The image of the object is 8cm to the left of the lens (D)
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</h3>
What is the image of an object?
The image of an object is said to be the location where light rays from that object intersect with a mirror by reflection.
It is calculated thus:
1÷v = 1÷f - 1÷u
<h3>How to calculate the image of an object</h3>
From the formula
1÷v = 1÷f - 1÷u
<h3>
Where </h3>
V = image distance fromthe object
U = object
f = focal length
Substitute the values
1÷v = 1÷8 - 1÷ 4
1÷v = - 1÷8
Make v the subject of formula
v = -8cm
Therefore, the image of the object is 8cm to the left of the lens (D)
Learn more on focal length here:
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If it;s a good insulator, there'll be no heat transfer warm to cold. So, over time, given the insulation ... nothing should happen ...
Answer:
Water
Explanation:
Because it does not conduct much energy.
The propagation errors we can find the uncertainty of a given magnitude is the sum of the uncertainties of each magnitude.
Δm = ∑
Physical quantities are precise values of a variable, but all measurements have an uncertainty, in the case of direct measurements the uncertainty is equal to the precision of the given instrument.
When you have derived variables, that is, when measurements are made with different instruments, each with a different uncertainty, the way to find the uncertainty or error is used the propagation errors to use the variation of each parameter, keeping the others constant and taking the worst of the cases, all the errors add up.
If m is the calculated quantity, x_i the measured values and Δx_i the uncertainty of each value, the total uncertainty is
Δm = ∑
| dm / dx_i | Dx_i
for instance:
If the magnitude is a average of two magnitudes measured each with a different error
m =
Δm = |
| Δx₁ + |
| Δx₂
= ½
= ½
Δm =
Δx₁ + ½ Δx₂
Δm = Δx₁ + Δx₂
In conclusion, using the propagation errors we can find the uncertainty of a given quantity is the sum of the uncertainties of each measured quantity.
Learn more about propagation errors here:
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Answer:

Explanation:
sin^2 60° = ( \|3 / 2 ) ^2 = 3 / 4.