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Marina86 [1]
2 years ago
12

What conclusions can be drawn about the existence of nitrogen-14 and nitrogen-15? A) They are isotopes of nitrogen and they cont

ain the same number of protons and electrons but each contains a different number of neutrons - 7 and 8 respectively. B) They are the same form of nitrogen and they include the same number of electrons and neutrons but each contains a different number of protons - 7 and 8 respectively. C) They are several allotropes of nitrogen and they accommodate the same number of neutrons but each contains a different numbers of protons and electrons - 7 and 8 respectively. D) They are different configurations of nitrogen and they consist of the same number of protons and neutrons but each contains a different number of electrons - 7 and 8 respectively. Eliminate
Physics
2 answers:
Stells [14]2 years ago
8 0

Answer:

A

Explanation:

AleksandrR [38]2 years ago
5 0
<span>Answer: A) They are isotopes of nitrogen and they contain the same number of protons and electrons but each contains a different number of neutrons - 7 and 8 respectively.

Isotopes are atoms of a chemical element whose nucleus has the same atomic number, Z, but different atomic mass, A. The atomic number corresponds to the number of protons in the atom, therefore the isotopes of an element contain the same number of protons and electrons (atoms have to be neutral particles). The difference in atomic masses arises from the difference in the number of neutrons in the atomic nucleus.
</span>
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As a mass on a spring moves farther from the equilibrium position, how do the velocity, acceleration, and force change
Umnica [9.8K]
Refer to the diagram shown below.

m =  the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A =  the amplitude ( the maximum distance) of the mass from the equilibrium
        position

The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω =  the circular frequency of the motion
T =  the period of the motion so that ω = (2π)/T

The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)

In the equilibrium position,
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At the farthest distance (A) from the equilibrium position,
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In the graphs shown, it is assumed (for illustrative purposes) that
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6 0
3 years ago
What is the period (in hours) of a satellite circling Mars 100 km above the planet's surface? The mass of Mars is 6.42 × 1023 kg
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To solve this problem it is necessary to apply the concepts related to the Centrifugal Force and the Gravitational Force. Since there is balance on the body these two Forces will be equal, mathematically they can be expressed as

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Where,

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Re-arrange to find the velocity we have,

v^2 = \frac{GM}{r}

At the same time we know that the period is equivalent in terms of the linear velocity to,

T = \frac{2\pi}{\frac{v}{r}}

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If our values are that the radius of mars is 3400 km and the distance above the planet is 100km more, i.e, 3500km we have,

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( \frac{2\pi r}{T})^2 =  \frac{GM}{r}

T = \sqrt{\frac{4\pi^2 r^3}{GM}}

Replacing we have,

T = \sqrt{\frac{4\pi^2 (3500*10^3)^3}{(6.67430*10^{-11})(6.42*10^23)}}

T = 6285.09s (\frac{1min}{60s})(\frac{1hour}{60min})

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Therefore the correct answer is C.

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Answer:

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