From the information given,
diameter of ornament = 8
radius = diameter/2 = 8/2
radius of curvature, r = 4
Recall,
focal length, f = radius of curvature/2 = 4/2
f = 2
Recall,
magnification = image d
Answer:
d. )directed upward.
Explanation:
As the electron has a negative charge, when under the influence of an electric field, is subject to an electric force, which direction is the opposite to the direction of the electric field.
This is because the electric field has the same direction that the force on a positive test charge at the same point.
As the electric field points vertically downward, the electric force on the electron (a negative charge) points vertically upward.
So, the statement d. is the one that results to be true.
Answer:
40 N/m
Explanation:
The diagram attached is used to answer the question
We know from Hooke's law that extension is directly proportional to the applied force hence
F=kx where x is extension, F is applied force and k is the spring constant. Making k the subject of the formula then
From the attached diagram extension is given by subtracting unstretched spring from stretched spring hence extension, x=1-0.5=0.5m
Substituting 20 N for F and 0.5 m for x then
The magnitude of the current in wire 3 is (I₃)= 0.33A
<h3>How to calculate the value of the magnitude of the current in wire 3 ?</h3>
To calculate the magnitude of the current in wire 3 we are using the Kirchhoff’s current law,
I₁ + I₂ + I₃ = 0
Where we are given,
I₁ = current in wire 1
=0.40 A.
I₂ = current in wire 2
= -0.73 A.
We have to calculate the magnitude of the current in wire 3, I₃
Now we put the known values in above equation, we get,
I₁ + I₂ + I₃ = 0
Or, I₃ = -.(I₁ + I₂)
Or, I₃ = -.(0.40 - 0.73)
Or, I₃ = 0.33 A
From the above calculation, we can conclude that the current in wire 3 is I₃ = 0.33 A
Learn more about current:
brainly.com/question/25537936
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