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Illusion [34]
2 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]2 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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A laser beam of wavelength 439.4 nm is inci- dent on two slits 0.306 mm apart.
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Constructive interference happens when this condition is satisfied:
dsin(\theta)=m\lambda; m=1,2,3,4,5,...
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If we say that the distance between two interference fringes is much smaller than the distance from the slit to the screen, we can use the following approximation:
sin(\theta)=\frac{y}{L}
Finally for the bright spots we have:
y=\frac{m\lambda L}{d}
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3 years ago
Which missing item would complete this alpha decay reaction ? ________—> 228/88Ra + 4/2He A.224/96 Rn B. 232/90 Th C. 230/92
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A.224/96 Rn I think i'm right if i'm wrong i'm srry


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3 years ago
A technician builds an RLC series circuit which includes an AC source that operates at a fixed frequency and voltage. At the ope
MariettaO [177]

Answer:

Xc = (0.467 - 0.427j)R

Explanation:

Since the resistance in the circuit is R, the reactance of the inductor is XL and the reactance of the capacitor is XC, then the impedance of the circuit is

Z = √[R² + (XL - XC)²]

Since the inductive reactance XL equals the resistance R, we have that

Z = √[R² + (XL - XC)²]

Z = √[R² + (R - XC)²]

Thus, the current in the circuit is thus I = V/Z = V/√[R² + (R - XC)²]

Now, when the plate separation of the parallel plate capacitor is reduced to one-half its original value, the current doubles. Also, when the plate separation is reduced to half, the capacitance doubles since C ∝ 1/d where C is capacitance and d separation between the plates. Since the capacitance doubles, the new reactance XC' is twice the initial reactance XC. So, XC' = 2XC. Thus the new impedance is thus

Z' = √[R² + (R - XC')²]

Z' = √[R² + (R - 2XC)²]

The new current is I' = V/Z' = V/√[R² + (R - 2XC)²]

Since the current doubles, I' = 2I.

V/√[R² + (R - 2XC)²] = 2V/√[R² + (R - XC)²]

1/√[R² + (R - 2XC)²] = 2/√[R² + (R - XC)²]

√[R² + (R - XC)²] = 2√[R² + (R - 2XC)²]

squaring both sides, we have

[R² + (R - XC)²] = 4[R² + (R - 2XC)²]

expanding the brackets, we have

[R² + R² - 2RXC + XC²] = 4[R² + R² - 4RXC + 4XC²]

[2R² - 2RXC + XC²] = 4[2R² - 4RXC + 4XC²]

2R² - 2RXC + XC² = 8R² - 16RXC + 16XC²

collecting like terms, we have

16RXC - 2RXC + XC² - 16XC² = 8R² - 2R²

14RXC - 15XC² = 6R²

15XC² - 14RXC + 6R² = 0

Using the quadratic formula to find XC, we have

XC = \frac{-(-14R) +/- \sqrt{(-14R)^{2} - 4 X 15 X 6R^{2} } }{2 X 15}\\= \frac{-(-14R) +/- \sqrt{196R^{2} - 360R^{2} } }{30}\\ \\= \frac{14R +/- \sqrt{- 164R^{2} } }{30}\\ \\= \frac{14R +/- 12.81Ri }{30}\\\\= 0.467R +/- 0.427Ri

Since it is capacitive, we take the negative part.

So, Xc = (0.467 - 0.427j)R

4 0
3 years ago
A sound wave has a frequency of 425 Hz. What is the period of this wave?
Serhud [2]
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8 0
3 years ago
What is the de Broglie wavelength for a proton with energy 50 keV? Due to the limitations of Canvas, please give the wavelength
ad-work [718]

Answer:

1.2826 x 10^-13 m

Explanation:

\lambda  = \frac{h}{\sqrt{2 m K}}

Here, k be the kinetic energy and m be the mass

K = 50 KeV = 50 x 1.6 x 10^-16 J = 80 x 10^-16 J

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\lambda  = \frac{6.63 \times  10^{-34}}{\sqrt{2 \times 1.67\times 10^{-27}\times 80\times 10^{-16}}}

λ = 1.2826 x 10^-13 m

6 0
2 years ago
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