A unit of angular velocity

In the international system of units, it is radians per second, which is read as radians per second. In the calculation process of circular motion, angular velocity must be used. In order to calculate with the international system of units, such as meter, meter / second, meter / second, kilogram, Newton, etc

Angular velocity and period
Period is equal to 2 π r divided by angular velocity, 2 π is a real number, angular velocity is a vector, then according to the algorithm of vector multiplication to get scalar, period should also be a vector, but period is a scalar, why (not to mention that the direction of angular velocity is not studied in middle school.)

First, what is a cycle?
Cycle is simply time
It's right to multiply a vector by a scalar, but only if it's a vector
In the process of movement and change, some features appear repeatedly, and the time of two consecutive occurrences is called "cycle"
Time is scalar and period belongs to time
Second, 2 / w = t, which is explained mathematically
And the direction of the cycle is meaningless

What is the angular velocity of the earth's rotation

The angular velocity of the earth rotation is ω = 2 π / (24x3600s) = 7.27x10 ^ - 5 rad / s

How to find the effective value of non sinusoidal alternating current?
How to calculate?

Average power = work / time, work is the area under the sinusoidal curve (which can be obtained by integration), and time is the abscissa. According to the above calculation, the average power of sinusoidal alternating current is 1 / √ 2 times of the peak value, and the effective current is proportional to the average power, that is, the effective value of sinusoidal alternating current is equal to 1 / √ 2 times of the peak value

Calculation of effective value of non sinusoidal alternating current
yuexiangxiyuehao

In fact, to calculate the effective value is to convert it into plane analytic geometry
Draw a curve in a two-dimensional rectangular plane coordinate system
The x-axis is the time t (unit s), and the y-axis is the instantaneous voltage value, finally forming a curve
Suppose that the area of the intersection of the unit period (T) and the abscissa (t axis) of this curve is s
Effective value = s / T;
How to calculate the area depends on your own

Given a certain sinusoidal current I = 100sin (314t-45 °) a, calculate the maximum value of alternating current I m, efficiency I, cycle T of alternating current, frequency f. initial phase of alternating current

AC maximum value im = 100A, effective value I = im / √ 2 = 100 / √ 2 = 70.7a, AC frequency f = ω / 2 π = 314 / 2 / 3.14 = 50 Hz, AC cycle T = 1 / F = 1 / 50 = 0.02 s, initial phase of AC = - 45 & # 186;

The effective value of alternating current is based on For a sinusoidal alternating current, its peak value is one percent of its effective value If the capacitor is connected to the AC circuit, it can play an important role__ Function

The effective value of alternating current is defined according to the thermal effect of current. For sinusoidal alternating current, its effective value is twice of its peak value. If the capacitor is connected to the AC circuit, it can play the role of connecting AC and DC. So the answer is: thermal effect of current, 2. Connecting AC and isolating DC

If a sinusoidal AC current I = 10sin (100 * 3.14 + 3.14 / 3) a is known, what is its effective value and frequency

Do you know the formula of sine alternating current? Do you know Baidu! The voltage and current of sine are sine wave. I = root sign, half of im (10) is: 5 times root sign, 2f = 100 / 2pai = 50 / Pai, initial phase: 3.14/3

Calculus proof of effective current value of sine alternating current

First of all, the effective value of sinusoidal alternating current refers to the value of direct current equivalent to its ability to do work. That is, in the same time, when the heat generated by a DC voltage and current and an AC voltage and current on the same resistance is equal, we call the DC voltage and current the effective value of corresponding AC voltage and current. Therefore, this problem should be deduced from the perspective of doing work
There is a pure resistance circuit, the resistance is r, the voltage instantaneous value of sinusoidal alternating current is u = um * sin ω T, the current instantaneous value is I = im * sin ω t
1. Calculate the instantaneous power,
Power P = u * U / r = um * um * sin ω t * sin ω T / r = 1 / 2 * um * um * (1-cos2 ω T) / R
2. Calculate the work of this formula in a period T
If the time changes by an infinitesimal increment DT, we obtain that the work done by alternating current in DT time is:
dw=p*dt=Um*Um*sinωt*sinωt*dt/R
=1/2*Um*Um/R*（1-cos2ωt)*dt;
Then calculate the work of alternating current in a period T
DW for time integral, the lower limit is 0 and the upper limit is t
Work w = int (0-t) P * DT = int (0-t) (1 / 2 * um * um * (1-cos2 ω T) * DT) / r = 1 / 2R * um * um * int (0-t) DT + 1 / 2R * um * um * int (0-t) Cos2 ω t * DT;
Where int is the integral and (0-t) is the integral limit
The integral on the right side of this equation is zero
W=1/2R * Um*Um*T.
The physical meaning of this formula is: 1 / 2 * um * um is a constant, which is a DC component in electricity, while the integral of AC component in a period is zero, because the waveform is symmetrical
3. By calculating the average power, we can get the equivalent direct current, i.e. voltage RMS and current RMS
P=W/T=1/2R * Um*Um.
Let: u be the corresponding equivalent DC voltage, i.e. U * U / r = 1 / 2 * um * um / R;
Similarly, P = I * I * r = im * im * r * sin ω t * sin ω t = 1 / 2 * im * im * r * (1-cos2 ω T);
W=int(0-t)P*dt=1/2*Im*Im*R*T
The average power P = w / T = 1 / 2 * im * im * r. let: I * I * r = 1 / 2 * im * im * r, we get:
I=√2/2 * Im.
Why can we only take one cycle? Because every cycle of alternating current is exactly the same, we can get the average value of one cycle, which is equal to the average value of all time
Enough details? If I don't understand, I won't move

How to calculate the current (alternating current) according to the power of electrical appliances

Current is related to power and voltage. Generally, 380V electricity is used. For motors, there are two currents per kilowatt. For 380 heating, there are 1.5 currents per kilowatt. For single-phase equipment, it is estimated that there are five currents per kilowatt

A unit of angular velocity