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zzz [600]
3 years ago
11

What is mass?Explain

Physics
1 answer:
Alexxx [7]3 years ago
8 0

Answer:

The term mass is used to refer to the amount of matter in any given object For instance, a person or object may be weightless on the moon because of the lack of gravity but that same person or object maintains the same mass regardless of location

Explanation:

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Feliz [49]
Force is found by multiplying mass (kg) and acceleration (m/s^2), so the metric unit of force is kg*m/s^2 or N (newtons)
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A beam of light has a wavelength of 4.5 x10^-7 meter in a vacuum. the frequency of this light is
valkas [14]
The basic relationship between frequency and wavelength for light (which is an electromagnetic wave) is
c= f \lambda
where c is the speed of light, f the frequency and \lambda the wavelength of the wave. 
Using \lambda=4.5 \cdot 10^{-7} m and c=3 \cdot 10^8 m/s, we can find the value of the frequency:
f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{4.5 \cdot 10^{-7} m}=6.7 \cdot 10^{14} Hz
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PLZ ANSWER FAST WILL GIVE BRAINILEST TO FIRST PERSON WHO ANSWERS:P
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4

Explanation:

6 0
3 years ago
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HELPPP PLEASEEE!!!!!
devlian [24]

Answer:

1) The speed of sound increases

2)  440 Hz

3)  29°C

4)  17°C

5) 434 Hz

6)  12 m/s

7)  17.3 m

Explanation:

1) The speed of sound increases

2) V = f×λ

f = V/λ = 343/0.78 = 439.744 ≈ 440 Hz

3) V = f×λ

512 × 0.68 = 348.16 m/s

348.16 - 331 = 17.16

T = 17.16/0.6 = 28.6 ≈ 29°C

4) Increase in speed = 350 - 340 = 10

Increase in temperature = 10/0.6 = 16.67° ≈ 17°C

5) f = V/λ = 343/0.79 = 434 Hz

6) 331 + 0.6×30 - (331 × 0.6 ×10) = 12 m/s

7) V = 331 + 0.6×25 = 346m/s

λ = 346/20 = 17.3 m

5 0
3 years ago
What is the frequency and wavelength, in nanometers, of photons capable of just ionizing nitrogen atoms?
nika2105 [10]

Answer:

The frecuency and wavelength of a photon capable to ionize the nitrogen atom are ν = 3.394×10¹⁵ s⁻¹  and λ = 88.31 nm.

Explanation:It is possible to know what are the frequency and wavelength of a photon capable to ionize the nitrogen atom using the equation of the energy of a photon described below.

E = hc/λ  (1)

Where h is the Planck constant, c is the speed of light and λ is the wavelength of the photon.

But first, it is neccesary to know the ionization energy of the nitrogen atom. The ionization energy is the energy needed to remove an electron from an atom, for the Nitrogen atom it will lose an electron of its outer orbit from the nucleus, farther snuff, so the electric force is weaker. Experimentally, it is known that it has a value of 14.04 eV. This value is easy to found in a periodic table.

So the nitrogen atom will need a photon with the energy of 14.04 eV to remove the electron from its outer orbit.

Replacing the Planck constant, the speed of light and the energy of the photon in the equation 1, the wavelength can be calculated:

λ = hc/E  (2)

Where h = 6.626×10⁻³⁴ J.s and c = 3.00×10⁸ m/s

But the Planck constant can be expressed in electron volts:

1 eV = 1.602 x 10⁻¹⁹ J

h = 6.626x10⁻³⁴ J/1.602x10⁻¹⁹ J . eV .s

h= 4.136x10⁻¹⁵ eV.s

Now, it is convenient to express the speed of light in nanometers:

1nm = 1x10⁻⁹ m

c = 3.00x10⁸ m/ 1x10⁻⁹ m

c = 3x10¹⁷ nm/s

Substituting in equation 2:

λ =  (4.136x10⁻¹⁵ eV.s)(3x10¹⁷ nm/s)/14.04 eV

λ = 1240 eV. nm/ 14.04 eV

λ = 88.31 nm

The frenquency is calculated using the equation 2 in the following way:

E = hν  (3)

Where ν is the frecuency

ν = E/h

ν = 14.04 eV/4.136×10⁻¹⁵ eV.s

ν = 3.394×10¹⁵ s-1

So the frecuency of a photon, capable to ionize the nitrogen atom, will be 3.394×10¹⁵ s⁻¹ and its wavelength 88.31 nm.

4 0
3 years ago
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