Answer:
A) the current in AC electricity varies in magnitude and direction.
C) the voltage in AC electricity varies in magnitude and direction.
Explanation:
In DC current and voltage the direction of current will not change with time and it always remains the same.
So here in DC voltage and DC current the magnitude may change with time but the direction will always remain same
While in AC voltage and AC current the direction of AC will change with time
periodically.
So here magnitude and direction both will change in AC current and AC voltage.
so the correct answer is
A) the current in AC electricity varies in magnitude and direction.
C) the voltage in AC electricity varies in magnitude and direction.
Answer:a) P = Po + rho×h×g
b) P = 5.4 × 10^9 pa
c) F = P/A = (Po + rho×h×g)/A
d) 1.174×10^11N
Explanation: Using the formula
P = Po + rho×h×g
P = 1.0 x 10^5 + 1000 × 5.5 × 9.81
P = 5.4 × 10^9pa
The magnitude of the force exerted by water on the top of the person's head F at the depth h in terms of P
F = P/A = (Po + rho×h×g)/A
Using the above formula
Where A = 0.046m^2
F = P/ A = 5.4×10^9/0.046
F = 1.174×10^11N
X - rays are highly penetrating electromagnetic radiations that are formed when the cathode rays strikes a dense metal.
It has many characteristics such as :
- travels with the speed of light.
Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V
Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:
h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.
In order to calculate the cutoff wavelength we have to consider that Ek=0
in this case h*ν=W
(h*c)/λ=4.52 eV
λ= (h*c)/4.52 eV
λ= (1240 eV*nm)/(4.52 eV)=274.34 nm
From this h*ν = Ek+W; we can calculate the kinetic energy for a radiation wavelength of 198 nm
then we have
(h*c)/(λ)-W= Ek
Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV
Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this acts to slow down the ejected electrons from the catode.
<span>Germanium
To determine which melts first, convert their melting temperatures so they're both expressed on same scale. It doesn't matter what scale you use, Kelvin, Celsius, of Fahrenheit. Just as long as it's the same scale for everything. Since we already have one substance expressed in Kelvin and since it's easy to convert from Celsius to Kelvin, I'll use Kelvin. So convert the melting point from Celsius to Kelvin for Gold by adding 273.15
1064 + 273.15 = 1337.15 K
So Germanium melts at 1210K and Gold melts at 1337.15K. Germanium has the lower melting point, so it melts first.</span>
