Answer:
A CC power source will maintain current at a relatively constant level, while a CV power source will maintain voltage at a relatively constant level.
Explanation:
A CC power source will maintain current at a relatively constant level, regardless of fairly large changes in voltage, while a CV power source will maintain voltage at a relatively constant level, regardless of fairly large changes in current.
Answer:
(3a + 4b)² / a²b²
Explanation:
This question is difficult to interpret, however the concept being tested in this question is the ability to simplify fractions with variables.
It is a basic rule that only like terms (as denominators) can be added and subtracted from each other, however if you need to add or subtract two fractions with different denominators, it is important to make sure you create a common denominator.
For example, in the beginning, the first fractions are:
16/a² + 24/ab + 9/b²
So it is important to make sure there is a common denominator between all 3 fractions. An easy way of doing this would be to multiply denominators together:
(16b²)/a²b² + (9a²)/a²b² + 24/ab
(9a²+16b²)/a²b² + 24/ab
So now it has gone from 3 fractions to 2, and then to 1:
(9a²+16b²)/a²b² + 24ab/a²b²
(9a² + 24ab + 16b²) / a²b²
Then all that needs to be done is simplify the numerator:
(3a + 4b)² / a²b²
Answer:
a) the inductance of the coil is 6 mH
b) the emf generated in the coil is 18 mV
Explanation:
Given the data in the question;
N = 570 turns
diameter of tube d = 8.10 cm = 0.081 m
length of the wire-wrapped portion l = 35.0 cm = 0.35 m
a) the inductance of the coil (in mH)
inductance of solenoid
L = N²μA / l
A = πd²/4
so
L = N²μ(πd²/4) / l
L = N²μ(πd²) / 4l
we know that μ = 4π × 10⁻⁷ TmA⁻¹
we substitute
L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)
L = 0.00841549 / 1.4
L = 6 × 10⁻³ H
L = 6 × 10⁻³ × 1000 mH
L = 6 mH
Therefore, the inductance of the coil is 6 mH
b)
Emf ( ∈ ) = L di/dt
given that; di/dt = 3.00 A/sec
{∴ di = 3 - 0 = 3 and dt = 1 sec}
Emf ( ∈ ) = L di/dt
we substitute
⇒ 6 × 10⁻³ ( 3/1 )
= 18 × 10⁻³ V
= 18 × 10⁻³ × 1000
= 18 mV
Therefore, the emf generated in the coil is 18 mV
Total heat rejected by compressor = Heat rejected by compression + Heat rejected by intercooler
Heat rejected by compression :
Q = (y - n)/(y - 1) • W
Q = (y - n)/(y - 1) • R(T2 - T1)/(n - 1) ... (i)
Heat rejected by intercooler :
Q = Cp(T2 - T1) ... (ii)
Sum of (i) and (ii) :
Q = (i) + (ii)
Q = (y - n)/(y - 1) • R(T2 - T1)/(n - 1) + Cp(T2 - T1) ... (iii)
Remember that :
R = Cp - Cv
y = Cp/Cv
So :
R = yCv - Cv
R = Cv(y - 1) ... (iv)
Now, subtitute (iv) to (iii) :
Q = (y - n)/(y - 1) • Cv(y - 1)(T2 - T1)/(n - 1) + Cp(T2 - T1)
Q = (y - n)/(n - 1) • Cv(T2 - T1) + Cp(T2 - T1)
Q = [Cp + Cv • (y - n)/(n - 1)] • (T2 - T1)
Hope this help