Question 7<br> (01.01 MC)<br> What is the difference between a hypothesis and a theory?

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

3 years ago

Two cannonballs are dropped from a second-floor physics lab at height h above the ground. Ball B has four times the mass of ball

denpristay [2]

Answer:

1:4

Explanation:

The formula for calculating kinetic energy is:

$KE=\dfrac{1}{2}mv^2$

If the mass is multiplied by 4, then, the kinetic energy must be increased by 4 as well. Since they will be travelling at the same speed when they are at the same point, the relation between KA and KB must be 1:4 or 1/4. Hope this helps!

3 0

3 years ago

A 1100 kg sports car accelerates from 0m/s to 32m/s in 10 seconds. What is the average power of the engine?

timurjin [86]

Answer:

The average power of the engine of the sports car is 56.32 kW

Explanation:

Given;

mass of the sports car, m = 1100 kg

initial velocity of the sports car, u = 0 m/s

final velocity of the sports car, v = 32 m/s

time of motion, t = 10 s

The kinetic energy of the car is given by;

K.E = ¹/₂m(v² - u²)

K.E = ¹/₂mv²

K.E = ¹/₂ x 1100 x 32²

K.E = 563200 J

The average power of the engine of the sports car is given by;

Pavg = Energy / time

Pavg = 563200 / 10

Pavg = 56320 W

Pavg = 56.32 kW

Therefore, the average power of the engine of the sports car is 56.32 kW

8 0

3 years ago

Need help in knowing what energy it’s using

ser-zykov [4K]

Answer:

8:P.E,9:gaining K.E,10:P.E,11:gaining K.E.

Explanation:

it gained potential energy at 8 because it was at its maximum height,and since potential energy is the energy that u posses due to ur height,thats the energy it possesed on the spot 8.as you can see on spot 9 its moving down so as its moving down,it gains kinetic energy,since its an energy possesed by a body due to its motion/movement,thats what it gained at spot 9, it gsained kineti energy because of its movement.and at 10 agian it would gain potential energy ecause of height and on point 11 as its heading upwards it gains kinetic energy because of movement.

4 0

2 years ago

A diver named Jacques observes a bubble of air rising from the bottom of a lake (where the absolute pressure is 3.50 atm) to the

Studentka2010 [4]

Answer:

A) the ratio of volumes of the bubble is Vs/Vb= 3.74

B) would not be safe since Vs/Vb== 3.5 and there is no lung capacity to store such amount of volume, and thus would generate an unsafe pressure over the lungs. would be better to release air accordingly to maintain safe conditions.

Explanation:

assuming the gas of the bubble behaves as ideal gas

P * V = n * R * T

where P= absolute pressure, V= volume occupied by the gas, n = number of moles , R = ideal gas constant , T = absolute temperature

if we assume that the mass of the bubble remains constant ( that is, it does not capture other bubbles during ascension of disaggregate into smaller ones and there is no mass transfer into the bubble due to diffusion)

inicial state) Pb * Vb = n * R * Tb

final state) Ps * Vs = n * R * Ts

dividing both equations

(Ps/Pb)(Vs/Vb) = Ts/Tb

therefore

Vs/Vb= (Ts/Tb) (Pb/Ps)

since Tb = 4°C = 277 K and Ts= 23°C = 296 K

Vs/Vb= (Ts/Tb) (Pb/Ps) = (296K/277K)*(3.5 atm/1 atm) = 3.74

B) if the T remains constant Ts=Tb and thus

Vs/Vb= (Ts/Tb) (Pb/Ps)= 1* (Pb/Ps) = 3.5 atm/1 atm = 3.5

it would not be safe since there is no lung capacity to store such amount of volume, and thus would generate an unsafe pressure over the lungs. would be better to release air accordingly to maintain safe conditions.

8 0

3 years ago

Question 7 (01.01 MC) What is the difference between a hypothesis and a theory?