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4 years ago

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Ludwig Boltzmann performed a simple, but powerful experiment to gather evidence concerning the velocity distribution of a sample

of gas particles. His experiment revealed that the velocities of gases:

1 answer:

sp2606 [1]4 years ago

5 0

Answer:

Ludwig Boltzmann was an Austrian Physicist and he performed a simple but powerful experiment to gather evidence concerning the velocity distribution of a sample of gas particles.

His experiment revealed that the way Gases are distributed in its normal form across a range of temperature is dependent on the molar mass amen temperature of the gas. Gases with high temperature move faster due to the high number of colliding particles when compared to those with low temperature. Gases with lower molar mass move faster than those with higher molar mass.

You might be interested in

How many coulombs of positive charge are there in 47.0 gm of plutonium, given its atomic mass is 244 and that each plutonium ato

enot [183]

Answer:

1.78×10⁶ C

Explanation:

Using the atomic mass of pluonium atoms (244 g/mol), you can calculate the number of atoms in 47.0 g. Then, knowing that each plutonium atom has 96 protons, you calculate the number of protons in the 47.0 g sample. Finally, using the positive charge of one proton, you calculate the total positive charge in the 47.0 g of plutonium.

<u>1. Number of atoms of plutonium in 47.0 g</u>

Number of moles = mass / atomic mass = 47.0 g / 244 = 0.1926 moles

Number of atoms = number of moles × 6.022 × 10²³ atoms/mol

Number of atoms = 0.1926 mol × 6.022 × 10²³ atoms/mol = 1.15998×10²³ atoms

<u>2. Number of protons</u>

Number of protons = 1.15998×10²³ atoms × 96 protons/atom = 1.11385×10²⁵ protons

<u>3. Charge</u>

<u />

Charge = charge of one proton × number of protons

Charge = 1.602×10⁻¹⁹ C/proton × 1.11385×10²⁵ protons = 1.78×10⁶C

7 0

3 years ago

A dart is inserted into a spring-loaded dart gun by pushing the spring in by a distance x. For the next loading, the spring is c

bezimeni [28]

Answer:

The second dart leaves the gun two times as faster than the first one.

Explanation:

Assuming no energy loss during the spring-dart energy transfer, we have by the conservation of energy principle

$U_s = K_d \\ \frac{1}{2} kx^2 = \frac{1}{2}mv^2 \\ v = \sqrt{\frac{k}{m}x^2}.$

Given an arbitrary $x$ and its double, $2x$ , launch velocities are

$v_1 = \sqrt{\frac{k}{m}x^2} \text{ and} \\ v_2 = \sqrt{\frac{k}{m}\left(2x\right)^2} = \sqrt{\frac{k}{m}4x^2} = 2\sqrt{\frac{k}{m}x^2} = \mathbf{2v_1}.$

7 0

4 years ago

During an experiment, Ellie records a measurement of 0.0034 m. How would

Goshia [24]

Answer:

(A) She needs to move the decimal point by 3 places

8 0

3 years ago

Is it possible to have a charge of 5 x 10-20 C? Why?

ruslelena [56]

1) No

2) Yes

3) No

4) Equal and opposite

5) 32400 N

6) Repulsive

7) The electric force is $2.3\cdot 10^{39}$ times bigger than the gravitational force

Explanation:

1)

In nature, the minimum possible charge that an object can have is the charge of the electron, which is called fundamental charge:

$e=1.6\cdot 10^{-19}C$

Electrons are indivisible particles (they cannot be separated), this means that an object can have at least the charge equal to the charge of one electron (in fact, it cannot have a charge less than $e$ , because it would meant that the object has a "fractional number" of electrons).

In this problem, the object has a charge of

$Q=5\cdot 10^{-20}C$

If we compare this value to $e$ , we notice that $Q$ , so no object can have a charge of $Q$ .

2)

As we said in part 1), an object should have an integer number of electrons in order to be charged.

