Answer:0.4 m
Explanation:
Given
Maximum displacement A=0.49
The sum of kinetic and elastic potential energy is 
where k=spring constant
U+K.E.=
when K.E.=U/2
K.E.=kinetic energy
U=Elastic potential Energy
Answer:
(B) Resistor only
Explanation:
Alternating Current: These are currents that changes periodically with time.
An LRC Ac circuit is an AC circuit that contains a Resistor, a capacitor and an inductor, connected in series.
In a purely resistive circuit, current and voltage are in phase.
In a purely capacitive circuit, the current leads the voltage by π/2
In a purely inductive circuit, the current lags the voltage by π/2.
Therefore when a alternating current is set up in LRC circuit, in the resistor, the current and the voltage are in phase.
The right option is (B) Resistor only.
Answer: Mechanical waves carry energy through a medium; electromagnetic waves carry energy through media and space.
Explanation:
<u>The waves which require medium to travel are known as mechanical waves. </u>The waves carry energy forward by compression and rarefaction that is by changing density /pressure of medium particles.
<u>Electromagnetic waves can transfer energy through space as well as medium. </u>In an electromagnetic wave, electric field generates magnetic field and magnetic field generates electric field. Electric field and magnetic field are perpendicular to each other and to the direction of motion.
Answer: 363 Ω.
Explanation:
In a series AC circuit excited by a sinusoidal voltage source, the magnitude of the impedance is found to be as follows:
Z = √((R^2 )+〖(XL-XC)〗^2) (1)
In order to find the values for the inductive and capacitive reactances, as they depend on the frequency, we need first to find the voltage source frequency.
We are told that it has been set to 5.6 times the resonance frequency.
At resonance, the inductive and capacitive reactances are equal each other in magnitude, so from this relationship, we can find out the resonance frequency fo as follows:
fo = 1/2π√LC = 286 Hz
So, we find f to be as follows:
f = 1,600 Hz
Replacing in the value of XL and Xc in (1), we can find the magnitude of the impedance Z at this frequency, as follows:
Z = 363 Ω
