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just olya [345]
2 years ago
12

I AM TIMED !! 30 POINTS Which is the smallest possible particle of an element?

Physics
2 answers:
zimovet [89]2 years ago
4 0

Answer:

atoms

Explanation:

they are the smallest

melamori03 [73]2 years ago
3 0
The correct answer is Atoms!
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Ross training for a half marathon which is 13.1 miles long. Within four months he has progressed to an 11 mile practice run if R
damaskus [11]

If Ross takes two months off from training, his fitness level will reduce in comparison to what it was two months ago.

  • In as little as 3–4 weeks after beginning strength training, Ross will probably experience weight increase, energy loss, diminished balance, diminished strength (making it tougher to carry out daily tasks), and overall fewer fitness levels.
  • Many people mistakenly believe they lose muscle mass far more quickly than they actually do because their muscles' ability to store water and glycogen is declining.
  • A decrease in strength and muscle mass, with beginners experiencing a smaller decline in strength than experienced lifters.
  • Ross will experience Increased VO2 Max from exercise. VO2 Max is almost completely lost in people who train at lower intensities.

learn more about fitness here: brainly.com/question/13490156

#SPJ10

4 0
2 years ago
Why are atoms in a covalent bond usually a certain distance away from each other?
ANTONII [103]
I think it is because the electrons repel each other

4 0
3 years ago
A camera takes a properly exposed photo with a 4.0 mm diameter aperture and a shutter speed of 1/1000 s. If the photographer foc
hoa [83]

Answer:

option E

Explanation:

given,

diameter = 4 mm

shutter speed = 1/1000 s

diameter of aperture = ?

shutter speed = 1/250 s

exposure time to the shutter time

E V = log_2(\dfrac{N^2}{t})

N is the diameter of the aperture and t is the time of exposure

now,

log_2(\dfrac{N^2}{t_1})=log_2(\dfrac{N^2}{t_2})

\dfrac{N_1^2}{t_1}=\dfrac{N_2^2}{t_2}

inserting all the values

\dfrac{4^2}{1000}=\dfrac{N_2^2}{250}

      N₂² = 4

      N₂ = 2 mm

hence , the correct answer is option E

4 0
3 years ago
The pupil of the human eye can vary in diameter from 2.00 mm in bright light to 8.00 mm in dim light. The eye has a focal length
Zarrin [17]

The indicated data are of clear understanding for the development of Airy's theory. In optics this phenomenon is described as an optical phenomenon in which The Light, due to its undulatory nature, tends to diffract when it passes through a circular opening.

The formula used for the radius of the Airy disk is given by,

y_r=1.22\frac{\lambda f}{d}

Where,

y_r = Range of the radius

\lambda = wavelength

f= focal length

Our values are given by,

State 1:

d=2.00mm = 2*10^{-3}m

f= 25mm = 25*10^{-3}m

\lambda = 750nm = 750*10^{-9}m

State 2:

d=8.00mm = 8*10^{-3}m

f= 25mm = 25*10^{-3}m

\lambda = 390nm = 390*10^{-9}

Replacing in the first equation we have:

y_{r1} = 1.22\frac{(750*10^{-9})(25*10^{-3})}{2*10^{-3}}

y_{r1}= 11.4\mu m

And also for,

y_{r2} =1.22\frac{(390*10^{-9})(25*10^{-3})}{8*10^{-3}}

y_{r2} = 1.49\mu m

Therefor, the airy disk radius ranges from 1.49\mu m to 11.4\mu m

7 0
3 years ago
When 597 J of heat are added to a gas, it expands. Its internal energy increases by 318 J. How much work does the gas do? (Unit=
damaskus [11]

The magnitude of work done by the gas is 279 J and the sign is negative so W = -279 J as work is done by the system.

<u>Explanation:</u>

According to first law of thermodynamics, the change in internal energy of the system is equal to the sum of the heat energy added or released from the system with the work done on or by the system. If the heat energy is added to the system to perform a certain work, then the heat energy is taken as positive, while it will be negative when the heat energy is released from the system.

Similarly, in this case, the heat energy of 597 J is added to the system. So the heat energy will be positive, while the gas expansion occurs means work is done by the system.

          ΔU = Q+W

Since ΔU is the change in internal energy which is given as 318 J and the heat energy added to the system is Q = 597 J.

Then the work done by the gas = ΔU - Q = 318 J - 597 J = - 279 J.

As the work is done by the system, so it will be denoted in negative sign and the magnitude of work done by the gas is 279 J.

7 0
3 years ago
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