This means that the charge of an object must be an integer multiple of the fundamental charge, so we can write it as:

$Q=ne$

where

Q is the charge of the object

n is an integer multiple

e is the fundamental charge

Here we have

$Q=2.4\cdot 10^{-18}C$

Substituting the value of e, we find n:

$n=\frac{Q}{e}=\frac{2.4\cdot 10^{-18}}{1.6\cdot 10^{-19}}=15$

n is integer, so this value of the charge is possible.

3)

We now do the same procedure for the new object in this part, which has a charge of

$Q=2.0\cdot 10^{-19}C$

Again, the charge on this object can be written as

$Q=ne$

where

n is the number of electrons in the object

Using the value of the fundamental charge,

$e=1.6\cdot 10^{-19}C$

We find:

$n=\frac{Q}{e}=\frac{2.0\cdot 10^{-19}}{1.6\cdot 10^{-19}}=1.25$

n is not integer, so this value of charge is not possible, since an object cannot have a fractional number of electrons.

4)

To solve this part, we use Newton's third law of motion, which states that:

"When an object A exerts a force on an object B (Action force), then object B exerts an equal and opposite force on object A (reaction force)".

In this problem, we have two objects:

- A charge Q

- A charge 5Q

Charge Q exerts an electric force on charge 5Q, and we can call this action force. At the same time, charge 5Q exerts an electric force on charge Q (reaction force), and according to Newton's 3rd law, the two forces are equal and opposite.

5)

The magnitude of the electric force between two single-point charges is

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the two charges

In this problem we have:

$q_1=+4.5\cdot 10^{-6}C$ is charge 1

$q_2=+7.2\cdot 10^{-6}C$ is charge 2

r = 0.30 cm = 0.003 m is the separation

So, the electric force between the two charges is

$F=(9\cdot 10^9)\frac{(4.5\cdot 10^{-6})(7.2\cdot 10^{-6})}{(0.003)^2}=32400 N$

6)

The electric force between two charged objects has direction as follows:

- If the two objects have charges of opposite signs (+ and -), the force between them is attractive

- If the two objects have charges of same sign (++ or --), the force between them is repulsive

In this problem, the two charges are:

$q_1=+4.5\cdot 10^{-6}C$ is charge 1

$q_2=+7.2\cdot 10^{-6}C$ is charge 2

We see that the two charges have same sign: therefore, the force between them is repulsive.

7)

The electric force between the proton and the electron in the atom can be written as

$F_E=k\frac{q_1 q_2}{r^2}$

where

$q_1 = q_2 = e = 1.6\cdot 10^{-19}C$ is the magnitude of the charge of the proton and of the electron

$r=5.3\cdot 10^{-11} m$ is the separation between them

So the force can be rewritten as

$F_E=\frac{ke^2}{r^2}$

The gravitational force between the proton and the electron can be written as

$F_G=G\frac{m_p m_e}{r^2}$

where

G is the gravitational constant

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

$m_e=9.11\cdot 10^{-27}kg$ is the electron mass

Comparing the 2 forces,

$\frac{F_E}{F_G}=\frac{ke^2}{Gm_p m_e}=\frac{(9\cdot 10^9)(1.6\cdot 10^{-19})^2}{(6.67\cdot 10^{-11})(1.67\cdot 10^{-27})(9.11\cdot 10^{-31})}=2.3\cdot 10^{39}$

8 0

3 years ago

If your friend drops a chocolate bar to you from a height of 5.0 m above your hands,

Sladkaya [172]

Answer:

<h3>1.01 s</h3>

Explanation:

Using the equation of motion S = ut+1/2gt² to solve the problem where;

u is the initial velocity of the chocolate = 0m/s

t is the time taken

g is the acceleration due to gravity = 9.81m/s²

S is the height of fall = 5.0m

Substituting the given parameter into the formula to get the time t we have;

5 = 0(t)+1/2(9.81)t²

5 = 4.905t²

t² = 5/4.905

t² = 1.019

t = √1.019

t = 1.009 secs

<em>Hence it will take 1.01 secs for me to catch the chocolate bar</em>

6 0

4 years ago