7B3.26%20%5Ctimes%2010%5E%7B-11%7D%20C%7D%7B8.85%20%5Ctimes%2010%5E%7B-12%7DC%5E%7B2%7DN%5E%7B-1%7Dm%5E%7B-2%7D%7D)
= 3.68 ![N m^{2}/C](https://tex.z-dn.net/?f=N%20m%5E%7B2%7D%2FC)
Hence, the electric flux through this cubical surface if its edge length is 4.00 cm is 3.68
.
(b) Similarly, we will calculate the electric flux when edge length is 16.8 cm as follows.
![q_{enclosed} = \rho V_{cube}](https://tex.z-dn.net/?f=q_%7Benclosed%7D%20%3D%20%5Crho%20V_%7Bcube%7D)
= ![(509 \times 10^{-9} C/m^{3}) \times (0.168 m)^{3}](https://tex.z-dn.net/?f=%28509%20%5Ctimes%2010%5E%7B-9%7D%20C%2Fm%5E%7B3%7D%29%20%5Ctimes%20%280.168%20m%29%5E%7B3%7D)
= ![509 \times 10^{-9} \times 4.74 \times 10^{-3}](https://tex.z-dn.net/?f=509%20%5Ctimes%2010%5E%7B-9%7D%20%5Ctimes%204.74%20%5Ctimes%2010%5E%7B-3%7D)
=
Now, according to Gauss's law
![\phi = \frac{q_{enclosed}}{\epsilon_{o}}](https://tex.z-dn.net/?f=%5Cphi%20%3D%20%5Cfrac%7Bq_%7Benclosed%7D%7D%7B%5Cepsilon_%7Bo%7D%7D)
= ![\frac{2.41 \times 10^{-11} C}{8.85 \times 10^{-12}C^{2}N^{-1}m^{-2}}](https://tex.z-dn.net/?f=%5Cfrac%7B2.41%20%5Ctimes%2010%5E%7B-11%7D%20C%7D%7B8.85%20%5Ctimes%2010%5E%7B-12%7DC%5E%7B2%7DN%5E%7B-1%7Dm%5E%7B-2%7D%7D)
= 2.72 ![N m^{2}/C](https://tex.z-dn.net/?f=N%20m%5E%7B2%7D%2FC)
Therefore, the electric flux through this cubical surface if its edge length is 4.00 cm is 2.72
.
A . <span>Increase from left to right</span>
To answer the problem we would be using this formula which isE = hc/L where E is the energy, h is Planck's constant, c is the speed of light and L is the wavelength
L = hc/E = 4.136×10−15 eV·s (2.998x10^8 m/s)/10^4 eV
= 1.240x10^-10 m
= 1.240x10^-1 nm