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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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The sun has a greater mass then the earrth
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Physicists often measure the momentum of subatomic particles moving near the speed of light in units of MeV/c, where c is the sp
maxonik [38]

Answer:

kg m/s

Explanation:

e = Charge = C

V = Voltage = \dfrac{N}{C}m

c = Speed of light = m/s

Momentum is given by

\dfrac{MeV}{c}=\dfrac{e\times V}{c}\\\Rightarrow \dfrac{MeV}{c}=\dfrac{C\times \dfrac{N}{C}\times m}{m/s}\\\Rightarrow \dfrac{MeV}{c}=Ns\\\Rightarrow \dfrac{MeV}{c}=kg\times \dfrac{m}{s}\times s\\\Rightarrow \dfrac{MeV}{c}=kg\cdot m/s

The unit of MeV/c in SI fundamental units is kg m/s

5 0
3 years ago
Which of these events is an example of the Doppler shift?
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The intensity of an electromagnetic wave increases with the field strength.
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3 years ago
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Based on the data given, in what direction will the car accelerate?
skad [1K]
Vertical forces:
There is a force of 579N acting upward, and a force of 579N
acting downward.
The vertical forces are balanced ... they add up to zero ...
so there's no vertical acceleration. 
Not up, not down.

Horizontal forces:
There is a force of 487N acting to the left, and a force of 632N
acting to the right.
The net horizontal force is

        (487-left + 632-right)  -  (632-right - 487-right) =  145N to the right.

The net force on the car is all to the right.
The car accelerates to the right.
7 0
3 years ago
A 2 kg rubber ball is thrown at a wall horizontally at 3 m/s, and bounces back the way it came at an equal speed. A 2 kg clay ba
Lyrx [107]

Answer:

THE RUBBER BALL

Explanation:

From the question we are told that

      The mass of the rubber ball is m_r   =  2 \ kg

      The  initial  speed of the rubber ball is  u =  3 \ m/s

      The final speed at which it bounces bank v  - 3 \ m/s

      The mass of the clay ball  is  m_c =  2  \ kg

       The  initial  speed of the clay  ball is u = 3 \ m/s

       The final speed of the clay ball is  v = 0 \  m/s

Generally Impulse is mathematically represented as

       I  =  \Delta p

where \Delta  p is the change in the linear momentum so  

       I  =  m(v-u)

For the rubber  is  

        I_r  =  2(-3 -3)

       I_r  = -12\ kg \cdot  m/s

=>     |I_r|  = 12\ kg \cdot  m/s

For the clay ball

       I_c  =  2(0-3)

        I_c =  -6 \ kg\cdot \ m/s

=>    | I_c| =  6 \ kg\cdot \ m/s

So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball

       

8 0
4 years ago
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