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3 years ago

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Question 2 Saved

1 answer:

Goshia [24]3 years ago

3 0

Protons and/or neutrons

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g The potential energy of a pair of hydrogen atoms separated by a large distance x is given by U(x)=−C6/x6, where C6 is a positi

Arisa [49]

Answer:

$F_x = -\frac{6 C_6}{2^7}$

Attractive

Explanation:

Data provided in the question

The potential energy of a pair of hydrogen atoms given by $\frac{C_6}{X_6}$

Based on the given information, the force that one atom exerts on the other is

Potential energy μ = $\frac{C_6}{X_6}$

Force exerted by one atom upon another

$F_x = \frac{\partial U}{\partial X} = \frac{\partial}{\partial X} (-\frac{C_6}{X^6})$

or

$F_x = \frac{\partial}{\partial X} (\frac{C_6}{X^6})$

or

$F_x = -\frac{6 C_6}{2^7}$

As we can see that the $C_6$ comes in positive and constant which represents that the force is negative that means the force is attractive in nature

5 0

3 years ago

Light is incident along the normal to face AB of a glass prism of refractive index 1.54. Find αmax, the largest value the angle

marusya05 [52]

To solve this problem it is necessary to use the concepts related to Snell's law.

Snell's law establishes that reflection is subject to

$n_1sin\theta_1 = n_2sin\theta_2$

Where,

$\theta =$ Angle between the normal surface at the point of contact

n = Indices of refraction for corresponding media

The total internal reflection would then be given by

$n_1 sin\theta_1 = n_2sin\theta_2$

$(1.54) sin\theta_1 = (1.33)sin(90)$

$sin\theta_1 = \frac{1.33}{1.54}$

$\theta = sin^{-1}(\frac{1.33}{1.54})$

$\theta = 59.72\°$

Therefore the $\alpha_{max}$ would be equal to

$\alpha = 90\°-\theta$

$\alpha = 90-59.72$

$\alpha = 30.27\°$

Therefore the largest value of the angle α is 30.27°

3 0

3 years ago

An 80-cm-long steel string with a linear density of 1.0 g/m is under 200 N tension. It is plucked and vibrates at its fundamenta

icang [17]

Answer:

Wavelength of the sound wave that reaches your ear is 1.15 m

Explanation:

The speed of the wave in string is

$v=\sqrt{\frac{T}{\mu} }$

where T= 200 N is tension in the string , $\mu$ =1.0 g/m is the linear mass density

$v=\sqrt{\frac{200}{1\times 10^{-3} }$

$v=447.2 m/s$

Wavelength of the wave in the string is

$\lambda =2L=2\times 0.8=1.6 m$

The frequency is

$f=\frac{v}{\lambda} \\f=\frac{447.2}{1.6}\\f=298.25 Hz$

The required wavelength pf the sound wave that reaches the ear is( take velocity of air v=344 m/s)

$\lambda=\frac{v_{air}}{f} \\\lambda=\frac{344}{298.25} \\\lambda=1.15 m$

8 0

3 years ago

Costal residents must do many things to prepare for hurricanes

tangare [24]

The answer is A it’s more safer that way

3 0

3 years ago

The most common form of angina is _______ angina. A. microvascular B. variant C. stable D. unstable

Paraphin [41]

The most common form of angina is stable angina.

C. stable

8 0

2 years ago

Read 2 more answers