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3 years ago

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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3 years ago

A woman is running around a playground while keeping an eye on her child who is on a swing set. The child is going back and fort

Nutka1998 [239]

Answer:

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Explanation:

Given that the time for one cycle of the swing is 1 s

The radius of the swing, R= 3.0m

The angle covered, \theta, by each swing is a quarter of the circle. i.e.

$\Theta=\frac {\pi}{2}$

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(b) As there are two swings in one cycle, so, the distance covered in one swing is the length of the circular shown in the figure.

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In 1 cycle, there are 2 swings, so distance covered in 1 cycle $= 3 \pi m$ .

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3 years ago

A net torque of magnitude 600 Nm is ex-

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2 years ago

At what minimum speed must a roller coaster be traveling when upside down at the top of a circle so that the passengers do not f

worty [1.4K]

Answer:

$v_{min} \approx 17.153\,\frac{m}{s}$

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$\frac{1}{2}\cdot m \cdot v_{min}^{2} = m\cdot g \cdot h$

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$v_{min} = \sqrt{2\cdot g \cdot h}$

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2 years ago