an increase in the total energy of the particles in an object results in an increase in the blank energy of the object

At a department store, you adjust the mirrors in the dressing room so that they are parallel and 6.2 ft apart. You stand 1.8 ft

barxatty [35]

Answer:

a) 3.6 ft

b) 12.4 ft

Explanation:

Distance between mirrors = 6.2 ft

difference from from the mirror you face = 1.8 ft

a) you stand 1.8 ft in front of the mirror you face.

According to plane mirror rules, the image formed is the same distance inside the mirror surface as the distance of the object (you) from the mirror surface. From this,

your distance from your first "front" image = 1.8 ft + 1.8 ft = 3.6 ft

b) The mirror behind you is 6.2 - 1.8 = 4.4 ft behind you.

the back mirror will be reflected 3.6 + 4.4 = 8 ft into the front mirror,

the first image of your back will be 4.4 ft into the back mirror,

therefore your distance from your first "back" image = 8 + 4.4 = 12.4 ft

8 0

3 years ago

30 points!

vlada-n [284]

Answer:

Since strong nuclear forces involve only nuclear particles (not electrons, bonds, etc) items 3 and 4 are eliminated.

Again item 2 refers to bonds between atoms and is eliminated.

This leaves only item 1.

Nuclear forces are very short range forces between components of the nucleus.

Weak nuclear forces are trillions of times smaller than strong forces.

Gravitational forces are much much smaller than the weak nuclear force.

6 0

2 years ago

"Force" can be defined as a push or a O pull O grab

Nady [450]

Answer:

pull

is your answer please give me some thanks

4 0

2 years ago

Read 2 more answers

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

How does what happens to the particles in a substance during melting differ from what happens during freezing?

sasho [114]

When in the melting process particles start to move more freely when in the freezing process particles tend to slow and vibrate in place

8 0

3 years ago