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Nutka1998 [239]

3 years ago

9

When the tube is filled with mercury vapor, as in this case, a sharp drop in the collected current is observed when the accelera

ting potential reaches 4.9 v. the current drops suddenly to a negligible value and slowly increases again when the accelerating potential is above 4.9 v. what is the energy e absorbed by the atomic electrons in the mercury atom when the accelerating potential is 4.9 v?

1 answer:

Nata [24]3 years ago

5 0

Answer:

The energy absorbed by the atomic electrons in the mercury atom is $7.84 \times 10^{-19}$ J

Explanation:

Given:

Potential $V = 4.9$ V

According to the conservation law,

Loss in kinetic energy = Gain in potential energy

Here, energy absorbed by the atomic electrons is given by,

$E = eV$

Where $e = 1.6 \times 10^{-19} C$ ( charge of electron )

$E = 1.6 \times 10^{-19 } \times 4.9$

$E = 7.84 \times 10^{-19}$ J

Therefore, the energy absorbed by the atomic electrons in the mercury atom is $7.84 \times 10^{-19}$ J

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alukav5142 [94]

A) d. 10T

When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by

$F=qvB$

where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.

This force acts as a centripetal force, keeping the particle in a circular motion - so we can write

$qvB = \frac{mv^2}{r}$

which can be rewritten as

$v=\frac{qB}{mr}$

The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):

$\frac{2\pi r}{T}=\frac{qB}{mr}$

So, we get:

$T=\frac{2\pi m r^2}{qB}$

We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.

B) $a. f/10$

The frequency of revolution of a particle in uniform circular motion is

$f=\frac{1}{T}$

where

f is the frequency

T is the period

We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:

T' = 10 T

Then its frequency of revolution will be:

$f'=\frac{1}{T'}=\frac{1}{(10T)}=\frac{f}{10}$

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3 years ago

Neglecting friction, of a pendulum bob has 100 joules of kinetic energy at the bottom of its swing, how much potential energy do

galben [10]

100 joule of kinetic energy

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3 years ago

What is the frequency of a monochromatic light used in a diffraction experiment that has a wavelength of 6.38 ✕ 10e-07 m?

Lady_Fox [76]

Answer:

$f=4.70\times 10^{14}\ Hz$

Explanation:

Given that,

The wavelength of light, $\lambda=6.38\times 10^{-7}\ m$

We need to find the frequency of the light. We know that,

$c=f\lambda\\\\f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{6.38\times 10^{-7}}\\\\f=4.70\times 10^{14}\ Hz$

So, the required frequency of light is equal to $4.70\times 10^{14}\ Hz$ .

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3 years ago

The displacement (in meters) of a particle moving in a straight line is given by the equation of motion:

lutik1710 [3]

Answer:

At $t = 1\; \rm s$ , the particle should have a velocity of $-8\; \rm m \cdot s^{-1}$ .
At $t = 2\; \rm s$ , the particle should have a velocity of $-1\; \rm m \cdot s^{-1}$ .
At $t = 3\; \rm s$ , the particle should have a velocity of $\displaystyle -\frac{8}{27}\; \rm m \cdot s^{-1}$ .

For $a > 0$ , at $t = a \; \text{second}$ , the particle should have a velocity of $\displaystyle -\frac{8}{a^3}\; \rm m \cdot s^{-1}$ .

Explanation:

Differentiate the displacement of an object (with respect to time) to find the object's velocity.

Note that the in this question, the expression for displacement is undefined (and not differentiable) when $t$ is equal to zero. For $t > 0$ :

$\begin{aligned}v &= \frac{\rm d}{{\rm d}t}\, [s] = \frac{\rm d}{{\rm d}t}\, \left[\frac{4}{t^2}\right] \\ &= \frac{\rm d}{{\rm d}t}\, \left[4\, t^{-2}\right] = 4\, \left((-2)\, t^{-3}\right) = -8\, t^{-3} =-\frac{8}{t^3}\end{aligned}$ .

This expression can then be evaluated at $t = 1$ , $t = 2$ , and $t = 3$ to obtain the required results.

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3 years ago

Jerome plays middle linebacker for souths varsity football team. In a game against cross-town rival north, he delivered a hit to

Makovka662 [10]

The momentum change of the running back is - 664.2 kg m/s or 664 west.

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Momentum is defined as the change in velocity of any object along with its mass. So mathematically, momentum can be derived using the product of mass with the change in velocity.

$\text { Momentum change }=m \times \Delta v$

As here mass is given as 82 kg and the initial velocity was 5.6 m/s and final velocity is 2.5 m/s.

Initial Momentum = $m \times \text { initial velocity }=82 \times 5.6=459.2 \mathrm{kg} \mathrm{m} / \mathrm{s}$

Final Momentum = $m \times \text { final velocity }=82 \times(-2.5)=-205 \mathrm{kg} \mathrm{m} / \mathrm{s}$

Momentum change = Final Momentum - Initial Momentum

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3 years ago