Answer: A
Explanation:
Well the high and lows effect the humidity the more humidity the more hot it is so the high brings higher temperatures.
Answer:
( a ) The specific volume by ideal gas equation = 0.02632 
% Error = 20.75 %
(b) The value of specific volume From the generalized compressibility chart = 0.0142 
% Error = - 34.85 %
Explanation:
Pressure = 1 M pa
Temperature = 50 °c = 323 K
Gas constant ( R ) for refrigerant = 81.49 
(a). From ideal gas equation P V = m R T ---------- (1)
⇒
= 
⇒ Here
= Specific volume = v
⇒ v = 
Put all the values in the above formula we get
⇒ v =
×81.49
⇒ v = 0.02632 
This is the specific volume by ideal gas equation.
Actual value = 0.021796 
Error = 0.02632 - 0.021796 = 0.004524 
% Error =
× 100
% Error = 20.75 %
(b). From the generalized compressibility chart the value of specific volume
= v = 0.0142 
The actual value = 0.021796 
Error = 0.0142 - 0.021796 = 
% Error =
× 100
% Error = - 34.85 %
Answer:
The electric field strength is 
Solution:
As per the question:
Area of the electrode, 
Charge, q = 50 nC = ![50\times 10^{- 9} C[/etx]Distance, x = 2 mm = [tex]2\times 10^{- 3} m](https://tex.z-dn.net/?f=50%5Ctimes%2010%5E%7B-%209%7D%20C%5B%2Fetx%5D%3C%2Fp%3E%3Cp%3EDistance%2C%20x%20%3D%202%20mm%20%3D%20%5Btex%5D2%5Ctimes%2010%5E%7B-%203%7D%20m)
Now,
To calculate the electric field strength, we first calculate the surface charge density which is given by:

Now, the electric field strength of the electrode is:

where



Answer:
halve the slit separation
Explanation:
As we know that
In YDS experiment, the equation of fringe width is as follows

where,
D denotes the separation in the middle of screen and slits
d denotes the distance in the middle of two slits
And to increase the Δx we have to decrease the d i.e, the distance between the two slits
Hence, the first option is correct
For a circular aperture, the first minima (n=1) as an angular separation from the peak of the central maxima given by
Sinθ = 1.22λ / d
Where,
d is the aperture or pupil diameter
d = 4.69 mm = 4.69 × 10^-3m
λ is the wavelength
λ = 545 nm = 545 × 10^-9 m
Then,
Sinθ = 1.22λ / d
Sinθ = 1.22 × 545 × 10^-9 / 4.69 × 10^-3
Sinθ = 1.418 × 10^-4 rad
Then, the head light sources have the same angular separation θ from the eye as the image have inside the eye.
For the headlight
Sinθ ≈ light separation / distantce for the eye
Light separation is give as x = 0.659 m
And let the distance of the eye be D
Then,
Sinθ = x / D
Make D subject of formula
D = x / Sinθ
D = 0.695 / 1.418 × 10^-4
D = 4902.316m
To km, 1km = 1000m
D ≈ 4.9 km