Bradley gets an x-ray at a radiology clinic that employs its own technologists and radiologists. Would the coder at the clinic r

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.

Answer:

The time required is 20 μs

Explanation:

Here is the data provided:

temperature = 20°C  = 293 K

radius = 1 cm

atomic mass of air = 29 u

To determine

the duration for air to refill the vacuum space

solution:

We calculate the root mean square velocity of air particles. This can be expressed as:

where m indicates mass, t is temperature, v is speed, and R is the ideal gas constant, which is approximately 8.3145 (kg·m²/s²)/K·mol.

v = ............................1

v =

Resulting in v = 501.99 m/s.

<pNow, to cover the distance of 1 cm,<pThe duration needed for air is calculated as:

time taken =

which gives us:

time taken =

so, time taken = 19.92 × seconds = 20μs.

Thus, the required time is 20 μs.

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.

The sheet's charge density calculates to be 1.384×10⁻⁷C/m².

Charge density is the ratio of charge per unit area.

The square sheet has dimensions  l=17 cm.

To compute the area A of the sheet.

The overall charge Q on the sheet is

The charge density σ is defined as

We substitute 4×10⁻⁹C for Q and 0.0289 m² for A.

Thus, the charge density for the sheet is 1.384×10⁻⁷C/m².

Answer:

The current flowing through the tube is 0.601 A

Explanation:

Given data;

the diameter of the fluorescent tube is d = 3 cm

the incoming negative charge in the tube is -e = 3 x 10¹⁸ electrons/second

the outgoing positive charge equals +e = 0.75 x 10¹⁸ electrons/ second

The current within the fluorescent tube results from both positive and negative charges which contribute to maintaining electrical neutrality in the conductor (fluorescent tube).

Q = It

I = Q/t

where;

I signifies current in Amperes (A)

Q represents charge measured in Coulombs (C)

t denotes time which is in seconds (s)

1 electron (e) accounts for 1.602 x 10⁻¹⁹ C

Thus, 3 x 10¹⁸ e/s can be computed as

= (3 x 10¹⁸ e/s  x 1.602 x 10⁻¹⁹ C) / 1e

= 0.4806 C/s

This reflects the negative charge per second (Q/t) = 0.4806 C/s

Meanwhile, the positive charge per second amounts to

(0.75 x 10¹⁸ e/s  x 1.602 x 10⁻¹⁹ C) / 1e

Thus, the positive charge per second is equal to 0.12015 C/s

The total charge per second in the tube is then obtained by summing both charges: Q / t = (0.4806 C/s + 0.12015 C/s)

                                                                I = 0.601 A

Therefore, the current within the tube computes to 0.601 A