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3 years ago

which part of the brain is represented by the highlighted area shown below? cerebrum cerebellum medulla oblongata pons

Biology

2 answers:

Feliz [49]3 years ago

8 0

Answer:

The ans: is cerebellum

Explanation:

Leviafan [203]3 years ago

7 0

Cerebrum is in the most anterior part of our brain, fronting of the skull. It functions in sensory and neural processes. Cerebellum is found at the back of the skull. It functions in coordination, balance and muscular activity. Medulla oblongata connects from the spinal cord which controls heart and lungs. Pons is part of the brainstem which controls breathing and communication.

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An owl has good night vision because its eyes can detect a light intensity as small as 5.0 × 10-13 W/m2. What is the minimum num

Lady bird [3.3K]

Answer:

Thus, the minimum number of photons per second is 77.34

Explanation:

Light intensity, $I_{min}$ = $5\times 10^{-13} W/m^{2}$

Pupil has a diameter, d = 8.5 mm

= 8.5 x $10^{-3}$ m

Radius of the eye, r = 4.25 x $10^{-3}$ m

∴ Area of the eye, A = $\pi .r^{2}$

= $3.14\times \left ( 4.25\times 10^{-3} \right )^{2}$

= $5.6\times 10^{-5} m^{2}$

Let $P_{min}$ be the minimum number of photons.

Therefore, $P_{min}$ = $I_{min}$ x A

= $5\times 10^{-13}$ x $5.6\times 10^{-5}$

= $2.8\times 10^{-17}$ W

Thus the minimum number of photons is given by

$N_{min}=P_{min}/E$

where E = $hc/\lambda$

= $\left (6.63\times 10^{-34}\times 3\times 10^{8} \right )/548\times 10^{-9}$

= $3.62\times 10^{-19} J$

Therefore, $N_{min}$ = $\frac{2.8\times 10^{-17}}{3.62\times 10^{-19}}$

= 77.34 photons per second

Thus, the minimum number of photons per second is 77.34

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