The energy transferred by the appliance using mains electricity is 17.3 KJ
<h3>Data obtained from the question </h3>
- Potential difference (V) = 230V
- Charge (Q) = 150 C
<h3>How to determine the energy transferred </h3>
The energy transferred can be obtained as follow:
E = ½QV
E = ½ × 150 × 230
E = 75 × 230
E = 17250 J
Divide by 1000 to express in kilojoules
E = 17250 / 1000
E = 17.3 KJ
Learn more about energy stored in a capacitor:
brainly.com/question/14739936
Answer:
0.546 ohm / μm
Explanation:
Given that :
N = 1.015 * 10^17
Electron mobility, u = 3900
Hole mobility, h = 1900
Ng = 4.42 x10^22
q = 1.6*10^-19
Resistivity = 1/qNu
Resistivsity (R) = 1/(1.6*10^-19 * 1.015 * 10^17 * 3900)
= 0.01578880889 ohm /cm
Resistivity of germanium :
R = 1 / 2q * sqrt(Ng) * sqrt(u*h)
R = 1 / 2 * 1.6*10^-19 * sqrt(4.42 x10^22) * sqrt(3900*1900)
R = 1 /0.0001831
R = 5461.4964 ohm /cm
5461.4964 / 10000
0.546 ohm / μm
Answer:
C = 4,174 10³ V / m^{3/4}
, E = 7.19 10² / ∛x, E = 1.5 10³ N/C
Explanation:
For this exercise we can calculate the value of the constant and the electric field produced,
Let's start by calculating the value of the constant C
V = C 
C = V / x^{4/3}
C = 220 / (11 10⁻²)^{4/3}
C = 4,174 10³ V / m^{3/4}
To calculate the electric field we use the expression
V = E dx
E = dx / V
E = ∫ dx / C x^{4/3}
E = 1 / C x^{-1/3} / (- 1/3)
E = 1 / C (-3 / x^{1/3})
We evaluate from the lower limit x = 0 E = E₀ = 0 to the upper limit x = x, E = E
E = 3 / C (0- (-1 / x^{1/3}))
E = 3 / 4,174 10³ (1 / x^{1/3})
E = 7.19 10² / ∛x
for x = 0.110 cm
E = 7.19 10² /∛0.11
E = 1.5 10³ N/C
Answer:
Measurement is called the process of finding exact quantity of a substances.
Answer:
Current = 10 Amperes.
Explanation:
Given the following dat;
Quantity of charge, Q = 36 kilocoulombs (KC) = 36 * 1000 = 36000C
Time = 1 hour to seconds = 60*60 = 3600 seconds
To find the current;
Quantity of charge = current * time
Substituting in the equation
36000 = current * 3600
Current = 36000/3600
Current = 10 Amperes.
