Two devices with capacitances of 25 μf and 5.0 μf are each charged with separate 120 v power supplies. calculate the total energ

Answer:

Consequently, we find that:

Thus, the most suitable answer would be:

a. vA = vB/4

Explanation:

In this situation, we can establish the following conditions:

both wires share the same length

both wires carry an identical current

Both wires are constructed of the same material, indicating that the electron density (n) remains constant across both wires

We also know that where r signifies the radius.

Given that wires are cylindrical in shape, we can determine the area for each case:

Thus, we conclude that

Now we are aware that the drift velocity of an electron in a wire can be described by:

Where I denotes the current, n is the electron density, e represents the electron charge, and A signifies the area.

By substituting, we arrive at:

So we observe that:

Thus, the most fitting answer is:

a. vA = vB/4

The question lacks complete details or specifications. Here is the missing information: 1. impossible to establish 2. half of Isaac's 3. identical to Isaac's 4. double Isaac's The angular velocity of Feng will match that of Isaac. Thus, the correct choice is option 3.

The light's wavelength absorbed during the transition is 459 nm. Energy difference between the 5-d and the 6-s sub-levels in gold is expressed as ΔE. Let the wavelength associated with the electron's transition from the 5-d to the 6-s state be λ. The relationship that describes the connection between energy and wavelength is defined as: E = hc/λ, where E stands for photon energy, h represents Planck's constant, c is the speed of light, and λ denotes the wavelength of the photon. Therefore, the absorption wavelength in this transition stands at 459 nm.

Answer:

1350 m/s

Clarification:

Bullet speed

The bullet travels 450 m

Sound travels a distance of 450 m

Using the equation S= V × t

==> t= S/V

Thus, the time for the bullet t1=450/vb

and the sound's travel time t2=450/vs

Given that there's a 1/2 sec interval from when the shot is fired to the moment the shooter hears the sound

==>  t1+t2= 1/2 sec

==> 450/vb+450/vs=0.5sec  ---- (1)

At a distance 'x' from both the gun and target, it takes 3 seconds for a person to hear the bullet sound from firing to impact.

Firing sound duration

n

Distance =x m

Speed of sound = V

time =T1

==> T1 = x/vs

During this time period, the bullet covers 450 m and the sound of impact travels a distance 'x'

The time taken for sound = 450/vb

The time it takes sound to travel distance 'x'= x/vs

therefore let T2= 450/vb + x/vs

However, all this occurs within 3 seconds, i.e., T = 3 sec

because firing takes place before hitting the target, implying the strike sound is heard in time T = T2-T1= 450/vb + x/vs -x/vs

Making T= 3sec

==> 3= 450/vb

==> vb= 1350 m/s

From equations 1 and 2

applying the same principle, in 3 seconds the observer sees the bullet travel 450 m and perceives the sound