Answer:
Frequency of oscillation, f = 4 Hz
time period, T = 0.25 s
Angular frequency, 
Given:
Time taken to make one oscillation, T = 0.25 s
Solution:
Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:
f = 
f = 
Frequency of oscillation, f = 4 Hz
The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.
Therefore, time period, T = 0.25 s
Angular frequency of oscillation is given by:
It's just asking you to sit down and COUNT the little squares in each sector.
It'll help you keep everything straight if you take a very sharp pencil and make a tiny dot in each square as you count it. That way, you'll be able to see which ones you haven't counted yet, and also you won't count a square twice when you see that it already has a dot in it.
(If, by some chance, this is a picture of the orbit of a planet revolving around the sun ... as I think it might be ... then you should find that both sectors jhave the same number of squares.)
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].
Ultraviolet light with a wavelength shorter than visible light and a higher radiant energy than visible light.
The shorter the wavelength, the higher the frequency and energy.
