Answer: The correct answer is B. "the electromagnetic waves appear more red in color".
Explanation:
Scientists use the Doppler effect to understand the universe by determining the motion of the object.
Red shift: The apparent frequency of light decreases as the source object in space moves away from us. The light is shifted to red end. The apparent frequency decreases and there is increase in the wavelength of the light.
Blue shift: The apparent frequency of light increases as the object in space moves towards us. The light is shifted to blue end. In this case, there is decrease in the wavelength of the light. The frequency of the wave increases.
In the given problem, when the light source moves further away from an observer, the electromagnetic waves appear red in color.
Therefore, the correct option is (B).
Explanation:
Area of ring 
Charge of on ring 
Charge on disk
![\begin{aligned}d v &=\frac{k d q}{\sqrt{x^{2}+a^{2}}} \\&=2 \pi-k \frac{a d a}{\sqrt{x^{2}+a^{2}}} \\v(1) &=2 \pi c k \int_{0}^{R} \frac{a d a}{\sqrt{x^{2}+a^{2}}} \cdot_{2 \varepsilon_{0}}^{2} R \\&=2 \pi \sigma k[\sqrt{x^{2}+a^{2}}]_{0}^{2} \\&=\frac{2 \pi \sigma}{4 \pi \varepsilon_{0}}[\sqrt{z^{2}+R^{2}}-(21)] \\&=\frac{\sigma}{2}(\sqrt{2^{2}+R^{2}}-2)\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Dd%20v%20%26%3D%5Cfrac%7Bk%20d%20q%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%7D%20%5C%5C%26%3D2%20%5Cpi-k%20%5Cfrac%7Ba%20d%20a%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%7D%20%5C%5Cv%281%29%20%26%3D2%20%5Cpi%20c%20k%20%5Cint_%7B0%7D%5E%7BR%7D%20%5Cfrac%7Ba%20d%20a%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%7D%20%5Ccdot_%7B2%20%5Cvarepsilon_%7B0%7D%7D%5E%7B2%7D%20R%20%5C%5C%26%3D2%20%5Cpi%20%5Csigma%20k%5B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%5D_%7B0%7D%5E%7B2%7D%20%5C%5C%26%3D%5Cfrac%7B2%20%5Cpi%20%5Csigma%7D%7B4%20%5Cpi%20%5Cvarepsilon_%7B0%7D%7D%5B%5Csqrt%7Bz%5E%7B2%7D%2BR%5E%7B2%7D%7D-%2821%29%5D%20%5C%5C%26%3D%5Cfrac%7B%5Csigma%7D%7B2%7D%28%5Csqrt%7B2%5E%7B2%7D%2BR%5E%7B2%7D%7D-2%29%5Cend%7Baligned%7D)
Note: Refer the image attached
Answer:
E=3 x 10^4 N/c
Explanation:
The electric field strength can be found out disk with a uniform positive surface charge density by
σ= charge density
r= radius of the disk
z= position in which we have to find electric field = 15 cm
ε_0= constant ( vacuum permitivity)
putting values we get
solving we get
E=30000 N/c
E=3 x 10^4 N/c
The net force acting on the car is
3 × 10^3
Newtons.
Force is defined as the product of the mass of the body and its aaceleration,
⇒ F = ma
Substituting the above given values we get,
F =(1500kg) (2.0 m/s^2) = 3000N = 3 × 10^3N .
<span>PV = nRT
moles of H2 = 1/2 = 0.5
moles of He = 1/4 =0.25
T = 273 + 27
partial pressure of H2
Px1 = 0.5x0.083x300
P=12.45 atmospheres
PP of He
px1 = 0.25x0.083x300
P =6 22 atmospheres
Totla pressure = 6.22 + 12.45 = 18.68 atm</span>
