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Illusion [34]
3 years ago
10

An experiment is designed to test what color of light will activate a photoelectric cell the best. The photocell is set in a cir

cuit that "clicks" in response to current. The faster the current, the more clicks per minute. In this experiment, the number of clicks in one minute is recorded for each color of light shining on the photocell. To change the color of light, a different color of cellophane is placed over the same flashlight and the flashlight is then located a specific distance from the photocell.
Which of the following was not kept constant?


the color of the light after it has passed through the cellophane
the source of the original light
the distance of the light from the photocell
the test for photocell activation
Physics
1 answer:
Mariana [72]3 years ago
6 0

Answer:

the color of the light after it has passed through the cellophane

Explanation:

Since in the given experiment, there is an impact of various colors of light on the cell i.e. photoelectric that should be measured. The photocell should be placed in a circuit when the current would passed. For every color that falls on the photocell, the value of the current that passed via the cell represent an idea.

In the given situation the color of light shows an independent variable and the dependent variable is clicks per minute or the current that passed through the cell

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bagirrra123 [75]

Answer:

(a) W = 1329.5 J = 1.33 KJ

(b) ΔU = 24.27 KJ

Explanation:

(a)

Work done by the gas can be found by the following formula:

W = P\Delta V

where,

W = Work = ?

P = constant pressure = (0.991 atm)(\frac{101325\ Pa}{1\ atm}) = 100413 Pa

ΔV = Change in Volume = 18.7 L - 5.46 L = (13.24 L)(\frac{0.001\ m^3}{1\ L}) = 0.01324 m³

Therefore,

W = (100413 Pa)(0.01324 m³)

<u>W = 1329.5 J = 1.33 KJ</u>

<u></u>

(b)

Using the first law of thermodynamics:

ΔU = ΔQ - W (negative W for the work done by the system)

where,

ΔU = change in internal energy of the gas = ?

ΔQ = heat added to the system = 25.6 KJ

Therefore,

ΔU = 25.6 KJ - 1.33 KJ

<u>ΔU = 24.27 KJ</u>

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3 years ago
Calculate the force on an object with mass of 50kg and gravity of 10​
GarryVolchara [31]
Answer: 500 N

Explanation:

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Hope this helps, brainliest would be appreciated :)
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Answer:

B: 980 Hz

Explanation:

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Explanation:

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<span>A. Comparison 

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