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2 years ago

The conservation of momentum is most closely related to Newton's?

Physics

1 answer:

AleksandrR [38]2 years ago

3 0

The conservation of momentum is most closely related to Newton's third law.

According to the conservation of momentum, momentum is neither created nor destroyed, although it can change from one form to another.

According to Newton's third law, for every action, there is an opposite reaction, and energy is neither created nor destroyed. Rather the energy is changed from the action to the reaction in a given system and the amount of energy remains the same.

Both the conservation of momentum and Newton's third law imply that energy in a system can be converted but the value remains the same in a system and hence is closely related.

To learn more about momentum, click here:

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A proton starting from rest travels through a potential of 1.0 kV and then moves into a uniform 0.040-T magnetic field directed

Hatshy [7]

Answer:

0.114m

Explanation:

From the general expression for the radius of the proton's resulting orbit, we have

$r=\frac{mv}{qB}$

where q is is the charge of the proton $1.6*10^{-19}C$

m is the mass of the proton $1.67*10^{-27}kg$

B is the magnetic field $0.040T$

and v i the speed.

to determine the speed, we use the expression

Kinetic Energy= $qV$

$1/2mv^{2}=qV$

where <em>V </em>is the voltage value i.e 1.0kv

and v is the speed

Hence, from simple rearrangement we have the speed v to be

$v=\sqrt{\frac{2Vq}{m}} \\$

if we substitute value, we have

$v=\sqrt{\frac{2*1000*1.6*10^{-19} }{1.67*10^{-27}}} \\$

carrying out careful arithmetic we arrive at

$v=4.38*10^{5} m/s$ .

using the value for the speed in the expression for the radius of the orbit as stated earlier, we have

$r=\frac{1.67*10^{-27}*4.38*10^{5}}{1.6*10^{-19}*0.04} \\$

$r=0.114m$

7 0

3 years ago

For atomic hydrogen, the Paschen series of lines occurs when nf = 3, whereas the Brackett series occurs when nf = 4 in the equat

zvonat [6]

Answer:

$(\lambda_{max} )_{brackett} < (\lambda_{min} )_{paschen}$

So the two wavelength range will over lap

Explanation:

The Rydberg equation is given by

$\frac{1}{\lambda} =\frac{2\pi mk^2e^4}{h^3c} t (\frac{1}{n^2_f}-\frac{1}{n^2_i} )$

m is the mass of electron

k = 1/4π∈₀

∈₀ = is the permitivity of free space

e is the charge of electron

h is the plank constant

c is the speed of light in vaccum

z is the atomic number = 1

$\frac{1}{\lambda} =R (\frac{1}{n^2_f}-\frac{1}{n^2_i} )$

where R is the Rydberg constant = 1.097373 × 10⁷m⁻¹

For Paschen series of H spectrum

$n_f = 3$

$n_i = 5,6,7 ...$

in Paschen series of H spectrum

The maximum wavelength occur for $n_i = 4$

$\frac{1}{\lambda_m_a_x } =(1.097373 \times 10^7)(\frac{1}{9} - \frac{1}{16} )\\\\\lambda_m_a_x=1874.6nm$

The minimum wavelength occur for $n_i$ = ∞

$\frac{1}{\lambda_m_i_n } =(1.097373 \times 10^7)(\frac{1}{9} - \frac{1}{_o_o} )\\\\\lambda_m_a_x=820.14nm$

The brackett series of H spectrum

The maximum wavelength occur for $n_i = 4$

$\frac{1}{\lambda_m_a_x } =(1.097373 \times 10^7)(\frac{1}{16} - \frac{1}{25} )\\\\\lambda_m_a_x=4050.05nm$

The minimum wavelength occur for $n_i$ = ∞

$\frac{1}{\lambda_m_i_n } =(1.097373 \times 10^7)(\frac{1}{16} - \frac{1}{_o_o} )\\\\\lambda_m_a_x=1458.03nm$

$(\lambda_{max} )_{brackett} < (\lambda_{min} )_{paschen}$

So the two wavelength range will over lap

8 0

3 years ago

Fill in the blanks

just olya [345]

Answer:

An atom that loses a electron is called a cation and has an overall positive charge.

4 0

4 years ago

What type of tissue in the heart pumps blood throughout the body?

slavikrds [6]

Answer:

Myocardium. That is the type. (srry i was in a rush hope this helps)

7 0

3 years ago

Determine the critical crack length for a through crack contained within a thick plate of 7150-T651 aluminum alloy that is in un

Mila [183]

Explanation:

Formula to determine the critical crack is as follows.

$K_{IC} = \gamma \sigma_{f} \sqrt{\pi \times a}$

$\gamma$ = 1, $K_{IC}$ = 24.1

[/tex]\sigma_{y}[/tex] = 570

and, $\sigma_{f} = 570 \times \frac{3}{4}$

= 427.5

Hence, we will calculate the critical crack length as follows.

a = $\frac{1}{\pi} \times (\frac{K_{IC}}{\sigma_{f}})^{2}$

= $\frac{1}{3.14} \times (\frac{24.1}{427.5})^{2}$

= $10.13 \times 10^{-4}$

Therefore, largest size is as follows.

Largest size = 2a

= $2 \times 10.13 \times 10^{-4}$

= $20.26 \times 10^{-4}$

Thus, we can conclude that the critical crack length for a through crack contained within the given plate is $20.26 \times 10^{-4}$ .

4 0

3 years ago