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qaws [65]
2 years ago
5

How are light waves used to bring far away objects into view and how does the eye translate them?

Physics
1 answer:
adelina 88 [10]2 years ago
5 0

Answer:

When focused light is projected onto the retina, it stimulates the rods and cones. The retina then sends nerve signals are sent through the back of the eye to the optic nerve. The optic nerve carries these signals to the brain, which interprets them as visual images.

Explanation:

Hope it will help u

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Define rectilinear propagation of light.??​
Tju [1.3M]

Answer:

it relates to the light propensity to travel over one straight line without having any interference in its trajectory

Explanation:

5 0
2 years ago
Read 2 more answers
A 295-kg object and a 595-kg object are separated by 4.10 m.
kodGreya [7K]

Answer:

a)F=3 x 10⁻⁷ N

b)x=2.405 m

Explanation:

Given that

m₁=295 kg

m₂=595 kg

d= 4.1 m

a)

m₃=63 kg

r=d/2 = 2.05 m

The force between the mass m₁ and m₃

F_{13}=\dfrac{Gm_1m_3}{r^2}

by putting the values

F_{13}=\dfrac{Gm_1m_3}{r^2}

F_{13}=\dfrac{6.67\times 10^{-11}\times 295\times 63 }{2.05^2}

F₁₃=2.94 x 10⁻⁷ N

The force  between the mass m₂ and m₃

by putting the values

F_{23}=\dfrac{Gm_2m_3}{r^2}

F_{23}=\dfrac{6.67\times 10^{-11}\times 595\times 63 }{2.05^2}

F₂₃=5.94 x 10⁻⁷ N

The net force F

F= F₂₃- F₁₃

F=5.94 x 10⁻⁷ N-2.94 x 10⁻⁷ N

F=3 x 10⁻⁷ N

b)

Lest take at distance x from mass m₂ net force is zero.

F_{23}=\dfrac{Gm_2m_3}{x^2}

F_{13}=\dfrac{Gm_1m_3}{(4.1-x)^2}

Form above two equation

\dfrac{Gm_1m_3}{(4.1-x)^2}=\dfrac{Gm_2m_3}{x^2}

\dfrac{m_1}{(4.1-x)^2}=\dfrac{m_2}{x^2}

\dfrac{295}{(4.1-x)^2}=\dfrac{595}{x^2}

x²=2.01(4.1-x)²

x=1.42 (4.1-x)

x=5.82 - 1.42x

x=2.405 m

4 0
3 years ago
What can make an inclined plane more efficient?
Sophie [7]

Answer:

Efficiency can be increase by using rollers in conjunction with the inclined plane. Wedge. The wedge is an adaptation of the inclined plane. It can be used to raise a heavy load over a short distance or to split a log

3 0
3 years ago
In the Rutherford model of hydrogen the electron (mass=9.11x10-31 kg) is in a planetary type orbit around the proton (mass=1.67x
Andrei [34K]

Answer:

(a) F_g=1.62*10^{-48}N

(b) F_e=3.68*10^{-9}N

Explanation:

(a) We use Newton's law of universal gravitation, in order to calculate the gravitational force between electron and proton:

F_g=-G\frac{m_1m_2}{r^2}

Where G is the Cavendish gravitational constant, m_1 and m_2 are the masses of the electron and the proton respectively and r is the distance between them:

F_g=-6.67*10^{-11}\frac{N\cdot m^2}{kg^2}\frac{(9.11*10^{-31}kg)(1.67*10^{-27}kg)}{(2.5*10^{-10}m)^2}\\F_g=-1.62*10^{-48}N

The minus sing indicates that the force is repulsive. Thus, its magnitude is:

F_g=1.62*10^{-48}N

(b) We use Coulomb's law, in order to calculate the electric force between electron and proton, here k is the Coulomb constant and e is the elementary charge:

F_e=-k\frac{e^2}{r^2}\\F_e=-8.99*10^{9}\frac{N\cdot m^2}{C^2}\frac{(1.6*10^{-19}C)^2}{(2.5*10^{-10}m)^2}\\F_e=-3.68*10^{-9}N

Its magnitude is:

F_e=3.68*10^{-9}N

6 0
3 years ago
How much work does it take to slide a box 37 meters along the ground by pulling it with a 217 N force at an angle of 19° from th
-Dominant- [34]

Answer:

W = 7591.56 J

Explanation:

given,

distance of the box, d = 37 m

Force for pulling the box, F = 217 N

angle of inclination with horizontal,θ = 19°

We know,

Work done is equal to product of force and the displacement.

W = F.d cos θ

W = 217 x 37 x cos 19°

W = 7591.56 J

Hence, the work done to pull the box is equal to W = 7591.56 J

8 0
3 years ago
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