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:
- Control variables
- Dependent variables
- 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.
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A. Internal. Most cars use that type of set up because it's more efficient, you can find more about it on this website, https://auto.howstuffworks.com/did-cars-ever-have-external-combustion-engines.htm
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~ Ria
Answer:
Coefficient of friction = 0.836
Explanation:
If v be the speed after one quarter of the circular path
v² = 2as = 2 x 1.85 x 2πr/4 ; v²/r = 1.85 x 3.14 = 5.8
tangential acceleration = 5.8 m/s²
radial acceleration = v² /r = 5.8
total acceleration = √2 x 5.8
m x√2 x 5.8 = m x g xμ
μ = √2 x 5.8 / 9.8 = 0.836
Answer: a) 0.315 (V/L)
Explanation:
From Conservation of angular momentum, we know that
L1 = L2 ,
Therefore MV L/2 = ( Irod + Ib) x W
M/4 x V x L/2 = (M (L/2)^2 + 1/3xMxL^2) x W
M/8 X VL = (ML^2/16 + ML^2 /3 )
After elimination we have,
V/8 = 19/48 x L x W
W = 48/8 x V/19L = 6/19 x V/L
Therefore W = (0.136)X V/L