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mars1129 [50]
3 years ago
15

What structure is located at the front edge of the retina and has a tooth like appearance?

Physics
1 answer:
pogonyaev3 years ago
3 0

Ora serrate is located at the front edge of the retina and has a tooth like appearance.

Our eye has many portions and parts, all of them are there for some function. The ora serrata is the most anterior extent of the retina or serrated junction between the retina and the ciliary body. The retina ends at the ora serrata, where the ciliary body begins.

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A student measures the speed of sound by echo destiny classes hands and then measures the time to hear the echo his distance to
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Explanation:

∆x=300 m×2

∆t=1.5 s

v=∆x/∆t → v=2×300/1.5 = 400 m/s

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The first mechanized industry was
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C. Textiles

It was the first thing mechanized in the Industrial Revolution

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What is the value of the equivalent resistance for the three resistors connected in series?
Harman [31]

The value of the equivalent resistance for the three resistors connected in series will be the sum of the three values.

To find the answer, we have to know more about the equivalent resistance.

<h3>What is meant by equivalent resistance?</h3>
  • equivalent resistance is the total value of the resistance connected in a circuit.
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                R_E=R_1+R_2+..........+R_n

  • In our question we have three resistors. Thus, the equivalent resistance will be,

               R_E=R_1+R_2+R_3

Thus, we can conclude that, the value of the equivalent resistance for the three resistors connected in series will be the sum of the three values.

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7 0
1 year ago
The larger the push, the larger the change in velocity. This is an example of Newton's Second Law of Motion which states that th
Paul [167]
According to Newton, an object will only accelerate if there is a net or unbalanced forceacting upon it. The presence of an unbalanced force will accelerate an object - changing its speed, its direction, or both its speed and direction.
6 0
3 years ago
Read 2 more answers
In classical physics, consider a 2 kg block hanging on a spring with a spring constant of 50 N/m. Ignore air resistance. The blo
RUDIKE [14]

Answer:

v = 0

Explanation:

This problem can be solved by taking into account:

- The equation for the calculation of the period in a spring-masss system

T = \sqrt{\frac{m}{k} }     ( 1 )

- The equation for the velocity of a simple harmonic motion

x = \frac{2\pi }{T}Asin(\frac{2\pi }{T}t)   ( 2 )

where m is the mass of the block, k is the spring constant, A is the amplitude (in this case A = 14 cm) and v is the velocity of the block

Hence

T = \sqrt{\frac{2 kg}{50 N/m}} = 0.2 s

and by reeplacing it in ( 2 ):

v = \frac{2\pi }{0.2s}(14cm)sin(\frac{2\pi }{0.2s}(0.9s)) = 140\pi  sin(9\pi ) = 0

In this case for 0.9 s the velocity is zero, that is, the block is in a position with the max displacement from the equilibrium.

5 0
3 years ago
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