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3 years ago

What determines how many electrons will leave a piece of metal in the photoelectric effect

2 answers:

5 0

The electrons in a piece of metal will leave when the polarized magnesis effect is reppeled and heated up to a certain tempeture.

GalinKa [24]3 years ago

4 0

Answer:

Work function is the minimum energy that is required by an electron to leave the metal target in the photoelectric effect

Explanation:

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HELP ASAP PLEASE!!!

Montano1993 [528]

Answer:

The answer is A, B, C, D

Explanation:

This is because gravity is the weakest force of the four fundamental forces, so it automatically cancels letter E

7 0

2 years ago

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At which location would you expect the LOWEST TEMPERATURE?

MissTica

Answer:

Explanation:

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2 years ago

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Two small nonconducting spheres have a total charge of 90.0 C.

valentina_108 [34]

Answer: (a) Smaller charge is $2.7 \times 10^{-5} C$ and larger charge is $11.7 \times 10^{-5} C$ .

(b) Smaller charge is $-11.4 \times 10^{-5}$ and larger charge is $9.1 \times 10^{-5}$ .

Explanation:

(a) When both the spheres have same charge then force is repulsive in nature as like charges tend to repel each other.

Therefore, total charge on the two non-conducting spheres will be calculated as follows.

$Q_{1} + Q_{2} = 90 \mu \times \frac{10^{-6}C}{1 \muC}$

= $9 \times 10^{-5} C$

Therefore, force between the two spheres will be calculated as follows.

F = $k\frac{Q_{1}Q_{2}}{r^{2}}$

12 N = $\frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}$

$Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}$

or, $Q_{1}(9 \times 10^{-5} - Q_{1}) = 0.104 \times 10^{-9} C^{2}$

$9 \times 10^{-5}Q_{1} - Q^{2}_{1} = 0.104 \times 10^{-9} C^{2}$

$Q^{2}_{1} - 9 \times 10^{-5}Q_{1} + 0.104 \times 10^{-9} = 0$

$Q_{1} = 11.7 \times 10^{-5} C, 2.7 \times 10^{-5} C$

This means that smaller charge is $2.7 \times 10^{-5} C$ and larger charge is $11.7 \times 10^{-5} C$ .

(b) When force is attractive in nature then it means both the charges are of opposite sign.

Hence, total charge on the non-conducting sphere is as follows.

$Q_{1} + (-Q_{2}) = 90 \mu \times \frac{10^{-6}C}{1 \muC}$

$Q_{1} - Q_{2} = 9 \times 10^{-5} C$

Now, force between the two spheres is calculated as follows.

F = $k\frac{Q_{1}Q_{2}}{r^{2}}$

12 N = $\frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}$

$Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}$

$Q_{1}(Q_{1} - 9 \times 10^{-5}) = 0.104 \times 10^{-9} C^{2}$

$Q^{2}_{1} - 9 \times 10^{-5}Q_{1} = 0.104 \times 10^{-9} C^{2}$

$Q_{1} = -11.4 \times 10^{-5}, 9.1 \times 10^{-5}$

Hence, smaller charge is $-11.4 \times 10^{-5}$ and larger charge is $9.1 \times 10^{-5}$ .

8 0

3 years ago

Unpolarizedlight of intensity I_0 is incident on three polarizingfilters. The axis of the first is vertical, that of the secondi

Marina86 [1]

To solve this problem it is necessary to apply the concepts related to the law of Malus which describe the intensity of light passing through a polarizer. Mathematically this law can be described as:

$I = I_0 cos^2\theta$

Where,

$I_0 =$ Indicates the intensity of the light before passing through the polarizer

I = Resulting intensity

$\theta$ = Indicates the angle between the axis of the analyzer and the polarization axis of the incident light

From the law of Malus when the light passes at a vertical angle through the first polarizer its intensity is reduced by half therefore

$I_1= \frac{I_0}{2}$

In the case of the second polarizer the angle is directly 60 degrees therefore

$I_2 = I_1 cos^2\theta$

$I_2 = (\frac{I_0}{2} ) cos^2(60)$

$I_2 = 0.125I_0$

In the case of the third polarizer, the angle is reflected on the perpendicular, therefore, its angle of index would be

$\theta_3 = 90-60 = 30$

Then,

$I_3 = I_2 cos^2\theta_3$

$I_3 = 0.125I_0 cos^2 (30)$

$I_3 = 0.09375I_0$

Then the intensity at the end of the polarized lenses will be equivalent to 0.09375 of the initial intensity.

5 0

3 years ago

A pressure cooker contains water and steam in equilibrium at a pressure greater than atmo-spheric pressure. How does this greate

Amanda [17]

Answer:

As the pressure inside the pressure cooker increases as it is heated, the temperature also increases.

Explanation:

A pressure cooker contains water and steam in equilibrium at a pressure greater than atmospheric pressure.

According to the ideal gas law, as the pressure increases, the temperature also increases.

As the pressure inside the pressure cooker increases as it is heated, the temperature also increases.

6 0

3 years ago