One coulomb of charge has the equivalent charge of 6.25x10^18 electrons. This is determined from the value of charge on one electron and the value of charge for 1 coulomb
Just as during formation, when the material contracts, the temperature and pressure increase. This newly generated heat temporarily counteracts the force of gravity, and the outer layers of the star are now pushed outward ( in not sure tho ) I hope this helps
Each 480-volt, three-phase, wye-connected feeder disconnect with a minimum rating of 100 amperes is required to be ground-fault protected.
A feeder disconnect is the main electrical disconnecting device in an electrical circuit. It provides an accessible location for disconnecting all conductors at one time and is usually located near the source of power. Feeders can consist of cables or busbars.
To be considered a feeder, it must have a voltage that exceeds 50 volts under normal conditions and must supply more than one receptacle or piece of equipment. A feeder disconnect must have a minimum rating of 100 amperes or less and may be rated up to 1000 amperes or higher depending on how much power it supplies to its load(s).
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Answer:
225nm
Explanation:
The Maxima of diffraction girating
dsin( θ)= m(λ)
Where θ)= angle between central axis and corresponding point
The wavelength is to corresponding sin( θ)
Then the maximum number of (λ) can be seen by setting sin( θ)= 1 then
d= m(λ)
At fifth order maximum m=1 then
d= 5(λ)
Where d= distance between the slit and with girating of 1/890rullig/mm
d= 1/890rullig/mm= 8.90×10^-6m
At maximum wavelength
d= 5(λ)
(λ)=5/d
= 8.90×10^-6m/5
=2.75*10^-7
= 275nm
Therefore, longest wavelength that forms an intensity maximum in the fifth order is 275nm
Answer:
.y₂ = 0.5704 m
, y2= 44.47 m,
Explanation:,
For this exercise we will use the kinematics relations
Ball
y₁ = y₀ + v₀₁ t
y₀ = 11.0 m
v₀₁ = 5.10 m / s
Pellet
y₂ = 0 + v₀₂ t - ½ g t²
V₀₂ = 39.0 m / s
At the meeting point the two bodies have the same height
y₁ = y₂
y₀ + v₀₁ t = v₀₂ t -1/2 g t²
11 + 5.1 t = 39 t - ½ 9.8 t²
4.9 t² - 33.9 t +11 = 0
.t2 - 6,918 + 2,245 = 0
t = [6,918 ±√ 6,918 2 - 4 2,245)] / 2
t = [6,918 ± 6.2356.] / 2
t₁ = 6.58 s
t₂ = 0.3412 s
Let's calculate the positions for each time
t₂ = 0.3412 s
y₂ = 39 t - ½ 9.8 t2
y₂ = 39 0.3112 - ½ 9.8 0.3412²
y₂ = 0.5704 m
.t1 = 6.58 s
.₂ = 39 6.58 - ½ 9.8 6.58 ^ 2
y2= 44.47 m