All categories

3 years ago

I need help!!! 60 POINTS!!!!

Physics

2 answers:

galina1969 [7]3 years ago

8 0

The answer is "A and B are different elements, while C is an isotope of B".

Scrat [10]3 years ago

3 0

Correct one is b
Good luck

You might be interested in

A gas occupies a volume of 20 cubic meters at 9,000 pascals. If the pressure is lowered to 5,000 pascals, what volume will the g

forsale [732]

We need to consider no change in the temperature of gas (isothermal transformation)

Volume and pressure are inversely proportional magnitudes, so we can write:

$P_1.V_1=P_2.V_2\\
\\
9.20=5.V_2\\
\\
V_2=\frac{180}{5}=36 \ m^3$

5 0

2 years ago

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a

S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, $KE_{t} = \frac{3}{2}k_{b}T$

$k_{b} = 1.38\times 10^{- 23} J/k$ = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, $\lambda_{e}$ :

$\lambda_{e} = \frac{h}{p_{e}}$

$\lambda_{e} = \frac{h}{m_{e}{v_{e}}$ (1)

where

h = Planck's constant = $6.626\times 10^{- 34}m^{2}kg/s$

$p_{e}$ = momentum of an electron

$v_{e}$ = velocity of an electron

$m_{e} = 9.1\times 10_{- 31} kg$ = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

$\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T$

$}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}$

$}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}$

$v_{e} = 2.86\times 10^{5} m/s$ (2)

Using eqn (2) in (1):

$\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm$

Now, to calculate the de-Broglie wavelength of proton, $\lambda_{e}$ :

$\lambda_{p} = \frac{h}{p_{p}}$

$\lambda_{p} = \frac{h}{m_{p}{v_{p}}$ (3)

where

$m_{p} = 1.6726\times 10_{- 27} kg$ = mass of proton

$v_{p}$ = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

$\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T$

$}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}$

$}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}$

$v_{p} = 6.674\times 10^{3} m/s$ (4)

Using eqn (4) in (3):

$\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm$

7 0

2 years ago

If the kinetic energy of an electron is 4.1e-18 j, what is the speed of the electron? (you can use the approximate (nonrelativis

arlik [135]

The kinetic energy of the electron is
$K= \frac{1}{2}mv^2$
where $m=9.1 \cdot 10^{-31} kg$ is the mass of the electron and v its speed. Since we know the value of the kinetic energy, $K=4.1 \cdot 10^{-18} J$ , we can find the value of the speed v:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2\cdot 4.1 \cdot 10^{-18}J}{9.1 \cdot 10^{-31}kg} } = 3\cdot 10^6 m/s$

3 0

3 years ago

The Red Sea is widening at a rate of 1.25 centimeters per year. How many years will it take to widen another 812.5 centimeters?

Pani-rosa [81]

E 812 Jak’s now fmfdkc

5 0

3 years ago

A torsional pendulum consists of a disk of mass 450 g and radius 3.5 cm, hanging from a wire. If the disk is given an initial an

Montano1993 [528]

To solve this problem we will use the kinematic equations of angular motion, starting from the definition of angular velocity in terms of frequency, to verify the angular displacement and its respective derivative, let's start:

$\omega = 2\pi f$