What sentence best supports the statement that hormones are involved in the regulation of homeostasis? A. The hormone cortisol s
1 answer:
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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:
To solve this problem we will start from the definition of energy of a spring mass system based on the simple harmonic movement. Using the relationship of equality and balance between both systems we will find the relationship of the amplitudes in terms of angular velocities. Using the equivalent expressions of angular velocity we will find the final ratio. This is,
The energy of the system having mass m is,
The energy of the system having mass 2m is,
For the two expressions mentioned above remember that the variables mean
m = mass
Angular velocity
A = Amplitude
The energies of the two system are same then,
Remember that
Replacing this value we have then
But the value of the mass was previously given, then
Therefore the ratio of the oscillation amplitudes it is the same.
Answer:
Check the explanation
Explanation:
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
The image is virtual
The image is upright
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
Kindly check the diagram in the attached image below.
