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KatRina [158]
3 years ago
7

Which characteristic gives the most information about what kind of element an atom is ?

Physics
2 answers:
Butoxors [25]3 years ago
7 0

Answer:

The atomic number

Explanation:

Salsk061 [2.6K]3 years ago
3 0
Answer :

The atomic number
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HELP PLEASE 20 POINTS SHOW WORK, ALL EQUATIONS
nataly862011 [7]

Answer:

s = 3 m

Explanation:

Let t be the time the accelerating car starts.

Let's assume the vehicles are point masses so that "passing" takes no time.

the position of the constant velocity and accelerating vehicles are

s = vt = 40(t + 2)  cm

s = ½at² = ½(20)(t)² cm

they pass when their distance is the same

½(20)(t)² = 40(t + 2)

10t² = 40t + 80

0 = 10t² - 40t - 80

0 = t² - 4t - 8

t = (4±√(4² - 4(1)(-8))) / 2(1)

t = (4± 6.928) / 2  ignore the negative time as it has not occurred yet.

t = 5.464 s

s = 40(5.464 + 2) = 298.564 cm

300 cm when rounded to the single significant digit of the question numerals.

7 0
2 years ago
HELP PLEAASE!!!You push against a steamer trunk with a force of 750 N at an angle of 25° with the horizontal . The trunk is on a
viva [34]
<span>your answer is m = 59 kg</span>
5 0
2 years ago
Which explicit definition matches this sequence? an={10,if n=1an−1−4,if n&gt;1 A. an = 10 + 4(n – 1) B. an = 10 – 4(n – 1) C. an
Step2247 [10]

Answer:

Añ=10+4(n-1)

Explanation:

3 0
2 years ago
Estimate the wavelength corresponding to maximum emission from each of the following surfaces: the sun, a tungsten filament at 2
igomit [66]

Answer

Applying Wein's displacement

\lamda_{max}\ T = 2898 \mu_mK

1) for sun T = 5800 K

      \lambda_{max} = \dfrac{2898}{5800}

      \lambda_{max} = 0.5 \mu_m

2) for tungsten T = 2500 K

      \lambda_{max} = \dfrac{2898}{2500}

      \lambda_{max} = 1.16 \mu_m

3) for heated metal T = 1500 K

      \lambda_{max} = \dfrac{2898}{1500}

      \lambda_{max} = 1.93 \mu_m

4) for human skin T = 305 K

      \lambda_{max} = \dfrac{2898}{305}

      \lambda_{max} = 9.50 \mu_m

5)  for cryogenically cooled metal T = 60 K

      \lambda_{max} = \dfrac{2898}{60}

      \lambda_{max} = 48.3 \mu_m

range of different spectrum

UV ----0.01-0.4

visible----0.4-0.7

infrared------0.7-100

for sun T = 5800

λ              0.01           0.4               0.7                 100

λT             58           2320            4060             5.8 x 10⁵

F                0             0.125             0.491                1

fractions

for UV = 0.125  

for visible = 0.441-0.125 = 0.366

for infrared = 1 -0.491 = 0.509  

8 0
3 years ago
One property that makes electromagnetic waves differ from other types of waves is that they can
Flauer [41]
travel through a vacuum at the speed of light. Other waves need a medium; sound waves need molecules that vibrate.
5 0
3 years ago
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