Which of these forces help protons and neutrons to stay at the center of the Atom

Which best describes a characteristic of an adiabatic process?

neonofarm [45]

Answer:

<h3>Can you please provide the choices? For now this is all I can give you.</h3>

Explanation:

In thermodynamics, an adiabatic process is a type of thermodynamic process which occurs without transferring heat or mass between the system and its surroundings.Unlike an isothermal process, an adiabatic process transfers energy to the surroundings only as work. It also conceptually supports the theory used to explain the first law of thermodynamics and is therefore a key thermodynamic concept.

5 0

3 years ago

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Assume a change at the source of sound reduces the wavelength of a sound wave in air by a factor of 3.

noname [10]

Explanation:

The speed of a wave is given by :

$v=f\lambda$ ......(1)

(i) Here, the wavelength of a sound wave in air reduces by a factor of 3. Equation (1) becomes :

$\lambda=\dfrac{v}{f}$

Wavelength and frequency are inversely proportional to each other. So, if wavelength of a sound wave in air reduces by a factor of 3, then the frequency will increases by a factor of 3.

(ii) It remains the same.

8 0

3 years ago

Which of the following is NOT a benefit of cool down activities?

quester [9]

B. Maximal lightheadedness

5 0

3 years ago

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A football player kicks a ball horizontally off a hill with an initial velocity of 42.0 m/s. It travels a horizontal distance of

damaskus [11]

Explanation:

.........................

7 0

3 years ago

A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm

andrew11 [14]

Answer:

2.8 cm

Explanation:

$y_1$ = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

$\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm$

Fringe width is also given by

$\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}$

For second order

$\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1$

Distance between two second order minima is given by

$y_2=2\beta_2$

$\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm$

The distance between the two second order minima is 2.8 cm

8 0

3 years ago