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AURORKA [14]
3 years ago
8

HELP... 3.00 amu = _____ Mev. 3.22 x 10-3 2.79 x 103 3.10 x 102

Physics
2 answers:
Nutka1998 [239]3 years ago
7 0

Answer:

2.79x103

Explanation:

DanielleElmas [232]3 years ago
5 0
The correct answer is:
2.79 \cdot 10^3 MeV
Let's see why.

1 amu corresponds to the mass of the proton, which is:
m_p = 1.66 \cdot 10^{-27} kg
if we convert this into energy, using Einstein equivalence between mass and energy, we find:
E=mc^2 = (1.66 \cdot 10^{-27} kg)(3\cdot 10^8 m/s)^2 = 1.49 \cdot 10^{-10} J
Now we can convert it into electronvolts:
E= \frac{1.49 \cdot 10^{-10}kg}{1.6 \cdot 10^{-19} J/eV} =9.34 \cdot 10^9 eV = 934 MeV

So, 1 amu = 934 MeV. Therefore, 3 amu corresponds to 3 times this value:
3 amu = 3 \cdot 934 MeV  \sim 2790 MeV = 2.79 \cdot 10^3 MeV
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8 mph

Explanation:

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Answer:

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Explanation:

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3 years ago
The amount of matter in an object is its _____.
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The amount of matter in an object ismass....anything that occupies spaca and has weight is called matter.....
7 0
3 years ago
A container is filled with an ideal diatomic gas to a pressure and volume of P1 and V1, respectively. The gas is then warmed in
lilavasa [31]

Answer:

Explanation:

The  change is as follows

P₁ V₁ to 3P₁, V₁ ( constt volume )  --- first process

3P₁,V₁ to 3P₁ , 5V₁ ( constt pressure ) ---- second process

In the first process Temperature must have been increased 3 times . So if initial temperature is T₁ then final temperature will be 3 T₁

P₁V₁ = n R T₁ , n is no of moles of gas enclosed.

nRT₁ = P₁V₁

Heat added at constant volume  = n Cv ( 3T₁ - T₁)

= n x 5/3 R X 2T₁ ( for diatomic gas Cv = 5/3 R)

= 10/3 x nRT₁

= 10/3x P₁V₁

In the second process,  Temperature must have been increased 5 times . So if initial temperature is 3T₁ then final temperature will be 15 T₁

Heat added at constant pressure in second case  

= n Cp ( 15T₁ - 3T₁)

= n x 7/3 R X 12T₁ ( For diatomic gas Cp = 7/3 R)

= 28 x nRT₁

= 28 P₁V₁

6 0
3 years ago
A particularly beautiful note reaching your ear from a rare Stradivarius violin has a wavelength of 39.1 cm. The room is slightl
USPshnik [31]

Given Information:  

Wavelength =  λ = 39.1 cm = 0.391 m

speed of sound = v = 344 m/s

linear density = μ = 0.660 g/m = 0.00066 kg/m

tension = T = 160 N

Required Information:

Length of the vibrating string = L = ?

Answer:

Length of the vibrating string = 0.28 m

Explanation:

The frequency of beautiful note is

f = v/λ

f = 344/0.391

f = 879.79 Hz

As we know, the speed of the wave is

v = √T/μ

v = √160/0.00066

v = 492.36 m/s

The wavelength of the string is

λ = v/f

λ = 492.36/879.79

λ = 0.5596 m

and finally the length of the vibrating string is

λ = 2L

L = λ/2

L = 0.5596/2

L = 0.28 m

Therefore, the vibrating section of the violin string is 0.28 m long.

3 0
2 years ago
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