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2 years ago

Importance of international bureau of weights and measures Inthe world

Physics

1 answer:

vaieri [72.5K]2 years ago

4 0

Answer:

The international bureau of weights and measures is an international organization to bring about the unification of different measurement systems to make or establish international standards and prototypes, etc.

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When a board with a box on it is slowly tilted to larger and larger angle, common experience shows that the box will at some poi

eimsori [14]

Answer: The force of kinetic friction is smaller than that of static friction, but F_g remains the same.

Explanation:

The situation is same as when a book is pushed with an increasing force on a table; When the force is low, book doesn't move, until that under a given force starts moving, and then it goes on movement even if the force decreases a bit.

The physical explanation for this, that friction force adopts any value needed to avoid to move the object, till a limit value is achieved, called static friction force, equal to the normal force times the static friction coefficient.

Once in movement, the kinetic friction coefficient replaces the static one , and in general is lower than the static one, so the force diminishes.

In the case of the box sliding down the board, the force that tries to move the object down the board, is the component of the weight parallel to the board, that can be showed that being equal to the weight times the sinus of the angle of the board with the horizontal, as follows:

F_g = m g sin θ

When θ increases, F_g does the same, so friction force always has the same magnitude than F_g (but opposite direction) so the box doesn't move, till that θ takes a value that produces a F_g equal to static friction force.

Beyond this limit angle, F_g (remaining the same for a given angle) is greater than the kinetic friction force, and the box slides.

In the limit, when θ=90º, sin θ =1⇒ F_g = mg, so the object is in free fall.

6 0

3 years ago

If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string

vovikov84 [41]

Complete Question

The speed of a transverse wave on a string of length L and mass m under T is given by the formula

$v=\sqrt{\frac{T}{(m/l)}}$

If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string

Answer:

$(m/l)=\frac{10}{V^2}$

Explanation:

From the question we are told that

Speed of a transverse wave given by

$v=\sqrt{\frac{T}{(m/l)}}$

Maximum Tension is $T=10.0N$

Generally making $(m/l)$ subject from the equation mathematically we have

$v=\sqrt{\frac{T}{(m/l)}}$

$v^2=\frac{T}{(m/l)}$

$(m/l)=\frac{T}{V^2}$

$(m/l)=\frac{10}{V^2}$

Therefore the Linear mass in terms of Velocity is given by

$(m/l)=\frac{10}{V^2}$

8 0

2 years ago

What are the characteristics of the radiation emitted by a blackbody? According to Wien's Law, how many times hotter is an objec

jasenka [17]

Answer:

a) What are the characteristics of the radiation emitted by a blackbody?

The total emitted energy per unit of time and per unit of area depends in its temperature (Stefan-Boltzmann law).

The peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase (Wien’s displacement law).

The spectral density energy is related with the temperature and the wavelength (Planck’s law).

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave length of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

Explanation:

A blackbody is an ideal body that absorbs all the thermal radiation that hits its surface, thus becoming an excellent emitter, as these bodies express themselves without light radiation, and therefore they look black.

The radiation of a blackbody depends only on its temperature, thus being independent of its shape, material and internal constitution.

If it is study the behavior of the total energy emitted from a blackbody at different temperatures, it can be seen how as the temperature increases the energy will also increase, this energy emitted by the blackbody is known as spectral radiance and the result of the behavior described previously is Stefan's law:

$E = \sigma T^{4}$ (1)

Where $\sigma$ is the Stefan-Boltzmann constant and T is the temperature.

The Wien’s displacement law establish how the peak of emission of the spectrum will be displace to shorter wavelengths as the temperature increase (inversely proportional):

$\lambda max = \frac{2.898x10^{-3} m. K}{T}$ (2)

Planck’s law relate the temperature with the spectral energy density (shape) of the spectrum:

$E_{\lambda} = {{8 \pi h c}\over{{\lambda}^5}{(e^{({hc}/{\lambda \kappa T})}-1)}}}$ (3)

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wavelength of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

It is need it to known the temperature of both objects before doing the comparison. That can be done by means of the Wien’s displacement law.

Equation (2) can be rewrite in terms of T:

$T = \frac{2.898x10^{-3} m. K}{\lambda max}$ (4)

Case for the object with the blackbody emission spectrum peak in the blue:

Before replacing all the values in equation (4), $\lambda max$ (450 nm) will be express in meters:

$450 nm . \frac{1m}{1x10^{9} nm}$ ⇒ $4.5x10^{-7}m$

$T = \frac{2.898x10^{-3} m. K}{4.5x10^{-7}m}$

$T = 6440 K$

Case for the object with the blackbody emission spectrum peak in the red:

Following the same approach above:

$700 nm . \frac{1m}{1x10^{9} nm}$ ⇒ $7x10^{-7}m$

$T = \frac{2.898x10^{-3} m. K}{7x10^{-7}m}$

$T = 4140 K$

Comparison:

$\frac{6440 K}{4140 K} = 1.55$

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

4 0

3 years ago

An interference pattern is produced by light with a wavelength 580 nm from a distant source incident on two identical parallel s

irakobra [83]

Answer:

Explanation:

1 )

Here

wave length used that is λ = 580 nm

=580 x 10⁻⁹

distance between slit d = .46 mm

= .46 x 10⁻³

Angular position of first order interference maxima

= λ / d radian

= 580 x 10⁻⁹ / .46 x 10⁻³

= 0.126 x 10⁻² radian

2 )

Angular position of second order interference maxima

2 x 0.126 x 10⁻² radian

= 0.252 x 10⁻² radian

3 )

For intensity distribution the formula is

I = I₀ cos²δ/2 ( δ is phase difference of two lights.

For angular position of θ1

δ = .126 x 10⁻² radian

I = I₀ cos².126x 10⁻²/2

= I₀ X .998

For angular position of θ2

I = I₀ cos².126x2x 10⁻²/2

= I₀ cos².126x 10⁻²

8 0

2 years ago

Consider the four quantum numbers of an electron in an atom, n, l, ml, and ms. The energy of an electron in an isolated atom dep

Effectus [21]

Answer:

The energy of an electron in an isolated atom depends on b. n only.

Explanation:

The quantum number n, known as the principal quantum number represents the relative overall energy of each orbital.

The sets of orbitals with the same n value are often referred to as an electron shell, in an isolated atom all electrons in a subshell have exactly the same level of energy.

The principal quantum number comes from the solution of the Schrödinger wave equation, which describes energy in eigenstates $E_n$ , and for the case of an hydrogen atom we have:

$E_n=-\cfrac{13.6}{n^2}\, eV$

Thus for each value of n we can describe the orbital and the energy corresponding to each electron on such orbital.

6 0

2 years ago

Importance of international bureau of weights and measures Inthe world​

Importance of international bureau of weights and measures Inthe world