Abnormal levels of __________ may lead to Alzheimer’s disease.
A.
endorphins
B.
serotonin
C.
GABA (gamma-aminobutyric acid)
D.
acetylcholine

Answer:

It is the core.

Answer:

Planets were like gods.

Explanation:

To the people of many ancient civilizations, the planets were thought to be deities. Our names for the planets are the Roman names for these deities. For example, Mars was the god of war and Venus the goddess of love.

Answer:

The correct answer is option A: 3.00 × 10^(-11) m.

Explanation:

To find the wavelength of a gamma ray with a frequency of about 10^19 s^(-1), we can use the equation:

wavelength = speed of light / frequency

Given:

Speed of light (c) = 3.00 × 10^8 m/s

Frequency (f) = 10^19 s^(-1)

Substituting the values into the equation:

wavelength = (3.00 × 10^8 m/s) / (10^19 s^(-1))

To simplify the expression, we can rewrite the denominator as (1 / 10^(-19)) s:

wavelength = (3.00 × 10^8 m/s) / (1 / 10^(-19)) s

To divide by a fraction, we multiply by its reciprocal:

wavelength = (3.00 × 10^8 m/s) × (10^(-19) s)

Applying the properties of exponents, we can add the exponents when multiplying with the same base:

wavelength = 3.00 × 10^(-11) m

Therefore, the wavelength of the gamma ray is approximately 3.00 × 10^(-11) m.

The wave speed is 840 cm/s and the fundamental frequency is 1120 Hz.

Frequency is the number of cycles of a periodic waveform that occur per unit of time. It is measured in Hertz (Hz).

We can use the equation λ=2L/n, where λ is the wavelength, L is the length of the string, and n is the harmonic number. Since the string has fixed ends, the harmonics must be odd-numbered, so we have n=1 for the fundamental frequency, n=3 for the second harmonic (315 Hz), and n=5 for the third harmonic (420 Hz).

Using n=1 and λ=2L/n, we get:

λ = 2L/1

λ = 2L

Using n=3 and λ=2L/n, we get:

λ = 2L/3

Using n=5 and λ=2L/n, we get:

λ = 2L/5

We can use the formula f=v/λ to relate the wave speed v, wavelength λ, and frequency f. For the two consecutive harmonics, we can write:

v/λ1 = f1

v/λ2 = f2

Since the two harmonics are consecutive, we can assume that they correspond to adjacent values of n, so we have:

λ1 = 2L/1 = 2L

λ2 = 2L/3

Substituting these values into the above equations and solving for v, we get:

v = f1λ1 = f2λ2 = (420 Hz)(2L) / (2L) = (315 Hz)(2L)/(2L/3) = 840 cm/s

To find the fundamental frequency, we use the formula f=v/λ1:

f = v/λ1 = 840 cm/s / 2L = (840 cm/s) / (0.75 m) = 1120 Hz

Therefore, the wave speed is 840 cm/s and the fundamental frequency is 1120 Hz.

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Answer:

Chemical symbols Al, O, H, O or a,c,f,h.

Coefficient 2  or e.

Number of atoms of the element - b,d ,g.

One atom of the element - i.

i.a,c,f and h are the chemical symbol.

A chemical symbol is a one- or two-letter designation of an element. Some examples of chemical symbols are O for oxygen, Zn for zinc, and Fe for iron.

For the diagram given above, the labelled parts which represent chemical symbol are: a, c, f and h

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ii.The labeled parts is a coefficient is e.

Coefficients are numbers written before the chemical symbol of elements or compound.

For the diagram given above, the labelled part which represent Coefficient is: e.

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iii.In the labeled parts is the number of atoms of the element are b,d,g and i.

Number of atoms of element present in a compound is simply obtained by taking note of the numbers written as subscript in the chemical formula of the compound.

For the diagram given above, the labelled part which represent the number of atoms of the element are: b, d, g and i.

iv.The labeled parts that indicates only one atom of the element is present in the substance is i

When no number is written as subscript in the formula of the element in the compound, it means the element has just 1 atom in the compound.

For the diagram given above, the labelled part which indicates that only 1 atom of the element is present is: i

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Answer:

It is because one cannot know exactly the position of the electron within the atom.

One formulation of Heisenberg's Uncertainty Principle tells us that one cannot know simultaneously  the position and momentum of the electron, so one cannot specify exactly either coordinate because the other would be infinite.

Bohr specified the most probable position of the electron at its lowest energy level in hydrogen and the product of the two would be about the Heisenberg value.  

The interval between the first and second echo is 7 seconds. This means that after the initial sound wave reaches the first cliff, it takes a total of 7 seconds for the sound to travel to the second cliff and then return to the student as the second echo.

To determine the interval between the first and second echo, we need to consider the time it takes for sound to travel from the student to the first cliff, and then from the first cliff to the second cliff, and finally back to the student.

Let's break down the distances and calculate the time for each part of the journey:

Distance from the student to the first cliff: 330 meters

Time taken: t₁ = distance / speed = 330 m / 330 m/s = 1 second

Distance from the first cliff to the second cliff: 990 meters

Time taken: t₂ = distance / speed = 990 m / 330 m/s = 3 seconds

Distance from the second cliff back to the student: 990 meters

Time taken: t₃ = distance / speed = 990 m / 330 m/s = 3 seconds

Now, we can calculate the total interval between the first and second echo by adding up the individual times:

Interval between first and second echo = t₁ + t₂ + t₃ = 1 s + 3 s + 3 s = 7 seconds

Therefore, the interval between the first and second echo is 7 seconds. This means that after the initial sound wave reaches the first cliff, it takes a total of 7 seconds for the sound to travel to the second cliff and then return to the student as the second echo.

