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bekas [8.4K]
3 years ago
5

What is the unit for atomic mass

Physics
2 answers:
nataly862011 [7]3 years ago
8 0

Answer:

<u><em>The answer is</em></u>: <u>The twelfth part (1/12) of the mass of an atom, neutral and unbound, of carbon-12.</u>

<u></u>

Explanation:

<em>The unit of unified atomic mass - symbol "u" -1 or dalton - symbol "Da" - is a standard unit of mass defined as</em> the twelfth part (1/12) of the mass of an atom, neutral and unbound , of carbon-12,<em> in its fundamental electrical and nuclear state, and is equivalent to 1,660 538 921 (73) × 10-27 kg (value recommended by CODATA). </em>

<u><em>The answer is</em></u>: <u>The twelfth part (1/12) of the mass of an atom, neutral and unbound, of carbon-12.</u>

Sholpan [36]3 years ago
3 0
An atomic mass unit (symbolized AMU or amu) is defined as precisely 1/12 the mass of an atom of carbon-12. The carbon-12 (C-12) atom has six protons and six neutrons in its nucleus.
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A device that uses electricity and magnetism to create motion is called a (motor magnet generator) . In a reverse process, a dev
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the answers are a. and c.

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3 years ago
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The efficiency of a modern bicycle is 95% if you exert 200 Joules of input work pedaling a bicycle on level ground what is the u
Delvig [45]

Answer:

95 J

Explanation:

You can calculate efficiency by dividing useful output by total input, then multiplying it to 100.

So the foumula goes like:

Efficiency= (Useful output/Total input)x100

In this question,

Efficiency= 95%

Useful output= x

Total input= 200

Therefore;

95=(x/200)x100

0.95=x/100

x=0.95x100

x=95 Joules

8 0
3 years ago
Find the angle for the third-order maximum for 591 nm wavelength light falling on a diffraction grating having 1460 lines per ce
Marat540 [252]

Answer:

15.32°

Explanation:

We have given the wavelength \lambda =591nm=591\times 10^{-9}m

Diffraction grating is 1460 lines per cm

So  d=\frac{10^{-2}}{1460}=6.71\times 10^{-6}m (as 1 m=100 cm )

For maximum diffraction

dsin\Theta =m\lambda here m is order of diffraction

So 6.71\times 10^{-6}sin\Theta =3\times 591\times 10^{-9}

sin\Theta =0.264

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6 0
3 years ago
An object is placed perpendicular to the the principal axis of convex lens of focal length 8,the distance of the object from the
Digiron [165]

Answer:

v = 24 cm and inverted image

Explanation:

Given that,

The focal length of the object, f = +8 cm

Object distance, u = -12 cm

We need to find the position &nature of the image​. Let v be the image distance. Using lens formula to find it :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

Put all the values,

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{8}+\dfrac{1}{(-12)}\\\\v=24\ cm

So, the image distance from the lens is 24 cm.

Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{24}{-12}\\\\m=-2

The negative sign of magnification shows that the formed image is inverted.

7 0
3 years ago
In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu
Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

Explanation:

The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

Distance traveled = 5.31 \times 10^{-13} m

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

  (K.E + P.E)_{initial} = (K.E + P.E)_{final}

 (\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)

where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

     U = (\frac{1}{2}m_{p}v^{2}_{p})

Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

            = (\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})

            = 5.218 \times 10^{-15} J

Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

4 0
3 years ago
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