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Salsk061 [2.6K]
3 years ago
8

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.
1. Which of the following was not kept constant?
the test for photocell activation
the color of the light after it has passed through the cellophane
the distance of the light from the photocell
the source of the original light

2. In the above experiment, which factor is the independent variable?
the photocell
the color of the light
the number of clicks
the original source of the light

3. Which of the following would be the best hypothesis for this experiment?
Photoelectric cells are affected by different wavelengths of light.
Red light will activate the photocell to the highest degree.
If photoelectric cells respond better to short wavelengths of light, the red light should activate it the least and blue the most.
Physics
2 answers:
Stels [109]3 years ago
5 0
..................”...........
barxatty [35]3 years ago
3 0

Answer:

1. The color of the light after it has passed through the Cellophane

2. The color of the light

3. If photoelectric cells respond better to short wavelengths of light, the red light should activate it the least and the blue the most.

Explanation:

1. The color of the light changes depending on what color cellophane is put in front of the flashlight

2. The color of the light is the independent variable because it is the only part of the experiment that is being altered.

3. In the description they say the purpose of the experiment is to determine what color of light will have the largest affect on the photoelectric cell. Therefore, this matches up best. The one where is says Photoelectric cells are affected by different wavelengths of light doesn't mention which one is better just that they are affected.

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A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th
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Answer:

Part a)

v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}

Part b)

v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

Mv_o = M v_1 + m v_2

here we also use angular momentum conservation

so we have

M v_o d = M v_1 d + \beta mL^2 \omega

also we know that the collision is elastic collision so we have

v_o = (v_2 + d\omega) - v_1

so we have

\omega = \frac{v_o + v_1 - v_2}{d}

also we know

M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})

also we know

v_1 = v_o - \frac{m}{M}v_2

so we have

M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})

mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}

now we have

(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}

v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}

Part b)

Now we know that speed of the ball after collision is given as

v_1 = v_o - \frac{m}{M}v_2

so it is given as

v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})

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3 years ago
The main water line enters a house on the first floor. The line has a gauge pressure of 1.94 x 105 Pa. (a) A faucet on the secon
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Answer

given,

gauge pressure =   1.94 x 10⁵ Pa

Pressure due to 4.90 m column of water

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= 1.94 x 10⁵Pa - 48020 Pa

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( b )

Let h = height of faucet from which no water can flow even if open

P = ρ g h

1.94 x 10⁵  = h x(1000) x (9.8)

h = 19.79 m

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3 years ago
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