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ira [324]
3 years ago
8

3.

Physics
1 answer:
Inga [223]3 years ago
7 0

Answer:

Iris

Explanation:

The iris seems to be the illuminated portion of the eyes which really covers the pupil. It controls the amount of light reaching the eye. The lens is indeed a translucent layer of the retina that serves to concentrate light and objects on the lens.

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Barney walks at a velocity of 1.7 meters/second on an inclined plane, which has an angle of 18.5° with the ground. What is the h
Viktor [21]
The velocities and the speed build a triangle, where the 1.7 m/s are the hypotenuse and the x-velocity and y-velocity are the other sides. 

<span>So the x-velocity is: speed*cos(angle) </span>

<span>now plug in </span>
<span>x=1.7 m/s * cos(18.5)=1.597 m/s </span>


3 0
3 years ago
A 40-W lightbulb is 1.7 m from a screen. What is the intensity of light incident on the screen? Assume that a lightbulb emits ra
Sonja [21]

Answer:

Intensity, I=1.101\ W/m^2

Explanation:

Power of the light bulb, P  = 40 W

Distance from screen, r = 1.7 m

Let I is the intensity of light incident on the screen. The power acting per unit area is called the intensity of the light. Its formula is given by :

I=\dfrac{P}{A}

I=\dfrac{P}{4\pi r^2}

I=\dfrac{40\ W}{4\pi (1.7\ m)^2}

I=1.101\ W/m^2

So, the intensity of light is 1.101\ W/m^2.

6 0
4 years ago
Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Give
Keith_Richards [23]

Complete Question

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).

Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Given the satellite specifications listed in the problem introduction, what is the amplitude E0 of the electric field vector of the satellite broadcast as measured at the surface of the earth? Use ϵ0=8.85×10^−12C/(V⋅m) for the permittivity of space and c=3.00×10^8m/s for the speed of light.

Answer:

The electric field vector of the satellite broadcast as measured at the surface of the earth is  E_o = 6.995 *10^{-6} \ V/m

Explanation:

From the question we are told that

     The height of the satellite is  r  = 35000 \ km  =  3.5*10^{7} \ m

      The power output of the satellite is P  = 1 \ KW  =  1000 \ W

       

Generally the intensity of the electromagnetic radiation of the satellite at the surface of the earth is  mathematically represented as  

     I  =  \frac{P}{4 \pi r^2}

substituting values

      I  =  \frac{1000}{4 * 3.142 (3.5*10^{7})^2}

      I  = 6.495*10^{-14} \  W/m^2

This intensity of the electromagnetic radiation of the satellite at the surface of the earth can also be   mathematically represented as  

          I  =  c * \epsilon_o * E_o^2

Where E_o is the amplitude of the electric field vector of the satellite broadcast so

         E_o =  \sqrt{\frac{2 * I}{c * \epsilon _o} }

substituting values

          E_o =  \sqrt{\frac{2 * 6.495 *10^{-14}}{3.0 *10^{8} * 8.85*10^{-12}} }

           E_o = 6.995 *10^{-6} \ V/m

 

   

4 0
4 years ago
Which of the following electron configurations for neutral atoms is correct?
liq [111]
Nitrogen: 1s2, 2s2, 2p3 is correct answer.
6 0
3 years ago
Read 2 more answers
Which statement is true? A Displacement can never be greater than distance. B Displacement can never be less than distance. C Di
qwelly [4]

Answer:

__________________

I think the answer is A.

__________________

3 0
3 years ago
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