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3 years ago

show that if a particle is in a stationary state at a given time it will always remain in a stationary state.

Physics

1 answer:

Lena [83]3 years ago

8 0

That statement is impossible to show, demonstrate, or prove.

A big part of the reason for the inability to show it is that the statement is ... prima facie, pro bono, and habeus corpus ... wrong, and as false as the day is long.

Although I am certainly far from what you might call a 'particle', I think I can show, using myself, that the statement is false:

As we speak, I am sitting, <u><em>in a stationary state</em></u>, before my computer. It is my intention, and I fully believe it will come to pass, that in a few seconds, my fingers may move to operate one of the letter keys, or my hand may reach out to snag my coffee cup, or my entire being may arise from my place and go into the kitchen for a bowl of oatmeal. At the INSTANT that any of these events takes place, the statement will have been proven false.

QED

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Why a pond of water appears shallower than it actually is?

GalinKa [24]

The refraction of light at the surface of water makes ponds look shallower then they really are

3 0

3 years ago

A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.10 m/s

yarga [219]

Answer:

$I=336.6kgm/s$

Explanation:

The equation for the linear impulse is as follows:

$I=F\Delta t$

where $I$ is impulse, $F$ is the force, and $\Delta t$ is the change in time.

The force, according to Newton's second law:

$F=ma$

and since $a=\frac{v_{f}-v_{i}}{\Delta t}$

the force will be:

$F=m(\frac{v_{f}-v_{i}}{\Delta t})$

replacing in the equation for impulse:

$I=m(\frac{v_{f}-v_{i}}{\Delta t})(\Delta t)$

we see that $\Delta t$ is canceled, so

$I=m(v_{f}-v_{i})$

And according to the problem $v_{i}=0m/s$ , $v_{f}=5.10m/s$ and the mass of the passenger is $m=66kg$ . Thus:

$I=(66kg)(5.10m/s-0m/s)$

$I=(66kg)(5.10m/s)$

$I=336.6kgm/s$

the magnitude of the linear impulse experienced the passenger is $336.6kgm/s$

6 0

3 years ago

Residential building codes typically require the use of 12-gauge copper wire (diameter 0.2053 cm) for wiring receptacles. Such c

AysviL [449]

Answer:

a) E = 4.26 W

b) E' = 6.724 W

c) copper wire is the safer option to use.

Explanation:

Given:

Diameter of the 12 gauge copper wire, d = 0.2053 cm

Thus, Radius of the 12 gauge copper wire, r = 0.2053 cm

/ 2 = 0.10265 cm = 0.10265 × 10⁻² m

Now.

the area (A) comes out as

A = π × (0.10265 × 10⁻²)²

A = 3.3103 × 10⁻⁶ m²

Length of the copper wire, L = 2.10 m

a) The resisitivity (ρ) of copper = 1.68 × 10⁻⁸ ohm m

Now,

the resistance of the copper , R = ρL/A

R = (1.68 × 10⁻⁸ × 2.1) / ( 3.3103 × 10⁻⁶)

R = 0.01065 ohm

The Energy (E) is given as,

E = I²R

where, I is the current

I = 20.0 A

on substituting the values, we get

E = 20.0² × 0.01065

E = 4.26 W

(b) For the aluminium

Resisitivity, ρ' = 2.65 × 10⁻⁸ ohm m

Now, the resistance of the aluminium wire, R' = (ρ' × L) / A

Since the cross-section of the aluminium wire is same as the copper wire

thus,

R = (2.65 × 10⁻⁸ × 2.1) / ( 3.3103 × 10⁻⁶)

R = 0.0168 ohm

Therefore,

The Rate of energy produced by the aluminium wire, E' = I²R'

E' = 20.0² × 0.0168

E' = 6.724 W

(c) From the above results, we can conclude that the power consumed or the rate of energy produced by the aluminium wire is more.

Hence, copper wire is the safer option to use.

8 0

4 years ago

When a low-pressure gas of hydrogen atoms is placed in a tube and a large voltage is applied to the end of the tube, the atoms w

FromTheMoon [43]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value of n is $n =7$

Explanation:

From the question we are told that

The value of m = 2

For every value of $m, n = m+ 1, m+2,m+3,....$

The modified version of Balmer's formula is $\frac{1}{\lambda} = R [\frac{1}{m^2} - \frac{1}{n^2} ]$

The Rydberg constant has a value of $R = 1.097 *10^{7} m^{-1}$

The objective of this solution is to obtain the value of n for which the wavelength of the Balmer series line is smaller than 400nm

For m = 2 and n =3

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{3^2} ]$

$\lambda = \frac{1}{1523611.1112}$

$\lambda = 656nm$

For m = 2 and n = 4

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{4^2} ]$

$\lambda = \frac{1}{2056875}$

$\lambda = 486nm$

For m = 2 and n = 5

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{5^2} ]$

$\lambda = \frac{1}{2303700}$

$\lambda = 434nm$

For m = 2 and n = 6

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{6^2} ]$

$\lambda = \frac{1}{2422222}$

$\lambda = 410nm$

For m = 2 and n = 7

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{7^2} ]$

$\lambda = \frac{1}{2518622}$

$\lambda = 397nm$

So the value of n is 7

7 0

3 years ago

A car is traveling north. can its acceleration vector ever Point South? explain

Scrat [10]

Yes.

The acceleration vector WILL point south when the car is slowing down while traveling north

7 0

3 years ago