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2 years ago

The atomic mass of carbon-13 is 13. It has six protons. How many neutrons does this isotope have?

2 answers:

Ber [7]2 years ago

7 0

Carbon-13 will have 6 protons (because it is carbon), 6 electrons, and 7 neutrons. Isotopes have a different number of neutrons.

IgorC [24]2 years ago

3 0

Atomic mass is the given by the sum of the number of protons and number of neutron in the nucleus of the atom.

A = P + N

where A = atomic mass of the atom

P = number of protons in the nucleus of the atom

N = Number of neutrons in the nucleus of the atom

Given that :

A = atomic mass = 13

P = number of protons in the nucleus = 6

N = ?

Using the equation

A = P + N

13 = 6 + N

N = 13 - 6

N = 7

hence the isotope has 7 neutrons

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2 years ago

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MA_775_DIABLO [31]

Answer:

Resolving horizontally. :

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2 years ago

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2 years ago

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3 years ago

Use the worked example above to help you solve this problem. An Alaskan rescue plane drops a package of emergency rations to str

Vlad [161]

Answer:

(a) The package lands 682 meters horizontally ahead from the point the package was dropped from the plane

(b) The horizontal component = 39.0 m/s

The vertical component = 171.55 m/s

Explanation:

The parameters given are;

Velocity of the plane, vₓ = 39.0 m/s

Height of the plane above the ground, h = 1.50 × 10² m = 1,500 m

(a) The time, t, before the package hits the ground when dropped from the plane is given by the relation;

h = u·t + 1/2×g×t²

Where:

g = Acceleration due to gravity = 9.81 m/s²

u = Initial vertical velocity = 0 m/s

Hence;

1500 = 0×t + 1/2 × 9.81 × t² = 4.905·t²

∴ t = √(1500/4.905) = 17.49 s

The horizontal distance the package travels before landing = 17.49 × 39 ≈ 682 m

The package lands 682 meters horizontally ahead from the point the package was dropped from the plane

(b) The vertical velocity, $v_y$ , of the package just before landing is given by the relation;

$v_y$ ² = u² + 2·g·h

u = 0 m/s

∴ $v_y$ ² = 0 + 2×9.81×1500 = 29430 m²/s²

$v_y$ = √29430 = 171.55 m/s

Hence the horizontal component = 39.0 m/s

The vertical component = 171.55 m/s

$tan \theta = \dfrac{v_y}{v_x} = \dfrac{171.55}{39.0 } = 4.4$

∴ θ = tan⁻¹(4.4) = 77.19°.

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3 years ago