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3 years ago

14

when you see something, your brain processes the visual information so that you are aware of it. Most of this processing occurs

in the ____ mind

2 answers:

aliina [53]3 years ago

7 0

Answer:

Occipital Lobe of mind

Explanation:

The information that is received by our eyes are sent to the brain through the optic nerves connected to retina. This information is sent to the visual cortex of the brain, part of cerebral cortex responsible for processing visual information.

This visual cortex is situated inside the occipital lobe of the brain which is one of the 4 major lobes in the brain.

Natasha_Volkova [10]3 years ago

5 0

Front love of the brain
it focused on learning

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Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated

Artemon [7]

Answer: Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN

Explanation: To find the answer we need to know more about the Newton's law of gravitation.

<h3>What is Newton's law of gravitation?</h3>

Gravitation is the force of attraction between any two bodies.
Every body in the universe attracts every other body with a force.
This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between these two masses.
Mathematically we can expressed it as,

$F=\frac{GMm}{r^2} \\where, G=6.67*10^-11Nm^2kg^-2$

<h3>How to solve the problem?</h3>

Here, we have given with the data's,

$M=8.22*10^9kg\\m=1.38*10^8 kg\\r=1.43*10^3m$

Thus, the force of attraction between these two bodies will be,

$F=6.67*10^-11*\frac{8.22*10^9*1.38*10^8}{1.43*10^3} =52.9kN$

Thus, if two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg and, If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN.

Learn more about the Newton's law of gravitation here:

brainly.com/question/28045318

#SPJ4

6 0

2 years ago

What is SI unit for distance

Dafna11 [192]

Answer:

meter

The SI unit of distance and displacement is the meter [m].

Explanation:

have advancedd

3 0

3 years ago

Read 2 more answers

A 150kg person stands on a compression spring with spring constant 10, 000 N/m and norminal.Length of 0.50.What is the togal len

Eva8 [605]

Answer:

Total length of spring 0.647 m

Explanation:

We have given mass of the person m = 150 kg

Acceleration due to gravity $g=9.8m/sec^2$

Spring constant k = 10000 N/m

Nominal length of spring = 0.50

According to hook's law

$mg=kx$

$150\times 9.8=10000\times x$

x = 0.147 m

So total length of spring = 0.50+0.147 = 0.647 m

4 0

3 years ago

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m

nekit [7.7K]

Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

$F_T \cdot r = I \cdot \alpha = m \cdot r^2 \cdot \alpha$

Where;

$F_T$ = The tangential force

I = The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

$I_m \cdot \alpha_m = I_m \cdot \dfrac{v_m}{r \cdot t}$

$I_m$ = The rotational inertia of the merry-go-round

$\alpha _m$ = The angular acceleration of the merry-go-round

$v _m$ = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

$I_b \cdot \alpha_b = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Where;

$I_b$ = The rotational inertia of the boy

$\alpha _b$ = The angular acceleration of the boy

$v _b$ = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

$I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Which gives;

$v_m = \dfrac{m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2 \cdot v_b}{I_m}$

From which we have;

$v_m = \dfrac{20 \times 3^2 \times 5}{600} = 1.5$

The velocity of the merry-go-round, $v_m$ , after the boy hops on the merry-go-round = 1.5 m/s.

5 0

3 years ago

Light of wavelength 441.1 nm falls on two slits spaced 0.29 mm apart. What is the required distance from the slits to the screen

pshichka [43]

Answer: R = 0.131 m

Explanation: The formulae for the distance between a fringe and the first (center) is given by

y = R×mλ/d

Where y = distance between first and nth fringe = 4mm = 4×10^-3m

λ = wavelength of light = 441.1nm = 441.1×10^-9m

R = distance between slits and screen =?

d = distance between slits = 0.29mm = 2.9×10^-4m

4×10^-3 = R ×2 ×441.4×10^-9/ 2.9×10^-5

Hence R = (4×10^-3) ×(2.9×10^-5)/2×441.4×10^-9

R = 1.16 × 10^-7/8.828×10^-7

R = 1.16/8.828

R = 0.131 m

8 0

3 years ago