Answer: 624 Hz
Explanation:
If the ratio of the inductive reactance to the capacitive reactance, is 6.72, this means that it must be satified the following expression:
ωL / 1/ωC = 6.72
ω2 LC = 6.72 (1)
Now, at resonance, the inductive reactance and the capacitive reactance are equal each other in magnitude, as follows:
ωo L = 1/ωoC → ωo2 = 1/LC
So, as we know the resonance frequency, we can replace LC in (1) as follows:
ω2 / ωo2 = 6.72
Converting the angular frequencies to frequencies, we have:
4π2 f2 / 4π2 fo2 = 6.72
Simplifying and solving for f, we have:
f = 240 Hz . √6.72 = 624 Hz
As the circuit is inductive, f must be larger than the resonance frequency.
By definition, we have to:
Where,
p: linear momentum
m: mass of the object
v: object speed
Therefore, knowing the mass of an object and its speed, we have the linear momemtum of the object.
In general, it is useful to calculate the momentum of objects in conservation problems to find the velocity before or after the collision of two objects.
Answer:
The measure of an object's mass and velocity is called momentum.
d. momentum
Answer:
Momentum is conserved.
All of the others are not conserved because of heat loss caused by deformation, etc.
Answer:
Using the log combination rules to reduce the famous Sakur-Tetrode equation, The change in entropy is given as:
∆S = NK*ln(V"V$/V").
Where V"V$ is final Volume (Vf) after constraint's removal,
V" is Initial Volume (Vi) before constraint's removal.
Temperature (T) is constant, Internal Energy, U is constant, N and K have their usual notations
Explanation:
Given in the question, the container is an adiabatic container.
For an adiabatic contain, it does not permit heat to the environment due to its stiff walls. This implies that the Internal Energy, U is kept constant(Q = U). The temperature is also constant (Isothermal). Thus, the famous Sakur-Tetrode equation will reduce to ∆S = NK* In(Vf/Vi).
Vf is the volume after the constraint is removed(Vf = V"V$). Vi is the volume occupied before the constraint is removed (Vi = V")
Answer:
The two rays, CY and DM are diverging rays and when extended behind the mirror, they appear to intersect each other at point M'. Therefore, the properties of the images formed here are formed behind the mirror, between the pole and principal focus (f), the images are diminished and are virtual and erect.
Explanation:
<h2>Spherical Mirrors</h2>
- There are two kinds of spherical mirrors, concave and convex.
- The focal point (F) of a concave mirror is the point at which a parallel beam of light is "focussed" after reflection in the mirror. ...
- The focal length (f) and radius of curvature (R) are defined in the diagram at the right.
<h3>hope it helps and thanks for following </h3><h2>please give brainliest </h2>
