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3 years ago

Where do organisms occur?

Physics

1 answer:

AVprozaik [17]3 years ago

7 0

Are you questioning when does evalution start..?

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Find the de Broglie wavelength of the following. (a) a 4-MeV proton 14.3 Correct: Your answer is correct. fm (b) a 40-GeV electr

kherson [118]

a) de Broglie wavelength of a 4-MeV proton: 14.3 fm

b) de Broglie wavelength of a 40-GeV electron: 0.031 fm

Explanation:

The de Broglie wavelength of an object is given by

$\lambda=\frac{h}{p}$ (1)

where

h is the Planck constant

p is the momentum of the particle

Here we want to find the de Broglie wavelength of a 4-MeV proton. The rest of mass of the proton in MeV is

$m_0 = 938 MeV$

And since $4MeV$ , this means that the proton is non-relativistic. So its kinetic energy is related to its momentum by

$E=\frac{p^2}{2m}$

which means

$p=\sqrt{2Em}$

where

$E=4 MeV \cdot 10^6 eV/MeV \cdot 1.6\cdot 10^{-19] J/eV=6.4\cdot 10^{-13} J$ is the kinetic energy

$m=1.67\cdot 10^{-27} kg$ is the proton mass

Substituting, we find

$\lambda=\frac{h}{\sqrt{2Em}}=\frac{6.63\cdot 10^{-34}}{\sqrt{2(6.4\cdot 10^{-13})(1.67\cdot 10^{-27})}}=14.3\cdot 10^{-15} m = 14.3 fm$

In this case, the electron has kinetic energy of 40 GeV, while the rest mass of an electron is

$m_0 = 0.511 MeV$

Since $40 GeV >> 0.511 MeV$ , the electron is ultra-relativistic: so we can rewrite its energy as

$E = pc$

The equation (1) can also be rewritten as

$\lambda = \frac{hc}{pc}$

where c is the speed of light. The quantity at the denominator is the energy, so

$\lambda=\frac{hc}{E}$

where:

$E=40 GeV = 40\cdot 10^9 eV \cdot (1.6\cdot 10^{-19})=6.4\cdot 10^{-9} J$ is the energy of the electron

And substituting, we find:

$\lambda=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{6.4\cdot 10^{-9}}=3.1 \cdot 10^{-17} m = 0.031 fm$

Learn more about de Broglie wavelength:

brainly.com/question/7047430

#LearnwithBrainly

4 0

3 years ago

Allen Aby sets up an Atwood Machine and wants to find the acceleration and the tension in the string. Please help him. Two block

Vitek1552 [10]

<u>Answers:</u>

In order to solve this problem we will use Newton’s second Law, which is mathematically expressed after some simplifications as:

This can be read as: The Net Force $F$ of an object is equal to its mass $m$ multiplied by its acceleration $a$ .

We will also need to <u>draw the Free Body Diagram of each block</u> in order to know the direction of the acceleration in this system and find the Tension $T$ of the string (<u>See figure attached). </u>

We already know<u> $m_{2}$ is greater than $m_{1}$ </u>, this means the weight of the block 2 $P_{2}$ is greater than the weight of the block 1 $P_{1}$ ; therefore <u>the acceleration of the system will be in the direction of $P_{2}$ </u>, as shown in the figure attached.

We also know by the information given in the problem that <u>the pulley does not have friction and has negligible mass</u>, and <u>the string is massless</u>.

This means that the tension will be the same along the string regardless of the difference of mass of the blocks.

Now that we have the conditions clear, let’s begin with the calculations:

1) Firstly, we have to find the weight of each block, in order to verify that block 2 is heavier than block 1.

This is done using equation (1), where the force of the weight $P$ is calculated using the <u>acceleration of gravity</u> $g=9.8\frac{m}{s^{2}}$ acting on the blocks:

<u>For block 1: </u>

$P_{1}=m_{1}g$ (3)

$P_{1}=1.5kg(9.8\frac{m}{s^{2}})$

<u>For block 2: </u>

$P_{2}=m_{2}g$ (5)

$P_{2}=2.4kg(9.8\frac{m}{s^{2}})$

Then, we are going to <u>find the acceleration $a$ of the whole system: </u>

$F_{r}=P_{1}+P_{2}$ (7)

Where the Resulting Force $F_{r}$ is equal to the sum of the weights $P_{1}$ and $P_{2}$ .

In the figure attached, note that $P_{1}$ is in opposite direction to the acceleration $a$ , this means it must <u>have a negative sing</u>; while $P_{2}$ is in the same direction of $a$ .

