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olchik [2.2K]
3 years ago
12

PLEASE HELP show the calculation of the number of neutrons of hydrogen​

Physics
1 answer:
xz_007 [3.2K]3 years ago
7 0
The answer is 2.3 hope this helps texted me and tell me if it’s right
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When the atmosphere is not quite clear, one may sometimes see colored circles concentric with the Sun or the Moon. These are gen
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Answer:

D) diffraction

Explanation:

Corona is an optical phenomenon produced by the diffraction of sunlight or moonlight, as light moves through water droplets in the atmosphere.

This phenomenon produces one or more diffuse concentric rings of light around the Sun or Moon, usually seen as colored circles.

Therefore, the explanation for these phenomena of colored concentric circles, sometimes seen with the Sun or the Moon involves diffraction.

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A bug crawls 3.0 mm east, 4.0mm north, and then 5.0 mm at 45 north of east. Draw a diagram showing its displacements and determi
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Explanation is given

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3 years ago
Solve the following system of equations by using either substitution or elimination.
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I believe that the answer to the question provided above are the following;

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3 years ago
in a railyard a train is being put together from freight cars.An empty freight car coasting 10 m/s strikes a loaded car that is
svetoff [14.1K]

Answer:m1v1 + m2v2 = (m1f + m2f)vf. 3000kg(10.0m/s) + (15000kg)(0.0m/s) = (18000kg)(vf).

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2 years ago
The component of the external magnetic field along the central axis of a 78-turn circular coil of radius 34.0 cm decreases from
grigory [225]

Answer:

Induced current, I = 18.88 A

Explanation:

It is given that,

Number of turns, N = 78

Radius of the circular coil, r = 34 cm = 0.34 m

Magnetic field changes from 2.4 T to 0.4 T in 2 s.

Resistance of the coil, R = 1.5 ohms

We need to find the magnitude of the induced current in the coil. The induced emf is given by :

\epsilon=-N\dfrac{d\phi}{dt}

Where

\dfrac{d\phi}{dt} is the rate of change of magnetic flux,

And \phi=BA

\epsilon=-NA\dfrac{dB}{dt}

\epsilon=-78\times \pi (0.34)^2\dfrac{(0.4-2.4)}{2}

\epsilon=28.32\ V

Using Ohm's law, \epsilon=I\times R

Induced current, I=\dfrac{\epsilon}{R}

I=\dfrac{28.32}{1.5}

I = 18.88 A

So, the magnitude of the induced current in the coil is 18.88 A. Hence, this is the required solution.

5 0
3 years ago
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