Suppose it turns out that one in 1 million stars has a planet that at some point in its history is home to an advanced civilizat
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Answer:
measuring the zero intensity point, we can deduce the movement of the screen.
The distance from the center of the pattern to the first zero is proportional to the distance to the screen,
Explanation:
The expression for the diffraction phenomenon is
a sin θ = m λ
for the case of destructive interference. In general the detection screen is quite far from the grid, let's use trigonometry to find the angles
tan θ = y / L
in these experiments the angles are small
tan θ = sin θ / cos θ = sin θ
sunt θ = y / L
we substitute
a = m λ
y = m L λ / a
therefore, by carefully measuring the zero intensity point, we can deduce the movement of the screen.
The distance from the center of the pattern to the first zero is proportional to the distance to the screen, so you can know where the displacement occurs, it should be clarified that these displacements are very small so the measurement system must be capable To measure quantities on the order of hundredths of a millimeter, a micrometer screw could be used.
The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.
The given parameters;
- <em>initial temperature of metals, = </em>
<em />
- <em>initial temperature of water, = </em>
<em> </em>
- <em>specific heat capacity of copper, </em>
<em> = 0.385 J/g.K</em>
- <em>specific heat capacity of aluminum, </em>
= 0.9 J/g.K
- <em>both metals have equal mass = m</em>
The quantity of heat transferred by each metal is calculated as follows;
Q = mcΔt
<em>For</em><em> copper metal</em><em>, the quantity of heat transferred is calculated as</em>;
<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>copper metal</em>;
<em>For </em><em>aluminum metal</em><em>, the quantity of heat transferred is calculated as</em>;
<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>aluminum metal </em><em>;</em>
Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.
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