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1 year ago

What is a limitation of the electron cloud model theory that a law about electrons would not have?

Physics

2 answers:

Drupady [299]1 year ago

8 0

Electrons couldn't orbit the nucleus like miniature planets~!

ch4aika [34]1 year ago

4 0

Explanation:

Electron cloud model describes the region in an atom which negatively charged and which has probability to find an electron. according to this model, an electron can move closer or away from the nucleus that is it can be inside the nucleus but according to Bohr's model, an electron is always at a fixed distance from the nucleus. Thus, it is the limitation of the electron cloud model but still it is a widely accepted model.

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An experiment consists of determining the speed of automobiles on a highway by the use of radar equipment. The random variable i

faust18 [17]

The random variable in this experiment is a Continuous random variable.

Option D

<u>Explanation</u>:

The continuous random variable is random variable where the data can take infinite variables. For example random variable is taken for measuring "speed of automobiles" on the highways. The radar instrument depicts time taken by automobile in particular what speed. They are the generalization of discrete random variables not the real numbers as a random data is created. It gives infinite sets of all possible outcomes. It is obvious that outcomes of the instrument depend on some "physical variables" those are not predictable as depends on the situation.

8 0

1 year ago

Read 2 more answers

Suppose Earth's mass increased but Earth's diame-

navik [9.2K]

Answer: It would increase.

Explanation:

The equation for determining the force of the gravitational pull between any two objects is:

$F = G \frac{m1m2}{r^2}$

Where G is the universal gravitational constant, m1 is the mass of one body, m2 is the mass of the other body, and r^2 is the distance between the two objects' centers squared.

Assuming the Earth's mass but not its diameter increased, in the equation above m1 (the term usually indicative of the object of larger mass) would increase, while the r^2 would not.

Thus, it goes without saying that, with some simple reasoning about fractions, an increasing numerator over a constant denominator would result in a larger number to multiply by G, thus also meaning a larger gravitational strength between Earth and whatever other object is of interest.

7 0

2 years ago

On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he

Liula [17]

Answer:

correct is d) a ’= g / 2

Explanation:

For this exercise let's use the kinematics equations

On earth

v = v₀ - a t

a = (v₀- v) / T

On planet X

v = v₀ - a' t’

a ’= (v₀-v) / 2T

Let's substitute the land values in plot X

a’= a / 2

Now let's use Newton's second law

W = ma

m g = m a

a = g

We substitute

a ’= g / 2

So we see that on planet X the acceleration is half the acceleration of Earth's gravity

4 0

1 year ago

You catch a volleyball (mass 0.270 kg) that is moving downward at 7.50 m/s. In stopping the ball, your hands and the volleyball

sesenic [268]

Explanation:

The work done equals the change in energy.

W = ΔKE

W = 0 − ½mv²

W = -½ (0.270 kg) (-7.50 m/s)²

W = -7.59 J

Work is force times displacement.

W = Fd

-7.59 J = F (-0.150 m)

F = 50.6 N

3 0

2 years ago

The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit

Margaret [11]

Let us consider two bodies having masses m and m' respectively.

Let they are separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as - $F = G\frac{mm'}{r^{2} }$ where G is the gravitational force constant.