Answer:
0.04
Explanation:
Fraction of power converted to sound = 80% = 0.08
Fraction incident upon each eardrum onstage = 0.08/2 = 0.04
Answer:
A damped oscillation means an oscillation that fades away with time while Forced oscillations occur when an oscillating system is driven by a periodic force that is external to the oscillating system.
Explanation:
Damping is the reduction in amplitude (energy loss from the system) due to overcomings of external forces like friction or air resistance and other resistive forces. ... When a body oscillates by being influenced by an external periodic force, it is called forced oscillation.
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Answer:
V = 9.682 × 10^(-6) V
Explanation:
Given data
thick = 190 µm
wide = 4.20 mm
magnitude B = 0.78 T
current i = 32 A
to find out
Calculate V
solution
we know v formula that is
V = magnitude× current / (no of charge carriers ×thickness × e
here we know that number of charge carriers/unit volume for copper = 8.47 x 10^28 electrons/m³
so put all value we get
V = magnitude× current / (no of charge carriers ×thickness × e
V = 0.78 × 32 / (8.47 x 10^28 × 190 × 1.602 x 10^(-19)
V = 9.682 × 10^(-6) V
Answer:
Refractive motion is the impact of a light wave that travels from medium to medium in an angle away from normal, where the direction of light varies. Light is refracted when it crosses the air-to-glass interface and moves slower.
Explanation:
Refractive motion is the impact of a light wave that travels from medium to medium in an angle away from normal, where the direction of light varies. Light is refracted when it crosses the air-to-glass interface and moves slower.
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Answer:
Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m.
Explanation:
The Coulomb force between two charges,
and
, separated by a distance,
, is given

<em>k</em> is a constant.
For the charge Z to be at equilibrium, the force exerted on it by charge X must be equal and opposite to the force exerted on it by charge Y.
It is to be placed along the <em>x</em>-axis. Hence, it is on the same line as charges X and Y.
Let the charge on Z be <em>Q</em>. It is positive.
Let the distance from charge X be <em>x m.</em> Then the distance from charge Y will be (0.60 - <em>x</em>) m.
Force due to charge X

Force due to charge Y

Since both forces are equal and opposite,







Applying the quadratic formula,

or 
Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m