Answer:
rounding
Explanation:
each of the digits of a number that are used to express it to the required degree of accuracy, starting from the first non-zero digit.
Answer is True.. hope I helped... pls mark brainliest
Answer:
For any string, we use
Explanation:
The pumping lemma says that for any string s in the language, with length greater than the pumping length p, we can write s = xyz with |xy| ≤ p, such that xyi z is also in the language for every i ≥ 0. For the given language, we can take p = 2.
Here are the cases:
- Consider any string a i b j c k in the language. If i = 1 or i > 2, we take and y = a. If i = 1, we must have j = k and adding any number of a’s still preserves the membership in the language. For i > 2, all strings obtained by pumping y as defined above, have two or more a’s and hence are always in the language.
- For i = 2, we can take and y = aa. Since the strings obtained by pumping in this case always have an even number of a’s, they are all in the language.
- Finally, for the case i = 0, we take , and y = b if j > 0 and y = c otherwise. Since strings of the form b j c k are always in the language, we satisfy the conditions of the pumping lemma in this case as well.
Answer:
86 mm
Explanation:
From the attached thermal circuit diagram, equation for i-nodes will be
Equation 1
Similarly, the equation for outer node “o” will be
Equation 2
The conventive thermal resistance in i-node will be
Equation 3
The conventive hermal resistance per unit area is
Equation 4
The conductive thermal resistance per unit area is
Equation 5
Since is given as 100, is 40 is 300 is 25
Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain
Equation 6
Equation 7
From equation 6 we can substitute wherever there’s with 3000L+40 as seen in equation 7 hence we obtain
The above can be simplified to be
-3000L=1.665-260
Therefore, insulation thickness is 86mm
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