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3 years ago

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

Physics

2 answers:

slava [35]3 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]3 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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3 years ago

A Cessna 150 aircraft has a lift-off speed of approximately 125 km/h. What minimum constant acceleration does this require if th

Marysya12 [62]

Answer:

Mininmum acceleration required = $2.74 m/s^{2}$

Explanation:

Given,

Lift off speed = 125km/hr.

We know km/hr conversion factor to m/s is $\frac{5}{18}$ ,

Therefore,

Lift off speed required = $125 \times\frac{5}{18}$ m/s

Lift off speed required = 34.72222 m/s

Length of Takeoff Run given = 220 m.

So,

the flight needs to achieve its takeoff speed at a constant acceleration before it reaches the end of the 220 m runway.

Consider the formula from kinematics,

$v^{2}-u^{2}=2as$ ,

where,

v - final velocity of particle,

u - initial velocity of particle

a - the constant acceleration at which the particle is moving with

s - distance travelled by particle

So at minimum acceleration,it reaches the takeoff speed at the end of the runway.

Therefore in the formula,

we use, v = Lift off speed = 34.72222 m/s;

u = 0 m/s (plane starts from rest);

a = minimum acceleration of the plane

s = 220 m;

Substituting these values in the formula, we get,

$(34.72222)^{2}-0 = 2\times a \times 220$

a = $\frac{(34.72222)^{2} }{2\times220}$

a = $2.74 m/s^{2}$ .

Therefore,the minimum acceleration of the plane required = $2.74 m/s^{2}$ .

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3 years ago

In about 5 billion years, at the end of its lifetime, our sun will end up as a white dwarf, having about the same mass as it doe

vitfil [10]

It is fine to use the equation given by Plitter, since we are told that the mass is about the same as it is now, and I seriously doubt the original question wants the student to go into relativistic effects, electron degeneracy pressure and magnetic effects that govern a real white dwarf star.
There is no need to make it unnecessarily complicated, when the question is set at high school level. The question asks, given a particular radius, and a given mass, what will the density be (which in this case will be the average density). To answer the question, one needs to know the mass of the sun (which is about 2×1030 Kg. One needs to convert the diameter to a radius, and then calculate the spherical volume of the white dwarf. Then one can use the formula given above, namely density=mass/volume

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3 years ago