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3 years ago

How can a small human retina detect objects larger than itself?

2 answers:

8 0

The best answer is C."the brain interprets the size of the images"
Because, light reflects off the things in the world, then is focussed by our eye's lens. The light changes the light image we are seeing from the distal stimulus(or 3 diminutional world) to the proximal stimulus(or 2 dimensional world), but the image is not actually interpreted until it passes through the retina and optic nerve. The retina converts light into electrical impulses and the optic nerve carries those impulses to the brain. The Brain then changes the electrical impulses into a visual image.

aliya0001 [1]3 years ago

7 0

Exactly the same way that you can photograph a mountain or a skyscraper
with the itty bitty camera in your smartphone.

Lenses are used to form a tiny image of a gigantic object.

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Two identical tiny spheres of mass m =2g and charge q hang from a non-conducting strings, each of length L = 10cm. At equilibriu

Citrus2011 [14]

Answer:

0.247 μC

Explanation:

As both sphere will be at the same level at wquilibrium, the direction of the electric force will be on the x axis. As you can see in the picture below, the x component of the tension of the string of any of the spheres should be equal to the electric force of repulsion. And its y component will be equal to the weight of one sphere. We can use trigonometry to find the components of the tensions:

$F_y: T_y - W = 0\\T_y = m*g = 0.002 kg *9.81m/s^2 = 0.01962 N$

$T_y = T_*cos(50)\\T = \frac{T_y}{cos(50)} = 0.0305 N$

$T_x = T*sin(50) = 0.0234 N$

The electric force is given by the expression:

$F = k*\frac{q_1*q_2}{r^2}$

In equilibrium, the distance between the spheres will be equal to 2 times the length of the string times sin(50):

$r = 2*L*sin(50) = 2 * 0.1m * sin(50) 0.1532 m$

And k is the coulomb constan equal to 9 *10^9 N*m^2/C^2. q1 y q2 is the charge of each particle, in this case, they are equal.

$F_x = T_x - F_e = 0\\T_x = F_e = k*\frac{q^2}{r^2}$

$q = \sqrt{T_x *\frac{r^2}{k}} = \sqrt{0.0234 N * \frac{(0.1532m)^2}{9*10^9 N*m^2/C^2} } = 2.4704 * 10^-7 C$

O 0.247 μC

8 0

3 years ago

A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.

MakcuM [25]

Answer:

b. 9.5°C

Explanation:

$m_i$ = Mass of ice = 50 g

$T_i$ = Initial temperature of water and Aluminum = 30°C

$L_f$ = Latent heat of fusion = $3.33\times 10^5\ J/kg^{\circ}C$

$m_w$ = Mass of water = 200 g

$c_w$ = Specific heat of water = 4186 J/kg⋅°C

$m_{Al}$ = Mass of Aluminum = 80 g

$c_{Al}$ = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

$m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C$