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2 years ago

Which is the correct symbol for an isotope of bromine with 35 protons and 38 neutrons?

Physics

2 answers:

slava [35]2 years ago

7 0

Answer:

73/35 Br

Explanation:

The symbol for a bromine isotope with 35 protons and 38 neutrons corresponds to the acronym used to represent bromine, with the atonic number on the left side of this circle in the upper corner and the mass number on the left side of this acronym in the lower corner.

The number is the number of protons, in this case the atomic number is 35. The number of the mass is the sum between the number of protons and the number of neutrons, which in this case is 35 + 38 = 73. The acronym that corresponds to bromine is "Br".

With that, we can conclude that the correct symbol would be 73/35 Br

HACTEHA [7]2 years ago

3 0

Answer:An atomic nucleus consists of protons and neutrons, collectively called nucleons. Although protons repel each other, the nucleus is held tightly together by a short-range, but very strong, force called the strong nuclear force.

Explanation:

An atomic nucleus consists of protons and neutrons, collectively called nucleons. Although protons repel each other, the nucleus is held tightly together by a short-range, but very strong, force called the strong nuclear force.

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Two metal balls have charges of 7.1 × 10-6 coulombs and 6.9 × 10-6 coulombs. They are 5.7 × 10-1 meters apart. What is the force

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2 years ago

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The cornea behaves as a thin lens of focal length approximately 1.80 {\rm cm}, although this varies a bit. The material of which

spayn [35]

Answer:

The height of the image will be "1.16 mm".

Explanation:

The given values are:

Object distance, u = 25 cm

Focal distance, f = 1.8 cm

On applying the lens formula, we get

⇒ $\frac{1}{v} -\frac{1}{u} =\frac{1}{f}$

On putting estimate values, we get

⇒ $\frac{1}{v} -\frac{1}{(-25)} =\frac{1}{1.8}$

⇒ $\frac{1}{v} =\frac{1}{1.8} -\frac{1}{25}$

⇒ $v=1.94 \ cm$

As a result, the image would be established mostly on right side and would be true even though v is positive.

By magnification,

$m=\frac{v}{u}$ and $m=\frac{h_{1}}{h_{0}}$

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3 years ago

Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they a

suter [353]

Answer:

(a) the force is 8.876 N

(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

coulomb's constant, K = 8.99 x 10⁹ N.m²/C²

distance between two charges, r = 10 cm = 0.1 m

force between the two charges, F = 15 N

when the distance between the charges changes to 13 cm (0.13 m)

force between the two charges, F = ?

Apply Coulomb's law;

$F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N$