Here we only have to isolate $a$ from equation (8) and substitute the values according to the conditions of the system:

$-14.7N+23.52N=(1.5kg+2.4kg)a$

$8.82N=(3.9kg)a$

Then:

$a=\frac{8.82N }{3.9kg}$

<h2> $a=2.26\frac{m}{ s^{2}}$ </h2><h2>This is the acceleration of the system. </h2>

2) For the second part of the problem, we have to find the tension $T$ of the string.

We can choose either the Free Body Diagram of block A or block B to make the calculations, <u>the result will be the same</u>.

Let’s prove it:

For $m_{1}$

we see in the free body diagram that the <u>acceleration is in the same direction of the tension of the string</u>, so:

$F_{r}=T-P_{1}$ (9)

$T-P_{1}=m_{1}a$ (10)

$T-14.7N=(1.5kg)( 2.26\frac{m}{ s^{2}})$

Then;

<h2> $T=18.09N$ This is the tension of the string </h2><h2> </h2>

For $m_{2}$

we see in the free body diagram that the acceleration is in opposite direction of the tension of the string and must <u>have a negative sign,</u> so:

$F_{r}=T-P_{2}$ (9)

$T-P_{2}=m_{2}a$ (10)

$T-23.52N=(2.4kg)(-2.26\frac{m}{ s^{2}})$

Then;

<h2> $T=18.09N$ This is the same tension of the string </h2>

6 0

3 years ago

A -4.00 nC point charge is at the origin, and a second -5.50 nC point charge is on the x-axis at x = 0.800 m.

mafiozo [28]

Answer:

a. $f=1.22*10^{-15} N$

b. $f=53.6*10^{-17} N$

Explanation:

The force existing between two charges is given as

$f=\frac{kq_{1}q_{2}}{r^{2}}$

where q= charge,

k=constant

r= distance between the two charges

Note: this force can either be repulsive or attractive force depending on the charges involve. it is repulsive if they are similar charge and it is attractive if it is opposite charges.

Also the charge of an electron is

$-1.602*10^{-19}$

A. we first determine the magnitude force between the -4nC and the electron

$f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{0.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{0.04}\\f_{21}=1.44*10^{-15}Ni$

this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis

for the -5.50nC the distance between them is 0.600m as can be seen in the diagram the magnitude of the force is

$f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.6^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.36}\\f_{23}=-(0.22*10^{-15})N i$

this this force will be repulsive force and it points away from the electron i.e points towards the -ve x-axis.

The total net force on the electron is thus

$f=f_{21}+f_{23}\\ f=1.44*10^{-15}-0.22*10^{-15}\\ f=1.22*10^{-15} N$

b. at distance of x=1.20m, this is shown on the diagram below (attachment 2)

we first determine the magnitude force between the -4nC and the electron

$f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{1.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{1.44}\\f_{21}=4.0*10^{-17}Ni$

this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis.

for the -5.50nC the distance between them is 0.4m as can be seen in the diagram the magnitude of the force is

$f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.4^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.16}\\f_{23}=49.6*10^{-17}Ni$

this this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis.

The total net force on the electron is thus

$f=f_{21}+f_{23}\\ f=4.0*10^{-15}+49.6*10^{-17}\\ f=53.6*10^{-17} N$

8 0

3 years ago

A screen is placed 1.60 m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40 c

Lunna [17]

Answer:

distance between the two second-order minima is 2.8 cm

Explanation:

Given data

distance = 1.60 m

central maximum = 1.40 cm

first-order diffraction minima = 1.40 cm

to find out

distance between the two second-order minima

solution

we know that fringe width = first-order diffraction minima /2

fringe width = 1.40 /2 = 0.7 cm

and

we know fringe width of first order we calculate slit d

β1 = m1λD/d

d = m1λD/β1

and

fringe width of second order

β2 = m2λD/d

β2 = m2λD / ( m1λD/β1 )

β2 = ( m2 / m1 ) β1

we know the two first-order diffraction minima are separated by 1.40 cm

y = 2β2 = 2 ( m2 / m1 ) β1

put here value

y = 2 ( 2 / 1 ) 0.7

y = 2.8 cm

so distance between the two second-order minima is 2.8 cm

6 0

3 years ago

A star is moving towards the earth with a speed at 90% the speed of light. It emits light, which moves away from the star at the

AnnyKZ [126]

Answer:

The speed of light measured in any frame is c = 3.00E8 m/s.

This is one of Einstein's postulates of special relativity.

5 0

3 years ago