Answer:
Electricity is most often generated at a power plant by electromagnetic generators.
Answer: charge Q = 40Coulombs
Explanation: The current in a circuit is the quantity of charge that flows through it per second.
Here the 2amp current is same as the amount of charge in coulomb that flows through the terminal in ONE second.
So, if 2 coulomb of charge flows in 1sec then in 20 secs we will have
Q(charge) = 2*20 = 40 coulombs
Answer:
the speed of the cruiser relative to the pursuit ship is 0.3846c
Explanation:
the solution is in the attached Word file
Answer:
The order of steps for writing a chemical formula is 3, 2, 1, 5, 4
Explanation:
Balance the charges. Identify the negative ion and charge. Identify the positive ion and charge. Use the information in the ratio to write the chemical formula. Write the ratio of ions that balances the charges.
The order of steps for writing a chemical formula from the options above is:
3.) Identify the positive in and charge.
2.) Identify the negative ion and charge.
1.) Balance the charges.
5.)Write the ratio of ions that balances the charges.
4.) Use the information in the ratio to write the chemical formula.
Answer:
Velocity = 20.3 [m/s]
Explanation:
This is a typical problem of energy conservation, where potential energy is converted to kinetic energy. We must first find the potential energy. In this way, we will choose as a reference point or point where the potential energy is zero when the carrriage rolls down 21 [m] from the top of the hill.
![E_{p} =m*g*h\\ where:\\m = mass = 25[kg]\\g = gravity = 9.81 [m/s^2]\\h = elevation = 21 [m]\\E_{p} =potential energy [J]\\E_{p} =25*9.81*21=5150[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3Dm%2Ag%2Ah%5C%5C%20where%3A%5C%5Cm%20%3D%20mass%20%3D%2025%5Bkg%5D%5C%5Cg%20%3D%20gravity%20%3D%209.81%20%5Bm%2Fs%5E2%5D%5C%5Ch%20%3D%20elevation%20%3D%2021%20%5Bm%5D%5C%5CE_%7Bp%7D%20%3Dpotential%20energy%20%5BJ%5D%5C%5CE_%7Bp%7D%20%3D25%2A9.81%2A21%3D5150%5BJ%5D)
Now this will be the same energy transformed into kinetic energy, therefore:
![E_{p}=E_{k} = 5150[J]\\E_{k} =0.5*m*v^{2} \\where:\\v=velocity [m/s]\\v=\sqrt{\frac{E_{k}}{0.5*25} } \\v=20.3[m/s]](https://tex.z-dn.net/?f=E_%7Bp%7D%3DE_%7Bk%7D%20%3D%205150%5BJ%5D%5C%5CE_%7Bk%7D%20%3D0.5%2Am%2Av%5E%7B2%7D%20%5C%5Cwhere%3A%5C%5Cv%3Dvelocity%20%5Bm%2Fs%5D%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7BE_%7Bk%7D%7D%7B0.5%2A25%7D%20%7D%20%5C%5Cv%3D20.3%5Bm%2Fs%5D)
