In order for a person to "see" an object, light waves pass through the ___________.

2 answers:

8 0

I think the correct answer from the choices listed above is the first option. In order for a person to "see" an object, light waves pass through the cornea. The cornea is the transparent layer forming at the front of the eye. Hope this answers the question. Have a nice day.

Marta_Voda [28]3 years ago

3 0

Answer: Cornea

Explanation:

Cornea is a transparent outer surface of the eye. It covers the inner parts of the eye ball. It is the first part of the eye through which the light enters the eye. The refractive power of the cornea bends the incident light rays, in such a way that they pass through the pupil. From pupil the light enters the opening of the iris through which the light enters the eye. The iris is like shutter of a camera. It regulates the amount of light entering the eye.

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GuDViN [60]

$A)F_0d$

Explanation

If you graph the force on an object as a function of the position of that object, then the area under the curve will equal the work done on that object, so we need to find the area under the function to find the work

Step 1

find the area under the function.

Area:

$\text{Area}=rec\tan gle_{green}+triangle_{gren}-triangle_{red}$ $\begin{gathered} \text{the area of a rectangle is given by} \\ A_{rec}=lenght\cdot widht \\ \text{and} \\ \text{the area of a triangle is given by:} \\ A_{tr}=\frac{base\cdot height}{2} \end{gathered}$

$\begin{gathered} \text{Area}=rec\tan gle_{green}+triangle_{gren}-triangle_{red} \\ \text{replace} \\ \text{Area}=(F_0\cdot d)+\frac{(F_0\cdot d)}{2}-\frac{(F_0\cdot d)}{2} \\ \text{Area}=(F_0\cdot d) \\ Area=F_0d \end{gathered}$

therefore, the answer is

$A)F_0d$

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1 year ago

What is the atomic number of this element?

Zina [86]

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3 years ago

Read 2 more answers

A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radiu

anyanavicka [17]

Answer:

1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

Explanation:

According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.

As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :

Q₁ = ∫ ρ dV

Here dV is the volume element of sphere of radius r.