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Stolb23 [73]
3 years ago
13

Match the author to their work. A Erwin Schrödinger B Albert Einstein C Niels Bohr D Max Plank E Louis de Broglie Developed the

Bohr model of the atom, in which he proposed that energy levels of electrons are discrete Discovery of energy quanta made him the originator of quantum theory Formulated the wave equation; stationary and time-dependent Schrödinger equation Best known to the general public for his "mass–energy equivalence formula" ; E = mc^2 Postulated the wave nature of electrons and suggested that all matter has wave properties
Physics
1 answer:
Klio2033 [76]3 years ago
5 0

Answer:Developed the Bohr model of the atom, in which he proposed that energy levels of electrons are discrete- Niels Bohr

Discovery of energy quanta made him the originator of quantum theory -Max Plank

Formulated the wave equation; stationary and time-dependent Schrödinger equation- Erwin Schrödinger

Best known to the general public for his "mass–energy equivalence formula" ; E = mc^2-B Albert Einstein

Postulated the wave nature of electrons and suggested that all matter has wave properties- Louis de Broglie

Explanation:

All the names mentioned above contributed towards the development of the wave mechanical model of the atom. Their names were matched with their contributions as required by the question.

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A piston is filled with gas. When the pressure is increasing, what is happening to the volume? Assume that all other properties
Kryger [21]
The correct answer is letter C. Volume is decreasing. For a fixed amount of an ideal gas kept at a fixed temperature, pressure and volume<span> are </span>inversely proportional<span>. </span>
3 0
4 years ago
Water has a density of 1g/cm. An object displaces 40ml of water (1ml= 1 cm^3)
vazorg [7]
Mass=density*volume
1*40=40g/0.04kg
3 0
4 years ago
a 20 of of ice at 0c is dropped into water at boiling point, specific heat capacity of water =4200 J/kg•c, sepesific latent heat
Contact [7]

Answer:

15 KJ

Explanation:

The quantity of heat (Q) required is given as:

Q = mcΔθ + mL

where m is the mass of ice, c is its specific heat capacity, L is its specific latent heat andΔθ is the change in temperature.

Given: m = 20g, temperature of ice = 0^{o} C, specific heat capacity of water = 4200 J/kg^{o} C, latent heat of fusion of ice = 3.3 x 10^(5) J/kg, temperature of water = 100^{o} C.

Q = m (cΔθ + L)

   = 0.02(4200 x (100) + 330000)

   = 0.02(420000 + 330000)

  = 0.02 (750000)

Q = 15000

Q = 15000 Joules

Q = 15KJ

The quantity of heat needed to complete the conversion is 15 KJ.

4 0
3 years ago
Two stationary positive point charges, charge 1 of magnitude 3.05 nC and charge 2 of magnitude 1.85 nC, are separated by a dista
luda_lava [24]

To solve this problem we will apply the concepts related to voltage as a dependent expression of the distance of the bodies, the Coulomb constant and the load of the bodies. In turn, we will apply the concepts related to energy conservation for which we can find the speed of this

V = \frac{kq}{r}

Here,

k = Coulomb's constant

q = Charge

r = Distance to the center point between the charge

From each object the potential will be

V_1 = \frac{kq_1}{r_1}+\frac{kq_2}{r_2}

Replacing the values we have that

V_1 =  \frac{(9*10^9)(3.05*10^{-9})}{0.41/2}+\frac{(9*10^9)(1.85*10^{-9})}{0.41/2}

V_1 = 215.12V

Now the potential two is when there is a difference at the distance of 0.1 from the second charge and the first charge is 0.1 from the other charge, then,

V_1 =  \frac{(9*10^9)(3.05*10^{-9})}{0.1}+\frac{(9*10^9)(1.85*10^{-9})}{0.41-0.1}

V_2 = 328.2V

Applying the energy conservation equations we will have that the kinetic energy is equal to the electric energy, that is to say

\frac{1}{2} mv^2 = q(V_2-V_1)

Here

m = mass

v = Velocity

q = Charge

V = Voltage

Rearranging to find the velocity

v = \sqrt{ \frac{2q(V_2-V_1)}{m}}

Replacing,

v = \sqrt{ \frac{-2(1.6*10^{-19})(328.2-215.12)}{9.11*10^{-3}}}

v = 6.3*10^6m/s

Therefore the speed final velocity of the electron when it is 10.0 cm from charge 1 is 6.3*10^6m/s

6 0
3 years ago
A box of spherical jawbreaker candies is 23 g and its volume is 32.3 cm3. If the average mass of a single jawbreaker is 0.94 g,
Alex777 [14]

The radius of each jawbreaker is approximately 0.68 cm.

<h3>Volume of a sphere;</h3>
  • v = 4 /3 πr³

where

r = radius

Therefore,

23 g  = 32.3 cm³

0.94 g  = ?

cross multiply

volume of a single jawbreaker = 32.3 × 0.94 / 23 = 30.362 / 23 = 1.32 cm³

Therefore,

volume of each jawbreaker = 4 /3 πr³

1.32 = 4 / 3 × 3.14 × r³

r³ = 1.32 /4.18666666667

r³ = 0.31533683707

r = ∛0.31533683707

r = 0.680651651 = 0.68

Therefore, the radius of each jawbreaker is approximately 0.68 cm.

learn more on radius here: brainly.com/question/19172427

5 0
3 years ago
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