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lana66690 [7]
3 years ago
10

Part D The surface area of a material is a factor that affects heat conductivity. Does heat flow faster through a large surface

area or through a small surface area? Why? (Hint: Think about the number of particles available to bump into one another)​
Physics
2 answers:
siniylev [52]3 years ago
3 0
Large surface area, as more particles are able to bump into one another and transfer heat.
kipiarov [429]3 years ago
3 0

Answer:

That guy is right i got a 100 as my score :)

Explanation:

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Use the image to determine the volume of the rock.
maria [59]

Answer:

12 ml

Explanation:

The initial volume in the cylinder is 20 ml

 adding the rock adds volume to the cylinder

       the new volume is 32 ml .....the increase in volume is the volume of the rock :   32 - 20 = 12 ml    volume of rock

6 0
2 years ago
During energy transformation, all energy systems: are
Tom [10]

B: Energy lose

i say this because in order to change they lose energy.



3 0
3 years ago
Read 2 more answers
What does Newton's third law describe?
Vesnalui [34]

Answer:

Newton's third law is: For every action, there is an equal and opposite reaction.

Explanation:

The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object.

3 0
3 years ago
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1. Which statement about subatomic particles is not true?
igomit [66]

1. Protons and neutrons have the same charge.

Protons have positive charge, equal to e=+1.6\cdot 10^{-19} C, while neutrons have zero charge.

2. mass number

The mass number of an atom is equal to the sum of protons and neutrons inside its nucleus.

3. Atoms are made up of smaller particles.

According to Dalton's theory, atoms are the smallest particles that make matter, and they are indivisible and indestructible, so they are NOT made up of smaller particles.

4. a solid sphere

In Dalton's theory, atoms are not made of smaller particles, so we can think them as solid spheres.

5. J. J. Thomson

In his experiment with cathode ray tubes, JJ Thomson demonstrated the existance of the electrons, which are negatively charged particles inside the atom. In his model of the atom (plum-pudding model), Thomson thought the atom consists of a uniform positive charge and the electrons are located inside this positive charge.

6. An electron has the same amount of energy in all orbitals.

In fact, each orbital corresponds to a different energy level: the farther the orbital from the nucleus, the higher the energy of the electrons contained in that orbital.

7. A hydrogen atom in heavy water has an extra neutron.

Heavy water is a type of water that contains deuterium, which is an isotope of the hydrogen consisting of one proton and one neutron (so, one extra neutron).

8. The glowing beam was always deflected by charged plates

In his cathode's ray tube experiment, Thomson shows that the beam of unknown particles (= the electrons) were deflected by charge plates, so the particles had to be also electrically charged.

9. electrons move to a lower energy level

When electrons move from a higher energy level to a lower energy, they emit a photon (light) of energy equal to the difference in energy between the two energy levels.

10. orbital

In quantum mechanics, electrons in the atom are not precisely located, since we cannot determine their exact position and velocity at the same time. Therefore, we can only describe regions of space where the electrons have a certain probability to be found, and these regions of space are called orbitals.

11. 14

According to Dalton's theory, the proportions of the reactants must be respected in order to form the same compound. Therefore, we can write:

2 g: 4 g = X : 28 g\\X=\frac{2 g \cdot 28 g}{4 g}=14 g

12. negative charge, found outside the nucleus

Electrons are particles with negative charge of magnitude e=-1.6\cdot 10^{-19}C that orbit around the nucleus. The nucleus, instead, consists of protons (positively charged, with charge opposite to the electron) and neutrons (neutrally charged).

13. move from higher to lower energy levels

When electrons move from a higher energy level to a lower energy inside a neon atom, they emit a photon (which is light) whose energy is equal to the difference in energy between the two energy levels.

14. atomic number from its mass number

In fact:

- the atomic number of an atom (Z) is equal to the number of protons inside the nucleus

- the mass number of an atom (A) is equal to the sum of protons+neutrons inside the nucleus

Therefore, we can find the number of neutrons in the nucleus by calculating the difference between A and Z:

Number of neutrons = A - Z

15. None of them

None of these examples is a good analogy to describe the location of an electron in an atomic orbital: in fact, the position of an electron in an orbital cannot be precisely described, we can only describe the probability to find the electron in a certain position, and none of these example is an analogy of this model.

8 0
3 years ago
At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m². A fan is 30 m from the loudspeak
Klio2033 [76]

To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.

The expression for sound power is,

P = AI

Here,

A = Area

I = Intensity

P = Power

At the same time the area can be written as,

A = \frac{\pi d^2}{4}

Now the intensity is inversely proportional to the square of the distance from the source, then

I \propto \frac{1}{r^2}

The expression for the intensity at different distance is

\frac{I_1}{I_2}= \frac{r^2_2}{r_1^2}

Here,

I_1 = Intensity at distance 1

I_2 = Intensity at distance 2

r_1 = Distance 1 from light source

r_2 = Distance 2 from the light source

If we rearrange the expression to find the intensity at second position we have,

I_2 = I_1 (\frac{r_1^2}{r_2^2})

If we replace with our values at this equation we have,

I_2 = (0.10W/m^2)(\frac{1.0m^2}{30.0m^2})

I_2 = 1.11*10^{-4} W/m^2

Now using the equation to find the area we have that

A = \frac{\pi (8.4*10^{-3}m)^2}{4}

A = 5.5*10^{-5}m^2

Finally with the intensity and the area we can find the sound power, which is

P = AI

P = (5.5*10^{-5}m^2)(1.11*10^{-4}W/m^2)

P = 6.1*10^{-9}J/s

Power is defined as the quantity of Energy per second, then

E = 6.1*10^{-9}J

8 0
4 years ago
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