How did Rutherford change the model of the atom? A.
He used an experiment with gold foil to prove that an atom had a positive nucleus in the middle and was surrounded by negative electrons.

B.
He used thought problems to determine that matter could be divided into smaller pieces until it got to the atomic level.

C.
He used an experiment with Cathode Ray tubes to prove electrons existed when they changed the color of a gas.

D.
He used various experiments to prove that atoms of the same element have the same mass.

Answers

Answer 1
Answer: A.) He used an experiment with gold foil to prove that an atom had apositive nucleus in the middle and was surrounded by negative electrons.

Hope this helps!

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Strong solar winds blew dust and gas out of the solar system during which phase of the development of the Sun?

Answers

The choices can be found elsewhere and as follows:

a. Alpha Centauri 
c. T-tauri 
b. The Big Bang 
d. Nebular

I believe the correct answer from the choices listed above is option D. Strong solar winds blew dust and gas out of the solar system during Nebular phase. This seems to be the most logical option from the choices. Hope this helps. Have a nice day.

its T-tauri on Edge nuity

how should the charge q be divided into two parts such that they experience maximum electrostatic repulsion

Answers

Answer:

To divide the charge q into two parts such that they experience maximum electrostatic repulsion, the charges should be equal in magnitude and opposite in sign. In other words, the charge q should be divided into two equal charges of -q/2 and +q/2. This arrangement will result in the maximum electrostatic repulsion between the charges, as like charges repel each other.

Explanation:

The sun is the ultimate source for energy on the Earth.
a. True
b. False

Answers

So we want to know is it true or false that the Sun is the ultimate source of energy on Earth. The answer is true because fossil fuels are solar energy trapped in the Earth's surface. Water cycle is dependent on solar radiation and we use water to produce electricity. We use solar panels to produce electrical energy. And also, all plant life harvests solar energy.

This statement is TRUE.

I need some help, anything will help, I just need a better understanding of this

Answers

Answer:

12 meters

Explanation:

to find this we look at the 8 on the horizontal line this is 8 seconds, you then look upwards to see the height at that point in time

Alpha particles \ beta particles / gamma radiation Put these different types of radiation in order from MOST to LEAST penetrating.

Answers

It goes: Gamma radiation > beta particles > alpha particles

Answer:

Gamma > Beta > Alpha

Explanation:

A 25.0-meter length of platinum wire with a cross-sectional area of 3.50 × 10^−6 meter^2 has a resistanceof 0.757 ohm at 20°C. Calculate the resistivity of the wire. [Show all work, including the equation and
substitution with units.]

Answers

The resistivity of the plantinum wire is 1.06 × 10⁻⁷Ωm

Calculating Resistivity

From the question,

We are to determine the resistivity of the wire

Resistivity can be calculated from the formula

\rho = (RA)/(l)

Where \rho is the resistivity of the wire

R is the resistance

A is the cross-sectional area

and l is the length of the wire

From the given information,

R = 0.757 ohm (Ω)

A = 3.50 × 10⁻⁶ m²

l = 25.0 m

Putting the parameters into the formula, we get

\rho = (0.757 * 3.50 * 10^(-6) )/(25)

\rho = 0.10598 * 10^(-6)

\rho = 1.0598 * 10^(-7)

\rho \approx 1.06 * 10^(-7) \ \Omega m

Hence, the resistivity of the plantinum wire is 1.06 × 10⁻⁷Ωm

Learn more on Calculating Resistivity here: brainly.com/question/5595373

R= (rou * L) / area
where R is the wire resistance
rou: resistivity of the wire material
L : wire length
A : cross section area of wire
by sub.
0.757= (rou*25)/ 3.5*10^-6
25*rou = 2.6495*10^-6
rou= 1.0598*10^-7 ohm.m