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allochka39001 [22]
1 year ago
6

Which of these discoveries contradict components of Dalton’s atomic theory? Check all of the boxes that apply. Atoms contain sma

ller particles: protons, neutrons, and electrons. Atoms of a given element can bond to other atoms only in specific ways. Atoms of a given element can have different numbers of neutrons. Nuclear reactions can change an atom of one element into an atom of another element. All atoms of a given element have the same number of protons.
Physics
2 answers:
Varvara68 [4.7K]1 year ago
7 0

Atoms of a given element can have different numbers of neutrons.

Nuclear reactions can change an atom of one element into an atom of another element.

Atoms contain smaller particles: protons, neutrons, and electrons.

EastWind [94]1 year ago
4 0

Explanation:

According to Dalton's atomic theory, all the atoms are individual, all the atoms of the same element are identical in properties and mass, the compound is formed from two or more kinds of the atoms, all the matter is made up of small atoms and the chemical reaction is a rearrangement of the atoms.

The discoveries which contradicts the components of Dalton's atomic theory from the given discoveries are:

Nuclear reactions can change an atom of one element into an atom of another element.

Atoms of a given element can have different numbers of neutrons.

Atoms contain smaller particles: protons, neutrons, and electrons.

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laiz [17]
L = 1.00 m, the original length
A = 0.5 mm² = 0.5 x 10⁻⁶ m², the cross sectional area
E = 2.0 x 10¹¹ n/m², Young's modulus
P = 1500 N, the applied tension

Calculate the stress.
σ = P/A = (1500 N)/(0.5 x 10⁻⁶ m²) = 3 x 10⁹ N/m²

Let δ =  the stretch of the string.
Then the strain is
ε = δ/L

By definition, the strain is
ε = σ/E = (3 x 10⁹ N/m²)/(2 x 10¹¹ N/m²) = 0.015
Therefore
δ/(1 m) = 0.015
δ = 0.015 m = 15 mm

Answer:  15 mm
4 0
2 years ago
Two wires are stretched between two fixed supports and have the same length. One wire A there is a second-harmonic standing wave
lina2011 [118]

(a) Greater

The frequency of the nth-harmonic on a string is an integer multiple of the fundamental frequency, f_1:

f_n = n f_1

So we have:

- On wire A, the second-harmonic has frequency of f_2 = 660 Hz, so the fundamental frequency is:

f_1 = \frac{f_2}{2}=\frac{660 Hz}{2}=330 Hz

- On wire B, the third-harmonic has frequency of f_3 = 660 Hz, so the fundamental frequency is

f_1 = \frac{f_3}{3}=\frac{660 Hz}{3}=220 Hz

So, the fundamental frequency of wire A is greater than the fundamental frequency of wire B.

(b) f_1 = \frac{v}{2L}

For standing waves on a string, the fundamental frequency is given by the formula:

f_1 = \frac{v}{2L}

where

v is the speed at which the waves travel back and forth on the wire

L is the length of the string

(c) Greater speed on wire A

We can solve the formula of the fundamental frequency for v, the speed of the wave:

v=2Lf_1

We know that the two wires have same length L. For wire A, f_1 = 330 Hz, while for wave B, f_B = 220 Hz, so we can write the ratio between the speeds of the waves in the two wires:

\frac{v_A}{v_B}=\frac{2L(330 Hz)}{2L(220 Hz)}=\frac{3}{2}

So, the waves travel faster on wire A.

7 0
1 year ago
How much energy is needed to change the speed of a 1600 kg sport utility vehicle from 15.0 m/s to 40.0 m/s?
podryga [215]

Answer:

Energy needed = 1100 kJ

Explanation:

Energy needed = Change in kinetic energy

Initial velocity = 15 m/s

Mass, m = 1600 kg

\texttt{Initial kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 15^2=180000J

Final velocity = 40 m/s

\texttt{Final kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 40^2=1280000J

Energy needed = Change in kinetic energy = 1280000-180000 = 1100000J

Energy needed = 1100 kJ

4 0
1 year ago
Alan wrote the following examples of changes in substance.
Yuliya22 [10]
A campfire being lighted and plants converting carbon-dioxide and water into glucose and oxygen are both forms of chemical change.

Therefore, the answer is:

B. Both are examples of chemical change.
5 0
2 years ago
Read 2 more answers
A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c
netineya [11]

Answer:

E=\frac{\lambda}{2\pi r\epsilon_0}

Explanation:

We are given that

Linear charge density of wire=\lambda

Radius of hollow cylinder=R

Net linear charge density of cylinder=2\lambda

We have to find the expression for the magnitude of the electric field strength inside the cylinder r<R

By Gauss theorem

\oint E.dS=\frac{q}{\epsilon_0}

q=\lambda L

E(2\pi rL)=\frac{L\lambda}{\epsilon_0}

Where surface area of cylinder=2\pi rL

E=\frac{\lambda}{2\pi r\epsilon_0}

8 0
1 year ago
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