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mrs_skeptik [129]
3 years ago
14

How many photons are absorbed during the dental x-ray?

Physics
1 answer:
MariettaO [177]3 years ago
8 0
<span>E=hc/wav. len
E = (6.62 x 10^-34 x 3 x 10^8)/0.0275 x 10^-9
E = 7.22182 x 10^-15 J
To convert to eV divide by 1.6 x 10^-19
E = 7.22182 x 10^-15/1.6 x 10^-19 eV
E =45.36 x 10^3 eV
Th energy, E, of a single x-ray photon in eV is = 45.36keV.

Number of photons, n = total energy/ energy of photon
n = 3.85 x 10^-6/7.22182 x 10^-15
n = 5.33 x 10^8 photons </span>
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A driver is traveling eastward on a dirt road when she spots a pothole ahead. She slows her car from 14.0 m/s to 5.5 m/s in 6.0
Ivahew [28]

Answer:

- 1.42m/s²

Explanation:

Acceleration is defined as the change in velocity of a body with respect to time.

Acceleration = change in velocity/time

Change in velocity = final velocity - initial velocity

Acceleration = final velocity - initial velocity/time

Since she slows her car from 14.0 to 5.5m/s in 6seconds,

Initial velocity = 14m/s

Final velocity = 5.5m/s

Time = 6seconds

Substituting in the given formula, we will have

Acceleration = 5.5 - 14/6

Acceleration = - 8.5/6

Acceleration = - 1.42m/s²

The negative acceleration shows that the car decelerates.

8 0
3 years ago
What is the relationship between the angle of an incline and the acceleration of an object moving down the incline? How would yo
iren2701 [21]

Answer:

See Explanation

Explanation:

The relationship between angle of an incline and the acceleration of an object moving down the incline.

As the angle of an incline increases, so does the acceleration of the body moving down the incline increases, resolving the force acting on an inclined object

Parallel force = mgsin, perpendicular = mgcosΘ

With th weigh component 'mg' of the parallel force accounting for the acceleration of the body down the incline.

mgsinΘ = ma

Fnet = ma

B.) From Fnet = ma

Fnet = ma

a = Fnet / m

Where Fnet = Net force = mgsinΘ, a = acceleration

5 0
3 years ago
4. An object is moving with an initial velocity of 9 m/s. It accelerates at a rate of 1.5
Mazyrski [523]

Answer:

11.87 ms⁻¹

Explanation:

You can use the kinematic equation

v² = u² + 2as

Where v = final velocity

          u = initial velocity

          a = acceleration

          s = displacement

v² = 9² + 2×1.5×20

So you get v = 11.87 ms⁻¹

7 0
3 years ago
What is the drivers displacement at the end of the lap
NeX [460]

Answer:

The distance that the racecar traveled is indeed 500m. But at the end of the lap, it is right back where it started. So overall, it has been displaced 0m.

Explanation:

4 0
3 years ago
PLEASE HELP!<br>I need help with this
melisa1 [442]

A)

At time = .003 hr

B)

90* (3*10^-3) = .27km

C)

.005 hr

D)

Car A = 150 * (.005) = .75km

Car B = (90 * (.005) = .45 km

E)

Car A = 150 * (0.008) = 1.2 km

Car B = 90 * (0.008) = 0.72 km

Car A is ahead



3 0
3 years ago
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