Answer:
170 W
Explanation:
Applying
P = VI.................... Equation 1
Where P = Power generated in watt, V = Voltage supplied to the circuit, I = Current running through the circuit.
From the question,
Given: V = 17 V, I = 10 A
Substitute these values into equation 1
P = (17×10)
P = 170 Watt.
Hence the power generated is 170 W.
The right option is A. 170 W
Answer:
This is True
Explanation:
I just did this exact unit in bio last week I hope this helps ;)
Answer:
- The energy that must be added to the electron to move it to the third excited state is -1.153 eV
- The energy that must be added to the electron to move it to the fourth excited state is -1.181 eV
Explanation:
Given;
Energy of electron in ground state (n = 1 ) = 1.23 eV
E₁ = 1.23 eV
Eₙ = E₁ /n²
where;
E₁ is the energy of the electron in ground state
n is the energy level,
For third excited state, n = 4
E₄ = E₁ /4²
E₄ = (1.23 eV) / 16
E₄ = 0.077 eV
Change in energy level, = E₄ - E₁ = 0.077 eV - 1.23 eV = -1.153 eV
The energy that must be added to the electron to move it to the third excited state is -1.153 eV
For fourth excited state, n = 5
E₅ = E₁ /5²
E₄ = (1.23 eV) / 25
E₄ = 0.049 eV
Change in energy level, = E₅ - E₁ = 0.049 eV - 1.23 eV = -1.181 eV
The energy that must be added to the electron to move it to the fourth excited state is -1.181 eV
It is D as it is the only answer referring to weather and climate.
Answer:
The charge on each plate is 0.0048 nC
Explanation:
for the distance between the plates d and given the area of plates, A, and ε = 8.85×10^-12 C^2/N.m^2, the capacitance of the plates is given by:
C = (A×ε)/d
=[(0.2304×10^-2)(0.2304×10^-2)×(8.85×10^-12))/(0.5974×10^-3)
= 7.86×10^-14 F
then if the plates are connected to a battery of voltage V = 61 V, the charge on the plates is given by:
q = C×V
= (7.86×10^-14)×(61)
= 4.80×10^-14 C
≈ 0.0048 nC
Therefore, the charge on each plate is 0.0048 nC.