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Taya2010 [7]
2 years ago
6

A photographer wants to determine the color of light he can use in the darkroom that will not expose the films he is processing.

In one trial, he used a blue incandescent bulb. Which bulb can he use for another trial?
A. Red incandescent bulb C. Red fluorescent bulb
B. Blue incandescent bulb D. Blue fluorescent bulb
Physics
1 answer:
Sonbull [250]2 years ago
4 0

For a photographer that wishes to determine the color of light that he can use in a dark room that will not expose the films he is processing, having used a Blue Incandescent bulb, he should proceed to use a Red Incandescent bulb for the next trial.

The photographer in question is performing an experiment. For these kinds of experiments it is important to identify the variables present, which can be of three kinds:

  1. Control variables
  2. Dependent variables
  3. Independent variables

For this experiment, the dependent variable is the exposure of the light onto the films, given that this is what we wish to measure. The independent variable will be the color of the light being used which is what will affect the dependent variable.

The remaining variable must be the control variable. Unlike the previous variables, we can have more than one of these. The control variable is there to make sure that only the dependent variable is affecting the outcome. We do this by keeping the control variable the same through each trial, which is why the photographer should not change the type of bulb in the second experiment, changing only the color of the light.

To learn more visit:

brainly.com/question/1549017?referrer=searchResults

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Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t
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Answer: 3.66(10)^{33}kg

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity \omega of the planet P1 with a period T=750years=2.36(10)^{10}s:

\omega=\frac{2\pi}{T}=\frac{V_{1}}{R} (1)

Where:

V_{1}=40.2km/s=40200m/s is the velocity of planet P1

R is the radius of the orbit of planet P1

Finding R:

R=\frac{V_{1}}{2\pi}T (2)

R=\frac{40200m/s}{2\pi}2.36(10)^{10}s (3)

R=1.5132(10)^{14}m (4)

On the other hand, we know the gravitational force F between the star S with mass M and the planet P1 with mass m is:

F=G\frac{Mm}{R^{2}} (5)

Where G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

In addition, the centripetal force F_{c} exerted on the planet is:

F_{c}=\frac{m{V_{1}}^{2}}{R^{2}} (6)

Assuming this system is in equilibrium:

F=F_{c} (7)

Substituting (5) and (6) in (7):

G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}} (8)

Finding M:

M=\frac{V^{2}R}{G} (9)

M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}} (10)

Finally:

M=3.66(10)^{33}kg (11) This is the mass of the star S

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3 years ago
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