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3 years ago

With which part of the brain is awareness typically associated?

Physics

2 answers:

Kitty [74]3 years ago

8 0

Answer:

Cerebral Cortex

Explanation:

The cerebral cortex has a left and a right hemisphere. Each hemisphere can be divided into four lobes: the frontal lobe, temporal lobe, occipital lobe, and parietal lobe. The lobes are functional segments. They specialize in various areas of thought and memory, of planning and decision making, and of speech and sense perception.

weeeeeb [17]3 years ago

4 0

Answer:cerebral cortex

Explanation:

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What is the relationship between dominant and recessive genes

Triss [41]

Dominant genes and recessive genes are both given to a parents offspring. However, not both can be expressed causing the difference of dominant and recessive. Dominant genes are more likely to be expressed and recessive genes are more likely to be repressed.

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3 years ago

How would you go about measuring the speed of a vehicle? What measurements would you have to take? What calculations would you h

White raven [17]

Answer:

For a body moving at a uniform velocity you can calculate the speed by dividing the distance traveled by the amount of time it took, for example one mile in 1/2 hour would give you 2 miles per hour. If the velocity is non-uniform all you can say is what the average speed is.

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2 years ago

In an experiment, a large number of electrons are fired at a sample of neutral hydrogen atoms and observations are made of how t

kakasveta [241]

Answer:

N = 1036 times

Explanation:

The radial probability density of the hydrogen ground state is given by:

$p(r) = \frac{4r^{2} }{a_{0} ^{3} } e^{\frac{-2r}{a_{0} } }$

$p(\frac{a_{0} }{2} ) = \frac{4(\frac{a_{0} }{2} )^{2} }{a_{0} ^{3} } e^{\frac{-2(\frac{a_{0} }{2} )}{a_{0} } }$

$p(2a_{0} ) = \frac{4(2a_{0}) ^{2} }{a_{0} ^{3} } e^{\frac{-4a_{0} }{a_{0} } }$

$N = 1300\frac{p(2a_{0}) }{p(\frac{a_{0} }{2} )}$

$N = 1300\frac{(2a_{0}) ^{2}e^{\frac{-4a_{0} }{a_{0} } } }{(\frac{a_{0} }{2} )^{2} e^{\frac{-a_{0} }{a_{0} } }}$

$N = 1300(16) e^{-3}$

N = 1035.57

N = 1036 times

6 0

3 years ago

Running at 1.55 m/s, Bruce, the 40.0 kg quarterback, collides with Biff, the 90.0 kg tackle, who is traveling at 7.0 m/s in the

Margarita [4]

Answer:Bruce is knocked backwards at

Explanation:

This is a problem of momentum (

→

) conservation, where

→

and because momentum is always conserved, in a collision:

→

We are given that

, and

The momentum of Bruce (

) before the collision is given by

→

(

)

(

)

→

Similarly, the momentum of Biff (

) before the collision is given by

→

(

)

(

)

630

The total linear momentum before the collision is the sum of the momentums of each of the football players.

→

∑

→

630

720

Because momentum is conserved, we know that given a momentum of

720

before the collision, the momentum after the collision will also be

720

. We are given the final velocity of Biff (

) and asked to find the final velocity of Bruce.

→

Solve for

→

−

Using our known values:

720

−

(

)

(

)

∴

Bruce is knocked backwards at

Explanation:

5 0

2 years ago

A day on a distant planet observed orbiting a nearby star is 21.5 hr. Also, a year on the planet lasts 69.3 Earth days. In other

serg [7]

Answer:

Part A

The angular speed of rotation of the plane is 8.11781 × 10⁻⁵ rad/s

Part B

The angular speed of orbit of the planet is 1.04938 × 10⁻⁶ rad/s

Explanation:

The parameters of the planet are;

The duration of a day on the distant planet = 21.5 hr.

The duration of a year on the distant planet = 69.3 Earth days

Part A

The duration of a day = The time to make one complete revolution of 2·π radians

∴ The average angular speed about its axis, $\omega_{rotation}$ = Angle turned/Time

∴ $\omega_{rotation}$ = 2·π/(21.5 × 60 × 60) s ≈ 8.11781 × 10⁻⁵ rad/s

The average angular speed of the planet about its own axis, $\omega_{rotation}$ = 8.11781 × 10⁻⁵ rad/s

The angular speed of rotation of the plane $\omega_{rotation}$ = 8.11781 × 10⁻⁵ rad/s

Part B

The time it takes the planet to revolve round the neighboring star once = 69.3 Earth days

Therefore, the average angular speed of the planet around its neighboring star, $\omega _{Star}$ , is given as follows;

$\omega _{Orbit}$ = 2·π/((69.3 × 24 × 60 × 60) s) = 1.04938 × 10⁻⁶ rad/s

The average angular speed of orbit, $\omega _{Orbit}$ = 1.04938 × 10⁻⁶ rad/s

The angular speed of orbit of the planet, $\omega _{Orbit}$ = 1.04938 × 10⁻⁶ rad/s.

3 0

2 years ago