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3 years ago

Definition of fluoresence

2 answers:

5 0

The visible or invisible radiation emitted by certain substances as a result of incident radiation of a shorter wavelength such as x-rays or ultraviolet light

denis23 [38]3 years ago

4 0

Answer:

the property of absorbing light of short wavelength and emitting light of longer wavelength.

Explanation:

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Explain why understanding bonding is important for understanding molecules.

matrenka [14]

Bond polarity is important because it helps to determine the polarity of molecules and hence their physical properties.

6 0

3 years ago

In the most common isotope of Hydrogen the nucleus is made out of a single proton. When this Hydrogen atom is neutral, a single

FinnZ [79.3K]

Answer:

The ratio of electric force to the gravitational force is $2.27\times 10^{39}$

Explanation:

It is given that,

Distance between electron and proton, $r=4.53\ A=4.53\times 10^{-10}\ m$

Electric force is given by :

$F_e=k\dfrac{q_1q_2}{r^2}$

Gravitational force is given by :

$F_g=G\dfrac{m_1m_2}{r^2}$

Where

$m_1$ is mass of electron, $m_1=9.1\times 10^{-31}\ kg$

$m_2$ is mass of proton, $m_2=1.67\times 10^{-27}\ kg$

$q_1$ is charge on electron, $q_1=-1.6\times 10^{-19}\ kg$

$q_2$ is charge on proton, $q_2=1.6\times 10^{-19}\ kg$

$\dfrac{F_e}{F_g}=\dfrac{kq_1q_2}{Gm_1m_2}$

$\dfrac{F_e}{F_g}=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{6.67\times 10^{-11}\times 9.1\times 10^{-31}\times 1.67\times 10^{-27}}$

$\dfrac{F_e}{F_g}=2.27\times 10^{39}$

So, the ratio of electric force to the gravitational force is $2.27\times 10^{39}$ . Hence, this is the required solution.

3 0

3 years ago

Si tu velocidad al caminar es 0,3 m/s. Determina el tiempo que demorarias en llegar al sol en horas (REALIZAR TRANSFORMACIÓN DE

Contact [7]

Answer:

138,516,546.9 horas.

Explanation:

Tenemos que usar la ecuación:

Velocidad = distancia/tiempo

Acá tenemos:

Velocidad = 0.3m/s

distancia = 149597870700 m

y queremos resolver la ecuación para el tiempo:

0.3m/s = 149597870700m/tiempo.

tiempo = 149597870700m/(0.3m/s) = 498,659,569,000 s

y sabemos que una hora tiene 3600 segundos, entonces si queremos transformar de segundos a horas tenemos:

498,659,569,000 s = (498,659,569,000/3600) h = 138,516,546.9 horas.

6 0

3 years ago

From the gravitational law calculate the weight W (gravitational force with respect to the earth) of a 89-kg man in a spacecraft

zhannawk [14.2K]

Answer:

$W=\frac{773}{4.45}=173.76 l b f$

Explanation:

$W=\frac{G \cdot m_{e} \cdot m}{(R+h)^{2}}$

The law of gravitation

$G=6.673\left(10^{-11}\right) m^{3} /\left(k g \cdot s^{2}\right)$

Universal gravitational constant [S.I. units]

$m_{e}=5.976\left(10^{24}\right) k g$

Mass of Earth [S.I. units]

$m=89 kg$

Mass of a man in a spacecraft [S.I. units]

$R=6371 \mathrm{~km}$

Earth radius [km]

Distance between man and the earth's surface

$h=261 \mathrm{~km} \quad[\mathrm{~km}]$

ESULT $W=\frac{6.673\left(10^{-11}\right) \cdot 5.976\left(10^{24}\right) \cdot 89}{\left(6371 \cdot 10^{3}+261 \cdot 10^{3}\right)^{2}}=773.22 \mathrm{~N}$

$W=\frac{773}{4.45}=173.76 l b f$

4 0

2 years ago

When the mass is removed, the length of the cable is found to be l0 = 4.66 m. After the mass is added, the length is remeasured

zaharov [31]

Answer:

$\gamma=6.07*10^5\frac{N}{m^2}$

Explanation:

For a linear elastic material Young's modulus is a constant that is given by:

$\gamma=\frac{F/A}{\Delta L/L_0}$

Here, F is the force exerted on an object under tensio, A is the area of the cross-section perpendicular to the applied force, $\Delta L$ is the amount by which the length of the object changes and $L_0$ is the original length of the object. In this case the force is the weight of the mass:

$F=mg\\F=55kg(9.8\frac{m}{s^2})\\F=539N$

Replacing the given values in Young's modulus formula:

$\gamma=\frac{F/\pi r^2}{(L_1-L_0)/L_0}\\\gamma=\frac{539N/\pi(0.045m)^2}{(5.31m-4.66m)/4.66m}\\\gamma=6.07*10^5\frac{N}{m^2}$

6 0

3 years ago