How is the unit EV converted into Joule in college physics How to calculate the energy level of EV?

How is the unit EV converted into Joule in college physics How to calculate the energy level of EV?

1eV=1.60217733×10^-19(C)* 1(V)=1.60217733×10^-19(A*s) *1(V)=1.60217733×10^-19(V*A*s)=1.60217733×10^-19(J)
1.60217733 × 10 ^ - 19 is an electronic charge, and the simplified calculation is 1.60 × 10 ^ - 19C
EV is called electron volt. It's a unit of energy. The energy of an electron volt is the product of the electric quantity of an electron and the voltage of one volt, resulting in joules
EV is usually expressed in the energy of the electron, read as 1 electron volt. Calculate the charge of 1eV = 1 electron * 1V = 1.602 * 10 ^ (- 19) (c) * 1 (V) = 1.602 * 10 ^ (- 19) (J). Generally, it is enough to keep four digits
How to convert the unit of capacitance (F, UF) to Joule (J)?
Please be clear!
W=1/2CU^2
W-energy in J
C-capacity in F
U-voltage in V
So the conversion relationship between F and j is j = FV ^ 2
How to convert the unit of measurement of natural gas in China from cubic meter to Joule?
It depends on the combustion value of natural gas. The combustion value of natural gas with good quality is higher
Cylinder area formula!
Rectangle, cone, cylinder all formulas!
Side area of cylinder = circumference of bottom circle × height
Surface area of cylinder = area of upper and lower bottom surface + side area
Volume of cylinder = bottom area × height
The resistance R1 = 6 ohm and R2 = 12 ohm are connected in parallel in the circuit, and the electric power consumed by R1 is 6 W
Ask: (1) what is the voltage at both ends of R1?
(2) What is the total electrical power consumed by R1 and R2?
1.P1=U^2/R1
U = root PR1 = root 6 * 6 = 6V
2. P2 = u ^ 2 / r2 = 6 ^ 2 / 12 = 3W
Ptotal = P1 + P2 = 6 + 3 = 9W
If an electric heater is connected to a 10V power supply and the electric power is 10W, then ()
A. The resistance of the electric heater is equal to 10 Ω B. the current through the electric heater is equal to 10 & nbsp; AC. the electric energy consumed by the electric heater is 10 & nbsp; JD every 1 min. the heat generated by the electric heater is 10 & nbsp; J every 1 min
A. The resistance of the electric heater R = u2p = (10V) 210W = 10 Ω, so a is correct; B. the current through the electric heater I = ur = 10v10 Ω = 1a, so B is wrong; C. the electric energy consumed by the electric heater every 1min w = uit = 10V × 1a × 60s = 600J, so C is wrong; D. the heat generated by the electric heater every 1min q = w = 600J, so D is wrong; so a
The Avogadro constant is Na, the molar mass of copper is m, and the density of copper is p
What is the number of atoms in 1kg copper and in 1m3 copper
Because I add physics. Chemistry is hard
I'm a senior three science student in chemistry,
Because the molar mass of copper is m, the amount of its substance is 1000g / m. therefore, its atomic number is the amount of substance × Avogadro constant, namely 1000g × Na / m
The international standard unit of density is kg / M & sup3;, so the atomic number of 1 cubic meter of copper is ρ × 1m & sup3; × 1000 (because the unit is kg at this time, × 1000 becomes g) × Na / m
Just know the amount of copper and multiply it by Na to get the atomic number
When there is 1kg of copper, the amount of copper substance = mass / molar mass = 1 / m, so the number of atoms = Na * the amount of copper substance = Na * (1 / M)
When there is 1 cubic meter, the mass of copper = density * volume = 1 * P = P, the mass of copper = mass / molar mass = P / m, so the number of atoms = Na * the mass of copper = Na * (P / M)... Expansion
Just know the amount of copper and multiply it by Na to get the atomic number
When there is 1kg of copper, the amount of copper substance = mass / molar mass = 1 / m, so the number of atoms = Na * the amount of copper substance = Na * (1 / M)
When there is 1 cubic meter, the mass of copper = density * volume = 1 * P = P, the mass of copper = mass / molar mass = P / m, so the number of atoms = Na * the mass of copper = Na * (P / M)
(1) The mass of 1mol Cu is m (g), so 1kg Cu has 1000 / mmol, the number of 1mol substance is Na, so the number of copper atoms is 1000na / m
(2) If you know the volume and the density, you can calculate the mass. The following process can imitate the first question. The result is p * 1000 * Na / m
Formula of cylinder side area
Expand the sides to get a rectangle
Area of cylinder side = perimeter of bottom surface x cylinder height
Bottom circumference × height = diameter × 3.14 × height = radius × 2 × 3.14 × height
The v-column is equal to s times H
s=πdh
Expand the sides to get a rectangle
Area of cylinder side = perimeter of bottom surface x cylinder height
Area of cylinder side = perimeter of bottom surface x cylinder height
S side = Ch
Resistor R1 = 8 ohm and R2 = 2 ohm are connected in parallel in the circuit. In order to make the electric power consumed by these two resistors equal, what is the feasible way?
Let X be the resistance to be connected in series. Because the voltages of parallel circuits are equal, then:
U^2/R1=[R2/(R2+X)]*U^2/(R2+X)
That is, (2 + x) ^ 2 = 16
I don't know what R2 / r2 + X is,
Because of the parallel connection, the voltages of the two resistors are the same, so that the power P of the two resistors is the same, and R2 is less than R1. Therefore, R1 should be connected in series with a resistor to share part of the voltage of R2 branch, so that the power P of the two resistors, i.e. U / R, is equal. So let the resistance value of the series resistor be xP1 = P2, and expand u ^ 2 / R1 = (U / R + X
Sorry, there is something wrong with your question. Please correct it before asking
If the square of voltage divided by resistance equals power, then the power consumed by 8-ohm resistance is u ^ 2 / 8, and 2-ohm resistance also consumes so much power, then the current passed by 2-ohm resistance is (u ^ 2 / 8) / 2 = the square root of u ^ 2 / 16, that is, U / 4, and the voltage at both ends should be (u ^ 2 / 8) / (U / 4) = u / 2. It can be seen that a 2-ohm resistance needs to be connected in series in the branch of 2-ohm resistance I don't know if you understand. ... unfold
If the square of voltage divided by resistance equals power, then the power consumed by 8-ohm resistance is u ^ 2 / 8, and 2-ohm resistance also consumes so much power, then the current passed by 2-ohm resistance is (u ^ 2 / 8) / 2 = the square root of u ^ 2 / 16, that is, U / 4, and the voltage at both ends should be (u ^ 2 / 8) / (U / 4) = u / 2. It can be seen that a 2-ohm resistance needs to be connected in series in the branch of 2-ohm resistance I don't know if you understand. Put it away
[U-UX/(2+X)]^2/2=U^2/8
4[1-X/(2+X)]^2=1
(X+6)(X-2)=0
X=2
The electric power of an electric heater connected in the 220 V circuit is 220 W
Use three hours a day, find (1) electric current, resistance and use of electric energy consumption of 30 days how many kW
1》 Current of electric heater:
I＝P/U＝220/220＝1(A)
2》 Resistance of electric heater:
R＝U/I＝220/1＝220(Ω)
Or:
R＝U×U/P＝220×220/220＝220(Ω)
3》 How much kW. H is the power consumption for three hours a day and 30 days
0.22×3×30＝19.8(KW.h)