Answer:
I=P/U=6/12=0.5(A)
Explanation: P=UI ( CÔNG SUẤT = HIỆU ĐIỆN THẾ NHÂN VỚI C Đ D Đ)
B.)from conductor to insulator
C.)from high potential to low potential
D.)from series circuit to parallel circuit
excretory system?
Answer:
excretion
Explanation:
The bodily process of discharging waste.
The coordinates of the event in system K can be found using the Lorentz transformation equations, and they are (10/3 m, 3.5 m, 3.5 m, 8/3 m/c).
The Lorentz transformation equations relate the coordinates of an event in one reference frame to the coordinates of the same event in another reference frame that is moving with a constant velocity relative to the first reference frame.
The equations are as follows:
x = γ(x' + vt')
y = y'
z = z'
t = γ(t' + vx'/c^2)
Where γ is the Lorentz factor, which is given by:
γ = 1/√(1-v^2/c^2)
In this case, v = 0.8c, so γ = 1/√(1-(0.8c)^2/c^2) = 5/3.
Plugging in the values for x', y', z', t', v, and γ into the Lorentz transformation equations, we get:
x = (5/3)(2 m + (0.8c)(0)) = 10/3 m
y = 3.5 m
z = 3.5 m
t = (5/3)(0 + (0.8c)(2 m)/c^2) = 8/3 m/c
Therefore, the coordinates of the event in system K are (10/3 m, 3.5 m, 3.5 m, 8/3 m/c).
To know more about Lorentz transformation equations, refer here:
brainly.com/question/29655824#
#SPJ11
Answer:
By the amount of energy they carry
Explanation:
Answer:
The average angular acceleration of the Earth, α = 6.152 X 10⁻²⁰ rad/s²
Explanation:
Given data,
The period of 365 rotation of Earth in 2006, T₁ = 365 days, 0.840 sec
= 3.1536 x 10⁷ +0.840
= 31536000.84 s
The period of 365 rotation of Earth in 2006, T₀ = 365 days
= 31536000 s
Therefore, time period of one rotation on 2006, Tₐ = 31536000.84/365
= 86400.0023 s
The time period of rotation is given by the formula,
Tₐ = 2π /ωₐ
ωₐ = 2π / Tₐ
Substituting the values,
ωₐ = 2π / 365.046306
= 7.272205023 x 10⁻⁵ rad/s
Therefore, the time period of one rotation on 1906, Tₓ = 31536000/365
= 86400 s
Time period of rotation,
Tₓ = 2π /ωₓ
ωₓ = 2π / T
= 2π /86400
= 7.272205217 x 10⁻⁵ rad/s
The average angular acceleration
α = (ωₓ - ωₐ) / T₁
= (7.272205217 x 10⁻⁵ - 7.272205023 x 10⁻⁵) / 31536000.84
α = 6.152 X 10⁻²⁰ rad/s²
Hence the average angular acceleration of the Earth, α = 6.152 X 10⁻²⁰ rad/s²
The average angular acceleration of the Earth from the year 1906 to 2006 would be -5.73 x 10^-20 rad/s^2. This value was obtained by finding the change in angular velocity and then dividing it by the elapsed time.
The question is asking for the average angular acceleration of the Earth from the year 1906 to 2006, during which the Earth's rotation rate decreased, causing the day to increase in duration by about 0.840 seconds.
To find the average angular acceleration, you first need to calculate the change in angular velocity, which can be found from the change in rotation time. One revolution (one day) is 2π radians, so the change in angular velocity is Δω = 2π/86400 s - 2π/(86400+0.840) s = -1.81 x 10^-10 rad/s.
The time interval from 1906 to 2006 is 100 years or about 3.16 x 10^9 seconds. Therefore, the average angular acceleration, α, which is the change in angular velocity divided by time, would be α = Δω/Δt = -1.81 x 10^-10 rad/s / 3.16 x 10^9 s = -5.73 x 10^-20 rad/s^2.
#SPJ3
B. grid.
C. anode.
D. deflecting plate.
Answer: B. GRID
Explanation: I JUST TOOK THE PF EXAM AND GOT IT CORRECT!!!!