How many kcal is 5000 kJ?

How many kcal is 5000 kJ?

1 cal = 4.2 Joule 1 joule equals 1 / 4.2 cal = 0.2391 cal
1 kJ is equal to 1 / 4.2 kcal = 0.2391 kcal
5000 kJ equals 1195.5 kcal
How many kilojoules is a large calorie?
1 cal = 4.18 J
1 kcal = 4.18 kJ
00411 kJ = 0
To be more precise:
1 cal = 4.184 J
4 kcal = 184
CAL in English, kcal in kilocalorie, j in Joule and kJ in kilojoule
Therefore, the units in the table of nutrients we see in processed products are all expressed in English
How many kilowatt hours is one kilowatt hour
One kilowatt hour is one kilowatt hour, and the electricity consumed by an electric appliance with one kilowatt hour is one kilowatt hour
The calculation formula is: power consumption (KWH) = power (kw) × service time (H)
How to calculate the three-phase current? How to calculate the single-phase current? Is the calculation formula of 220 V power divided by voltage the current value?
Is 380V power divided by voltage × 1.732 equal to current?
As the saying goes, is the root 3 1.732? P = power, u = voltage, I = what
380V three phase: power (apparent) = root 3 * u * I
220 V single phase: power (apparent) = root 2 * u * I
380V is the power divided by (voltage × 1.732) equal to the current
Root 3 is 1.732
I = current
Calculation of cable cross section area: the last 4 * 1 is the diameter? Isn't the diameter the square of 1.13?
The formula for calculating the cross-sectional area of wire and cable is as follows: first, we need to know the radius of the conductor of wire and cable
The square number of wires and cables (i.e. square millimeter) = the square of the circumference (3.14) × the conductor radius (mm) of wires and cables
If we know the square of the wire and cable, we can calculate the wire diameter in the same way, as follows:
The wire diameter of 4-square wire is: 4 △ 3.14 = 1.27, then 1.13MM is obtained by square, so the wire diameter of 4-square wire is: 4 × 1.13MM = 4.52mm
Because the area formula is: S = 3.14 × R & # 178;
If s = 4, then 4 = 3.14 × R & # 178;
R=√(4÷3.14)=1.13
Then wire diameter = 2 × r = 2 × 1.13 = 2.26 mm
It is wrong to calculate "so the line diameter of 4 square lines is 4 × 1.13MM = 4.52mm"
The resistance of a copper wire is 8 ohm. When its length is doubled, its resistance becomes the same as before_______
Consider the cross-sectional area
The resistance of a copper wire is 8 ohm. When its length is doubled, its resistance becomes the same as before___ 4 times____
Because after stretching twice, the cross-sectional area becomes half of the original, and the length becomes twice of the original
SO 2 × 2 = 4 times
You can ask me if you don't understand my answer!
R=pL/S
R'=p*2L/S'
According to the volume of the wire unchanged
LS=2LS' -> S'=1/2*S
R '= 4PL / S = 4R = 32 Ω
The resistance of a copper wire is 8 ohm. When its length is doubled, its resistance becomes 32 ohm
Resistance is directly proportional to length and inversely proportional to cross-sectional area
R=pL/S
When the conductor is lengthened to twice the original length, the cross-sectional area will be 1 / 2 of the original,
The resistance is quadrupled.
Therefore, the resistance of a copper wire is 8 ohm. When its length is doubled, its resistance becomes the same as before__ 4X (32 Euro)_____ ... unfold
Resistance is directly proportional to length and inversely proportional to cross-sectional area
R=pL/S
When the conductor is lengthened to twice the original length, the cross-sectional area will be 1 / 2 of the original,
The resistance is quadrupled.
Therefore, the resistance of a copper wire is 8 ohm. When its length is doubled, its resistance becomes the same as before__ 4X (32 Euro)_____ Put it away
Look up information Newton watt Zhang Heng what are their inventions or discoveries?
Newton gravitation, watt electric power, Zhang Heng seismograph
The relationship between power, voltage and current
For direct current, power is equal to current times voltage
Power (DC) = current * voltage
For the common alternating current, it is also multiplied by the power factor
Power (single phase AC) = voltage * current * power factor
If three-phase alternating current is used, multiply by 1.732
Power (three-phase AC) = voltage * current * power factor * 1.732
P=IU P=I^2*R P=U^2/R
I=P/U I=U/R
U=IR
I = U/R
U = IR
P = IU = I^2*R = U^2/R
Power = voltage × current.
P = UI, that is, power = voltage at both ends of the effective circuit * current passing through the circuit
p=w/t
P=IU P=I^2*R P=U^2/R
I = U/R
U = IR
P = IU = I^2*R = U^2/R
I=U/R U=IR R=U/I
W=IUt
P=W/t=IUt/t=IU
P=IU P=I^2*R P=U^2/R
I=P/U I=U/R
U=IR
Voltage = current * resistance
U=IR
How to calculate cable diameter according to power
The diameter of the cable is generally selected according to the current (the current can be calculated by volt ampere method, for example, if the power of the appliance is 2000W and the connecting voltage is 220kV, the current = 2000W / 220V = 9a)
GB 4706.1-19921/1998: load current 6a-8a for 1 square copper wire
1.5.8A-15A
2.5.16A-25A
4.25A-32A
6.32A-40A
Hope to be useful to you!
The resistance of the wire is 4 ohm. If the wire is evenly elongated to twice the original resistance, the resistance will be?
R=ρL/S
L is increased by two times, s is reduced by two times, so the resistance is increased by four times to 16 ohm