Answer:
500 N
Explanation:
Since the work done on the spring W = Fx where F = force applied and x = compression length = 0.170 m (since the spring will be compressed its full length when the force is applied)
Since W = 85.0 J and we need to find F,
F = W/x
= 85.0 J/0.170 m
= 500 N
So, the magnitude of force must you apply to hold the platform stationary at the final distance given above is 500 N.
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:
Explanation:
Given
Ball of mass m
maximum Bearable Tension in string is F
Let length of the cord be L m and moving at a speed of v m/s
Here Tension will Provide Centripetal Force
T=Centripetal Force
Answer:
The nodes and anti nodes would reverse roles.
Explanation:
I believe it has to do with the path differences. If waves are in phase, then the path differences are such that the waves reach the screen with crests superimposing crests and troughs superimposing troughs. This happens when the periods of each wave are equal or the paths themselves differ by a whole number multiple of the wavelength (λ, 2λ, 3λ, ...).
Now make these waves out of phase. Then half of the waves will travel half a wavelength farther than the rest. So the path difference will be 0.5λ, 1.5λ, 2.5λ, ....
