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2 years ago

The change in momentum that occurs when a 1. 0 kg ball traveling at 4. 0 m/s strikes a wall and bounces back at 2. 0 m/s is.

Physics

1 answer:

Doss [256]2 years ago

7 0

Answer:

The change is momentum is given by ∆p=p(inital) - p(final) =4-2=2 kg.m/s

Explanation:

momentum is the product of mass and velocity (speed)

So it's initial momentum would be:

p=mv=(1)(4)=4 kg.m/s

It's final momentum is given by:

p=mv=(1)(2)=2 kg.m/s

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Why has the atomic model changed over time

inysia [295]

The first model of the atom was developed by JJ Thomson in 1904, who thought that atoms were composed purely of negatively charged electrons. This model was known as the 'plum pudding' model.

This theory was then disproved by Ernest Rutherford and the gold foil experiment in 1911, where Rutherford shot alpha particles at gold foil, and noticed that some went through and some bounced back, implying the existence of a positive nucleus.

In 1913, Niels Bohr proposed a model of the atom where the electrons were contained within quantized shells that orbited the nucleus. This was because it was impossible for the cloud of negative electrons proposed by Rutherford to exist, as the negative electrons would be drawn to the positive nucleus, and the atom would collapse in on itself.

In 1926, the Austrian physicist Erwin Schrödinger created a quantum mechanical model of the atom by combining the equations for the behavior of waves with the de Broglie equation to generate a mathematical model for the distribution of electrons in an atom.

However the model used today is closest to the Bohr model of the atom, using the quantized shells to contain the electrons.

For more info:

http://chemistry.about.com/od/chemistryglossary/a/debroglieeqdef.htm

7 0

3 years ago

What determines how the different moon phases appear from earth

pychu [463]

Answer:

The position of the earth and moon.

Explanation:

The position of the earth and moon are the two main factors.

Hope this helps, have great day/night!

7 0

3 years ago

Read 2 more answers

An electron is pushed into an electric field where it acquires a 1-V electrical potential. Suppose instead that two electrons ar

sleet_krkn [62]

Answer:

0.5 V

Explanation:

The electric potential distance between different locations in an electric field area is unaffected by the charge that is transferred between them. It is solely dependent on the distance. Thus, for two electrons pushed together at the same distance into the same field, the electric potential will remain at 1 V. However, the electric potential of one of the two electrons will be half the value of the electric potential for the two electrons.

6 0

3 years ago

Does specific heat of a substance depend on its temperature?

lara [203]

Answer:

temperature

Explanation:

In general, the specific heat also depends on the temperature. The table below lists representative values of specific heat for various substances. Except for gases, the temperature and volume dependence of the specific heat of most substances is weak.

4 0

3 years ago

The op amp in this circuit is ideal. R3 has a maximum value of 100 kΩ and σ is restricted to the range of 0.2 ≤ σ ≤ 1.0. a. Calc

Firlakuza [10]

I have attached the circuit image missing in the question.

Answer:

A) The range of vo is; -6.6V≤ vo ≤-1V

B) σ = 0.1861

Explanation:

A) First of all, Let VΔ be the voltage from the potentiometer contact to the ground.

Thus; [(0 - vg)/(2000)] +[(0 - vΔ)/(50,000)] = 0

So, [(- vg)/(2000)] +[(- vΔ)/(50,000)] = 0

Simplifying further; -25 vg - vΔ = 0

From the question, vg = 40mV = 0.04 V

So - 25(0.04) = vΔ

So: vΔ = - 1 V

Now, [vΔ/(σRΔ)] + [(vΔ - 0)/(50,000)] + [(vΔ - vo)/((1 - σ)RΔ))] = 0

So, multiplying each term by RΔ to get; [vΔ/(σ)] + [(vΔ x RΔ)/(50,000)] + [(vΔ - vo)/((1 - σ))] = 0

So RΔ = 100kΩ or 100,000Ω from the question.

So, substituting for RΔ, we get,

[vΔ/(σ)] + [2vΔ] + [(vΔ - vo)/((1 - σ))] = 0

Let's put the value of - 1 for vΔ as gotten before.

So, ( - 1/σ) - 2 + [(-1 - vo)/(1 - σ)] = 0

Now let's make vo the subject of the equation to get;

-1 - vo = (1 - σ)[2 + (1/σ)]

-1 - vo = 2 - 2σ + (1/σ) - 1

-vo = 1 + 2 - 2σ + (1/σ) - 1

-vo = 2 - 2σ + (1/σ)

vo = - 1 (2 - 2σ + (1/σ))

When σ = 0.2; vo = - 1(2 - 0.4 + 5) =

- 1 x 6.6 = - 6.6V

Also when σ = 1;

vo = - 1(2 - 2 + 1) = - 1V

Therefore, the range of vo is;

- 6.6V ≤ vo ≤ - 1V

B) it will saturate at vo = - 7V

So, from;

vo = - 1 (2 - 2σ + (1/σ))

-7 = - 1 (2 - 2σ + (1/σ))

Divide both sides by (-1)

7 = (2 - 2σ + (1/σ))

Now, subtract 2 from both sides to get; 5 = - 2σ + (1/σ)

Multiply each term by α to get;

5σ = - 2σ^(2) + 1

So 2σ^(2) + 5σ - 1 = 0

Solving simultaneously and picking the positive value , we get σ to be approximately 0.1861

8 0

3 years ago