Answer:
![I_{rms}=2.12\ A](https://tex.z-dn.net/?f=I_%7Brms%7D%3D2.12%5C%20A)
Explanation:
Given that,
The instantaneous current in the circuit is giveen by :
![I=3\sin\theta\ A](https://tex.z-dn.net/?f=I%3D3%5Csin%5Ctheta%5C%20A)
We need to find the rms value of the current.
The general equation of current is given by :
![I=I_o\sin\theta](https://tex.z-dn.net/?f=I%3DI_o%5Csin%5Ctheta)
It means, ![I_o=3\ A](https://tex.z-dn.net/?f=I_o%3D3%5C%20A)
We know that,
![I_{rms}=\dfrac{I_o}{\sqrt2}\\\\=\dfrac{3}{\sqrt2}\\\\=2.12\ A](https://tex.z-dn.net/?f=I_%7Brms%7D%3D%5Cdfrac%7BI_o%7D%7B%5Csqrt2%7D%5C%5C%5C%5C%3D%5Cdfrac%7B3%7D%7B%5Csqrt2%7D%5C%5C%5C%5C%3D2.12%5C%20A)
So, the rms value of current is 2.12 A.
1. All the relevant resistors are in series, so the total (or equivalent) resistance is the sum of the resistances of the resistors: 20 Ω + 80 Ω + 50 Ω = 150 Ω [choice A].
2. The ammeter will read the current flowing through this circuit. We can find the ammeter reading using Ohm's law in terms of the electromotive force provided by the battery: I = ℰ/R = (30 V)(150 Ω) = 0.20 A [choice C].
3. The voltmeter will measure the potential drop across the 50 Ω resistor, i.e., the voltage at that resistor. We know from question 2 that the current flowing through the resistor is 0.20 A. So, from Ohm's law, V = IR = (0.20 A)(50 Ω) = 10. V, which will be the voltmeter reading [choice F].
4. Trick question? If the circuit becomes open, then no current will flow. Moreover, even if the voltmeter were kept as element of the circuit, voltmeters generally have a very high resistance (an ideal voltmeter has infinite resistance), so the current moving through the circuit will be negligible if not nil. In any case, the ammeter reading would be 0 A [choice B].
Answer:
The distance from Witless to Machmer is 438.63 m.
Explanation:
Given that,
Machmer Hall is 400 m North and 180 m West of Witless.
We need to calculate the distance
Using Pythagorean theorem
![D = \sqrt{(d_{m})^2+(d_{w})^2}](https://tex.z-dn.net/?f=D%20%3D%20%5Csqrt%7B%28d_%7Bm%7D%29%5E2%2B%28d_%7Bw%7D%29%5E2%7D)
Where,
=distance of Machmer Hall
=distance of Witless
Put the value into the formula
![D = \sqrt{(400)^2+(180)^2}](https://tex.z-dn.net/?f=D%20%3D%20%5Csqrt%7B%28400%29%5E2%2B%28180%29%5E2%7D)
![D=438.63\ m](https://tex.z-dn.net/?f=D%3D438.63%5C%20m)
Hence, The distance from Witless to Machmer is 438.63 m.
Answer: The following statement is true about squall line thunderstorm development: <em><u>These often form ahead of the advancing front but rarely behind it because lifting of warm, humid air and the generation of a squall line usually occur in the warm sector ahead of an advancing cold front. Behind a cold front, the air motions are usually downward, and the air is cooler and drier.</u></em>
<em>An upper-level wave, accountable for the fabrication of a squall line, extend in front of and backside a cold front, the air backside the front is cold, steady and settling while the air ahead of the front is hot and co-seismic.</em>