Which type of wave does NOT have a crest?

An inductor in an LC circuit has a maximum current of 2.4 A and a maximum energy of 56 mJ.

Harrizon [31]

Answer:

The energy stored in the capacitor, when the current in the inductor is 1.2 A, is 41.6 mJ.

Explanation:

In a LC oscillating circuit, the energy is stored in the electric field (between the plates of the capacitor) and in the magnetic field (surrounding the wires of the inductor).

At any time, the sum of both energies can be expressed as follows:

E = 1/2 Q² / C + 1/2 L I²

In this type of circuit, energy oscillates, which means that it is exchanging between both fields all time.

When the capacitor is completely discharged, all the energy is stored in the magnetic field, and at that time, the current is maximum.

The total energy, when I is maximum, can be written as follows:

E = 1/2 L I² (1)

In our case, when I= 2.4A, E= 56 mJ.

So, we can find out the value of L, which will allow us to know the value of the magnetic energy at any time, having the value of the instantaneous current.

Solving for L in (1):

L = 2 *.56 mJ / (2.4)² A² = 20 mH

The next step is getting the value of the energy stored in the inductor, when I = 1.2 A, as follows:

Em = 1/2 *20 mH.* (1.2)² A² = 14.4 mJ

As the total energy must be always the same, i.e., 56 mJ, the energy stored in the capacitor, assuming no losses, must be the difference between the total energy and the one stored in the magnetic field:

Ec = 56 mJ - 14.4 mJ = 41.6 mJ

3 0

3 years ago

A plane traveled west for 4.0 hours and covered a distance of 4,400 meters what’s the velocity

tatuchka [14]

0.31m/s

Explanation:

Given parameters:

Time of travel = 4hrs = 4 x 60 x 60 = 14400s

Displacement = 4400m due west

Unknown:

Velocity = ?

Solution:

Velocity is defined as the displacement per unit of time. It is expressed in m/s or km/hr:

Velocity = $\frac{displacement}{time}$

Velocity = $\frac{4400}{14400}$ = 0.31m/s

Learn more:

Velocity brainly.com/question/10883914

#learnwithBrainly

8 0

3 years ago

The ratio of output force to input force of a machine it its

GarryVolchara [31]

Mechanical advantage is the ratio of output force to input force of a machine.

hope this helps and have a great day :)

7 0

3 years ago

Read 2 more answers

Four solid plastic cylinders all have radius 2.41 cm and length 5.94 cm. Find the charge of each cylinder given the following ad

Paladinen [302]

Answer:

Check explanation

Explanation:

QUICK NOTE: THE QUESTION IS NOT COMPLETE. Although it is not, we can make assumptions, since we only need values for the UNIFORM CHARGE DENSITY.

SO, LET US BEGIN;

To solve this question we are to use the equation (1) below;

Charge,Q = uniform charge density,p × Total area of the cylinder,A ------------------------------------------------------------------------(1).

From the question, we are given radius, R to be 2.41 cm and length, L to be 5.94 cm.

Step one: calculate for the total area of the cylinder, A.

Total area of the cylinder, A= area of the top surface + area of the buttom + area of the curved surface of the cylinder.

Hence, total area of the cylinder,A is;

==> πR^2 + πR^2 + 2πRL. -------------------------------------------------------------------------(2).

Then, total area of the cylinder,A is;

==> (L + R)2πR.

Step two: find the charge of each cylinder.

===> For the first cylinder; we have the uniform charge density to be 35 nC/m^2.

Therefore, the combination of equation (1) and (3) gives;

Charge Q= p × (L + R)2πR...----------------------------(4)

Hence, Q= 35 × [(5.94 + 2.41) 2× 3.143 × 5.94].= 10912.615 coulumb.

====> For the second cylinder, we have a uniform charge density of 50 nC/m^2.

Using equation (4), charge,Q= 15,589.45 Coulumb

=====> For THE third cylinder, the uniform charge density is 600, we make use of equation (4);

Charge,Q= 600×311.789.

Charge,Q= 187,073.4 coulumb.

====> For THE fourth cylinder, the uniform charge density is 750 nC/m^2.., we make use of equation (4);

Charge,Q= 233,841.75 coulumb.

7 0

3 years ago

1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x

allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

(A) Considering a uniform linear density $λ_{0}$ on the ring, then:

$dQ=\lambda dl$ (1)⇒ $Q=\lambda_{0} 2\pi R$ (2)⇒ $\lambda_{0}=\frac{Q}{2\pi R}$ (3)

Applying the technique of charge integration for finite charges:

$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

Integrating results:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$ (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

$|E| =-\nabla V$ (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

$|E| =-\frac{dV(z)}{dz}$ (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$ (S_b)

4 0

3 years ago