If a measurement is 4.2 cm and its possible error is 0.04 cm, the relative error is?
1 answer:
Given the value of possible error, the relative error of the measurement is 0.0095.
Given the data in the question;
- Measurement taken;
- Absolute or possible erorr;
- Relative error;
Relative error is simply the ratio of the absolute or possible error of a measurement to the measurement its self.
Relative error is expressed as;
Where e is the absolute or possible error and m is the measurement being taken.
To determine the relative error, we substitute our given values into the expression above.
%
Therefore, given the possible error, the relative error of the measurement is 0.0095.
Learn more about relative error: brainly.com/question/17016021
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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.
Answer:
∆h = 0.071 m
Explanation:
I rename angle (θ) = angle(α)
First we are going to write two important equations to solve this problem :
Vy(t) and y(t)
We start by decomposing the speed in the direction ''y''
Vy in this problem will follow this equation =
where g is the gravity acceleration
This is equation (1)
For Y(t) :
We suppose yi = 0
This is equation (2)
We need the time in which Vy = 0 m/s so we use (1)
So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)
2.236 m is the maximum height from the shell (in which Vy=0 m/s)
Let's calculate now the height for t = 0.555 s
The height asked is
∆h = 2.236 m - 2.165 m = 0.071 m
