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Brrunno [24]
2 years ago
14

According to the article, which pattern of brain waves are most conducive to studying new information?

Physics
1 answer:
sashaice [31]2 years ago
7 0
Alpha brain waves are those most conducive to studying new information.

When consciously alert, we generally function along a beta brain rhythm. In diminishing this rhythm to alpha, we transition into a state of physical and mental relaxation that is ideal for learning new information and storing facts and  data. Studies have shown that the effect of decreasing brain rhythm is linked to feelings of increased mental clarity and remembrance. As it is a prime condition for synthetic thought and creativity, it becomes easier to visualize and create associations (information is better learned and absorbed by using such study methods). 

Hope this helps! :)
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A team of astronauts is on a mission to land on and explore a large asteroid. In addition to collecting samples and performing e
Blizzard [7]
<h2>Answer: 117.626m/s</h2>

Explanation:

The escape velocity V_{esc} is given by the following equation:

V_{esc}=\sqrt{\frac{2GM}{R}}   (1)

Where:

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M  is the mass of the asteroid

R  is the radius of the asteroid

On the other hand, we know the density of the asteroid is \rho=3.84(10)^{8}g/m^{3} and its volume is V=2.17(10)^{12}m^{3}.

The density of a body is given by:

\rho=\frac{M}{V}  (2)

Finding M:

M=\rhoV=(3.84(10)^{8} g/m^{3})(2.17(10)^{12}m^{3})  (3)

M=8.33(10)^{20}g=8.33(10)^{17}kg  (4)  This is the mass of the spherical asteroid

In addition, we know the volume of a sphere is given by the following formula:

V=\frac{4}{3}\piR^{3}   (5)

Finding R:

R=\sqrt[3]{\frac{3V}{4\pi}}   (6)

R=\sqrt[3]{\frac{3(2.17(10)^{12}m^{3})}{4\pi}}   (7)

R=8031.38m   (8)  This is the radius of the asteroid

Now we have all the necessary elements to calculate the escape velocity from (1):

V_{esc}=\sqrt{\frac{2(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(8.33(10)^{17}kg)}{8031.38m}}   (9)

Finally:

V_{esc}=117.626m/s This is the minimum initial speed the rocks need to be thrown in order for them never return back to the asteroid.

6 0
2 years ago
What principles of science (like facts, laws, and theories) might help explain why similar investigations conducted in many part
natulia [17]

Answer:

Reproducibility of research

Explanation:

The principle of science that explains why similar experimental investigations conducted in different parts of the world could result in the same outcome is referred to as reproducibility.

<em>A good research or experiment in science must be reproducible, otherwise, the outcome of such an experiment might become inadmissible within the scientific community. It is a core principle of the scientific method that similar results should be obtained when an experiment or observational study conducted in one place is repeated in another place with the same procedure. Hence, an experiment must be reproducible in science in order for the outcome of such an experiment to be part of the general scientific knowledge. </em>

7 0
2 years ago
Calculate the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s.
Aleonysh [2.5K]

De broglie wavelength, \lambda = \frac{h}{mv}, where h is the Planck's constant,  m is the mass and v is the velocity.

h = 6.63*10^{-34}

Mass of hydrogen atom,  m = 1.67*10^{-27}kg

v = 440 m/s

Substituting

   Wavelength \lambda = \frac{h}{mv} = \frac{6.63*10^{-34}}{1.67*10^{-27}*440} = 0.902 *10^{-9}m = 902 *10^{-12}m

1 pm = 10^{-12}m\\ \\ So , \lambda =902 pm

So  the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s is 902 pm

7 0
2 years ago
The famous cliff divers of Acapulco leap from a perch 35 m above the ocean. How fast are they moving when they reach the surface
Rus_ich [418]

1) 26.2 m/s

The mechanical energy of the divers at any point of their vertical motion is sum of the kinetic energy and the gravitational potential energy:

E=K+U = \frac{1}{2}mv^2 + mgh

where

m is the mass of the diver

v is the speed

g = 9.8 m/s^2 is the acceleration due to gravity

h is the height above the water

When the diver is on the cliff, v = 0 (he is at rest), so K=0 and the initial mechanical energy is just potential energy:

E_i = mgh

where h=35 m is the height of the cliff.

When the diver hits the water above, h = 0, so U=0 and the final mechanical energy is just kinetic energy:

E_f = \frac{1}{2}mv^2

since the total mechanical energy is conserved, we have

E_i = E_f\\mgh = \frac{1}{2}mv^2

And solving the equation for v, we find the speed when they reach the surface of the water:

v=\sqrt{2gh}=\sqrt{2(9.8 m/s^2)(35 m)}=26.2 m/s

2) It is converted into thermal energy of the water

When the diver enters the water, he suddenly feels another force acting against the motion of the diver: the resistance of the water. The resistance of the water acts upward, slowing down the diver until he stops.

In this process, the speed of the diver (v) decreases, and therefore the kinetic energy of the diver decreases as well, until it becomes zero.

However, this does not mean that the conservation of energy has been violated. In fact, the kinetic energy of the diver has been converted into thermal energy of the molecules of water surrounding the diver.

8 0
2 years ago
A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod
ivann1987 [24]

Answer:

The inducerd emf is 1.08 V

Solution:

As per the question:

Altitude of the satellite, H = 400 km

Length of the antenna, l = 1.76 m

Magnetic field, B = 8.0\times 10^{- 5}\ T

Now,

When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

e = - l(vec{v}\times \vec{B})

Here, velocity v is perpendicular to the rod

Thus

e = lvB           (1)

For the orbital velocity of the satellite at an altitude, H:

v = \sqrt{\frac{Gm_{E}}{R_{E}} + H}

where

G = Gravitational constant

m_{e} = 5.972\times 10^{24}\ kg = mass of earth

R_{E} = 6371\ km = radius of earth

v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}}{6371\times 1000 + 400\times 1000} = 7670.018\ m/s

Using this value value in eqn (1):

e = 1.76\times 7670.018\times 8.0\times 10^{- 5} = 1.08\ V

5 0
2 years ago
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