Based on the article “Will the real atomic model please stand up?,” why did J.J. Thomson experiment with cathode ray tubes? to s
1 answer:
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<h2>Complete Question:</h2>
You are on an aluminum ladder that is standing on the ground, trying to fix an electrical connection with a metal screwdriver having a metal handle. Your body is wet because you are sweating from the exertion; therefore, it has a resistance of 1.60 kΩ .
(a) If you accidentally touch the "hot" wire connected to the 120 V line, how much current will pass through your body?
(b) How much electrical power is delivered to your body?
<h2>Answer:</h2>(a) 0.075A
(b) 9W
<h2>Explanation:</h2>The voltage (V) passing across or supplied to a body is directly proportional to the current (I) flowing through the body as stated by Ohm's law. i.e
V ∝ I
=> V = I x R ----------------------(i)
Where;
R = constant of proportionality called resistance of the body
(a) As stated in the question;
The body is wet and thus will conduct electricity and has the following;
V = voltage supplied = 120V
R = resistance of the wet body = 1.60kΩ = 1.6 x 1000Ω = 1600Ω
Substitute these values into equation(i) as follows;
120 = I x 1600
Solve for I;
I =
I = 0.075A
Therefore the amount of current that will pass through your body is 0.075A
(b) Electrical power(P), which is commonly measured in Watts(W), delivered to a body is the product of the current(I) and voltage (V) supplied to the body. i.e
P = I x V ---------------------(ii)
Where;
I = 0.075A [as calculated above]
V = 120V [given in the question]
Substitute these values into equation (ii) as follows;
P = 0.075 x 120
P = 9W
Therefore, the electric power delivered to your body is 9W
We get the clearest image if there is no magnification. When we have no magnification the image and real object have the same size.
If we look at the diagram that I attached we can see that:
Two triangles that I marked are similar and from this we get:
The image and the object must have the same height so we get:
This tells how far the screen should be from the lens.
The position of the screen on the optical bench is:
If we look at the diagram that I attached we can see that:
Two triangles that I marked are similar and from this we get:
The image and the object must have the same height so we get:
This tells how far the screen should be from the lens.
The position of the screen on the optical bench is:
Answer:
the expected distance is 4.32 m
Explanation:
given data
half life time = 1.8 × s
speed = 0.8 c = 0.8 × 3 ×
to find out
expected distance over
solution
we know c is speed of light in air is 3 × m/s
we calculate expected distance by given formula that is
expected distance = half life time × speed .........1
put here all these value
expected distance = half life time × speed
expected distance = 1.8 × × 0.8 × 3 ×
expected distance = 4.32
so the expected distance is 4.32 m