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3 years ago

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 small ball is attached to one end of a rigid rod with negligible mass. The ball and the rod revolve in a horizontal circle wit

castortr0y [4]

Answer:

mass of the ball will be 1 kg

Explanation:

We have given force F = 0.5 N

This force must be equal to the centripetal force for the ball to revolve in the circle

So $F=F_C=0.5N$

Centripetal acceleration is given as $a_c=0.5m/sec^2$

We have to calculate the mass of the ball

We know that force is given by

$F_C=ma_c$

So $0.5=m\times 0.5$

m = 1 kg

So mass of the ball will be 1 kg

5 0

3 years ago

What happens when a star exhausts its core hydrogen supply?

Allushta [10]

Answer:

It becomes a giant or supergiant.

Explanation:

Once all the hydrogen supply is gone, fusion of hydrogen into helium stops. The core starts to contract and liberates energy, which heats the superior layer until it becomes hot enough to start the fusion of hydrogen into helium.

6 0

3 years ago

The downsprue leading into the runner of a certain mold has a length of 175 mm. The cross-sectional area at the base of the spru

adoni [48]

Answer:

(a) Velocity at bottom is 1.85 m/s

(b) Volume flow rate is 7.4 x 10⁻⁴ m³/s.

Explanation:

(a)

Applying Bernoulli's Equation on both ends of the down sprue, with the assumptions that every point is at atmospheric pressure and the liquid metal at the pouring basin is at zero velocity. The equation then becomes:

V = √2gh

where,

V = velocity at bottom of down sprue

h = height of down sprue = 175 mm = 0.175 m

V = √2(9.8 m/s²)(0.175 m)

V = 1.85 m/s

(b)

The volume flow rate is given as:

Volume Flow Rate = (V)(A)

where,

V = velocity at bottom = 1.85 m/s

A = Area of bottom = 400 mm² = 0.0004 m²

Therefore,

Volume Flow Rate = (1.85 m/s)(0.0004 m²)

Volume Flow Rate = 7.4 x 10⁻⁴ m³/s = 740 cm³/s

(c)

The time required to fill the cavity is given as:

Volume Flow Rate = V/t

where,

V = Volume of mold Cavity = 0.001 m³

t = time required to fill the cavity = ?

Therefore,

t = V/Volume Flow Rate

t = 0.001 m³/7.4 x 10⁻⁴ m³/s

t = 1.35 s

5 0

3 years ago

How do you calculate final velocity

serg [7]

Try looking google..

3 0

3 years ago

Read 2 more answers

An electron is confined to a one dimensional region, bounded by an infinite potential. If the energy of the electron in its firs

OLga [1]

Answer:

The energy in its ground state is 10 meV.

Explanation:

It is given that,

The energy of the electron in its first excited state is 40 meV.

Energy of the electron in any state is given by :

$E=\dfrac{n^2\pi^2h^2}{8mL^2}$

For ground state, n = 1

$E_1=\dfrac{\pi^2h^2}{8mL^2}$ .............(1)

For first excited state, n = 2

$40=\dfrac{2^2\pi^2h^2}{8mL^2}$ .............(2)

Dividing equation (1) and (2), we get :

$\dfrac{E_1}{40}=\dfrac{1}{4}$

$E_1=10\ meV$

So, the energy in its ground state is 10 meV. Hence, this is the required solution.

4 0

3 years ago