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DaniilM [7]
3 years ago
11

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

Physics
2 answers:
pantera1 [17]3 years ago
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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Does the voltage of a battery affect the strength of an electromagnet?
swat32

I'm trying to make an electromagnet that's strength is constantly getting incremented by small amounts every second. I need to know, which would have a greater effect on the electromagnet's strength, amps or volts? (I know increasing the turns and/or density of the magnet wire will increase the strength, but I am looking for answers other than that particular one.)

7 0
2 years ago
The question is on the picture.
Jet001 [13]
Between noon and 2 pm, the amount of water in the rain gauge decreased.
This can be caused by evaporation, which turns water into water vapor.
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Runoff is when water on land drains into water sources such as lakes, rivers, oceans, etc. 
So the answer is A. evaporation.
5 0
3 years ago
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Polly is pushing a box across the floor with a force of 30 N. The force of gravity is -8 in, and the normal force is eight in. W
aleksandr82 [10.1K]

Answer:

A. -30 N

Explanation:

not sure but if the box isn't moving then the force opposite of Polly would be equal to the force she's exerting.

5 0
2 years ago
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How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe
Rus_ich [418]

Explanation:

(a)  For an isothermal process, work done is represented as follows.

             W = -nRT ln(\frac{V_{2}}{V_{1}})

Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

             = -12280.82 \times ln (0.09)

             = -12280.82 \times -2.41

             = 29596.78 J

or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = \frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}

              = \frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}

              = \frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}

              = 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c)   We know that for an isothermal process,

               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})

                    = 11 atm

Hence, the required pressure is 11 atm.

(d)   For adiabatic process,  

          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

or,       P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}

                    = 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0
3 years ago
What units do we use to measure volume
Sonbull [250]
I think it is liters, cubic meters, or milliliters.<span />
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