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2 years ago

Rutherford's atomic theory included which idea?

Physics

1 answer:

Murljashka [212]2 years ago

4 0

Your answer would be C Atoms are composed of a positively charged sphere that contains electrons.

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What do you understand by waves. what is the equation?

kipiarov [429]

A wave is a perturbation in a material from the point the perturbation was produced, to the sorrounding area

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2 years ago

Frequency is the ____________ that move past a point during a specific amount of time. Frequency is measured in ____________ , a

yulyashka [42]

Answer:

Frequency is the number of waves that move past a point during a specific amount of time. Frequency is measured in Hertz, and is classified as high, medium, or low. Frequency is interpreted as the pitch of a sound. Intensity refers to the loudness of a sound and is measured in decibels. Louder sounds increase the rate of nerve signals relayed to the brain.

Explanation:

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2 years ago

A 30.0 g mass of iron at 24.5°C is heated to 45.0°C. The theoretical specific

diamong [38]

Answer:

276.135 J

Explanation:

Given that:

mass of Fe = 30.0 g

initial temperature = 24.5°C

final temperature = 45.0°C

specific heat of Fe = 0.449 J/g°C

We can determine the thermal energy added by using the formula;

Q = mcΔT

Q = 30.0g × 0.449 J/g°C × (45.0 - 24.5)°C

Q = 276.135 J

8 0

2 years ago

An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com

solong [7]

Answer:

(a) $3.81\times 10^5\ Pa$

(b) $4.19\times 1065\ Pa$

Explanation:

Given:

$T_1$ = The first temperature of air inside the tire = $10^\circ C =(273+10)\ K =283\ K$

$T_2$ = The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$

$T_3$ = The third temperature of air inside the tire = $85^\circ C =(273+85)\ K=358 \ K$

$V_1$ = The first volume of air inside the tire

$V_2$ = The second volume of air inside the tire = $30\% V_1 = 0.3V_1$

$V_3$ = The third volume of air inside the tire = $2\%V_2+V_2= 102\%V_2=1.02V_2$

$P_1$ = The first pressure of air inside the tire = $1.01325\times 10^5\ Pa$

Assume:

$P_2$ = The second pressure of air inside the tire

$P_3$ = The third pressure of air inside the tire
n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

Part (a):

Using the above equation for this part of compression in the air, we have

$\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa$

Hence, the pressure in the tire after the compression is $3.81\times 10^5\ Pa$ .

Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

Hence, the pressure in the tire after the car i driven at high speed is $4.19\times 10^5\ Pa$ .

8 0

2 years ago

The driving force behind an efficient supply chain is __________.

Doss [256]

Customer satisfaction is considered to be the "driving force" in order to achieve an efficient supply chain. An efficient supply chain takes place when the organization, or the company itself, meets with the demands of the consumers to improve and provides services that satisfies the people.

5 0

2 years ago