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SIZIF [17.4K]
3 years ago
13

Which of these sensors would be least likely used as an input sensor?

Physics
1 answer:
Alex787 [66]3 years ago
8 0

Answer:

C. Fuel

Explanation:

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TRANSFORMA: 765 mm Hg a atm
-Dominant- [34]

Presión

765

=

ATMÓSFERA

1,00658

Answer:

Explanation:

5 0
3 years ago
The photon energies used in different types of medical x-ray imaging vary widely, depending upon the application. Single dental
pav-90 [236]

A) 5.0\cdot 10^{-11} m

The energy of an x-ray photon used for single dental x-rays is

E=25 keV = 25,000 eV \cdot (1.6\cdot 10^{-19} J/eV)=4\cdot 10^{-15} J

The energy of a photon is related to its wavelength by the equation

E=\frac{hc}{\lambda}

where

h=6.63\cdot 10^{-34}Js is the Planck constant

c=3\cdot 10^8 m/s is the speed of light

\lambda is the wavelength

Re-arranging the equation for the wavelength, we find

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4\cdot 10^{-15}J}=5.0\cdot 10^{-11} m

B) 2.0\cdot 10^{-11} m

The energy of an x-ray photon used in microtomography is 2.5 times greater than the energy of the photon used in part A), so its energy is

E=2.5 \cdot (4\cdot 10^{-15}J)=1\cdot 10^{-14} J

And so, by using the same formula we used in part A), we can calculate the corresponding wavelength:

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1\cdot 10^{-14}J}=2.0\cdot 10^{-11} m

4 0
2 years ago
an object is placed 20cm from a converging lens. If the real image is formed 80cm from the object, what is the focal length of t
masya89 [10]

Explanation:

Here,

object distance(u)=20cm

image distance(v)=80cm-20cm=60cm

focal length(f)=?

we know ,

(1/f)=(1/v)+(1/u)

1/f = 1/60 + 1/20

1/f = (1+3)/60

60 =4f

f=15cm

8 0
2 years ago
What is the following correct way to write 2,330,000 In a scientific notation
NISA [10]

Answer:

2.33 × 10^6

hope this helps.

8 0
3 years ago
Read 2 more answers
How many miles must you travel to make the Earth fall the same amount?
Ede4ka [16]

Let's take the analogy of the baseball pitcher a step farther. When a baseball is thrown in a straight line, we already said that the ball would fall to Earth because of gravity and atmospheric drag. Let's pretend again that there is no atmosphere, so there is no drag to slow the baseball down. Now, let's assume that the person throwing the ball throws it so fast that as the ball falls towards the Earth, it also travels so far, before falling even a little, that the Earth's surface curves away from the ball's path.

In other words, the baseball falls as it did before, but the ball is moving so fast that the curvature of the Earth becomes a factor and the Earth "falls away" from the ball. So, theoretically, if a pitcher on a 100 foot (30.48 m) high hill threw a ball straight and fast enough,the ball would circle the Earth at exactly 100 feet and hit the pitcher in the back of the head once it circled the globe! The bad news for the person throwing the ball is that the ball will be traveling at the same speed as when they threw it, which is about 8 km/s or several times faster than a rifle bullet. This would be very bad news if it came back and hit the pitcher, but we'll get to that in a minute.

4 0
3 years ago
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