It's important to note that this calculation assumes a straight path for the sound waves and neglects factors such as air temperature and wind that can affect the speed of sound. Additionally, it assumes perfect reflection of sound waves off the cliffs, which may not be the case in real-world scenarios.

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Note the complete questions is:

A student stands 330m from a tall cliff which is 990m from another tall cliff. If the speed of sound between the cliffs is 330m/s.What is the interval between the first and second echo?

The reflection in a clear window of a store is a(n) image.

Images are seen as reflections because of the behavior of light. When light strikes a smooth, reflective surface such as a mirror or still water, it bounces off the surface at the same angle at which it hit it. This process is called reflection. The reflected light rays then travel to our eyes, creating an image.

The angle of incidence (the angle at which the light strikes the surface) is equal to the angle of reflection (the angle at which the light bounces off the surface). This causes the reflected image to be a mirror image of the original object.

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Answer:

1. Power = 5000 Watts

2. Workdone = 11185.5 Joules

Explanation:

Given the following data;

1. Distance = 18 m

Energy = 100 KJ = 100,000 Joules

Time = 20 seconds

To find the average power of the elevator;

Power = energy/time

Power = 100000/20

Power = 5000 Watts

2. Power = 0.4 HP

Time = 15 seconds

Conversion:

1 horsepower = 745.7 Watts

0.4 horsepower = 0.4 * 745.7 = 298.28 Watts

To find the amount of work done by the electric mixer;

Work done = power * time

Workdone = 745.7 * 15

Workdone = 11185.5 Joules

The average speed of hair growth is 9.7994×10⁻⁹ m/s. So I cut 1 inch per month to achieve this speed.

If a number is too big or too small to be conveniently represented in decimal form, it can be expressed using scientific notation. Scientific form or standard index form are two terms that might be used to describe it.

Let us assume that the growth of our hair is synonymous with the velocity of the tip of the hair. First, convert the estimated growth rate to m/s.

Let's assume the growth of the hair occurs at a rate of 1 inch/month. Then, converting this, we get,

\(\begin{aligned}\mathrm{\frac{1\;inch}{month}}&=\mathrm{\left(\frac{1\;inch}{month}\right)\left(\frac{1\;month}{30\;days}\right)\left(\frac{1\;day}{86400\;seconds}\right)\left(\frac{0.0254\;m}{1\;inch}\right)}\\&=\mathrm{\frac{0.0254\;m}{30\times86400\;seconds}}\\&=\mathrm{9.7994\times10^{-9}\;m/s}\end{aligned}\)

Therefore, the average speed of hair growth is 9.7994×10⁻⁹ m/s.

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Answer:

350 N

Explanation:

F=ma

\(f = force \\ m = mass \\ a = acceleration\)

\(m = 2(55kg) + 240kg \\ a = 1.0 \frac{m}{ {s}^{2} } \)

Force = 350 Newtons

The net force acting on the elevator would be 350 Newtons as it accelerates upward at 1.00 m/s2.

Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.

The mathematical expression for Newton's second law is as follows

F = ma

As given in the problem two people each have a mass of 55 kg. They are both in an elevator that has a mass of 240 kg. When the elevator begins to move, the people and the elevator have an upward acceleration of 1.00 m/s2, then we have to find the net force acting on the elevator,

The net force acting on the elevator,

F = ma

F =(2×55 + 240)×1

  = 350 Newtons

Thus, the net force acting on the elevator would be 350 Newtons as it accelerates upward at 1.00 m/s2

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Answer:

You would have to do four times the work W₁ to make the box go the same distance in half the time.

Explanation:

By definition work is:

\( W = F*d \)    (1)

Where:

F: is the force and d: is the distance

On the other hand, the force is given by:              

\( F = ma \)    (2)

By entering equation (2) into (1):

\( W = F*d = m*a*d \)     (3)  

Also, we can express the acceleration in terms of distance and time by using the following kinematic equation:

\( d_{f} = d_{0} + v_{0}t + \frac{1}{2}at^{2} \)    (4)

Where:

\(d_{f}\): is the final distance

\(d_{0}\): is the initial distance = 0

\(v_{0}\): is the initial speed = 0 (the box starts at rest)

t: is the time  

\( d_{f} = \frac{1}{2}at^{2} \)  

Solving for a:

\( a = \frac{2d_{f}}{t^{2}} \)   (5)  

Initially we have:

t = T₁, W = W₁, \(d_{f}\) = d      

\( a_{1} = \frac{2d}{T_{1}^{2}} \)

\( W_{1} = m*a_{1}*d = m*d*\frac{2d}{T_{1}^{2}} = 2m(\frac{d}{T_{1}})^{2} \)

And when the box goes the same distance in half the time:

\( t = \frac{T_{1}}{2} \)    

The acceleration is (from equation 5):

\(a_{2} = \frac{2d_{f}}{t^{2}} = \frac{2d}{(\frac{T_{1}}{2})^{2}} = \frac{8d}{T_{1}^{2}}\)    (6)        

Finally, the work W₂ in terms of W₁ is:

\(W_{2} = m*a_{2}*d = m*d*\frac{8d}{T_{1}^{2}} = 8m(\frac{d}{T_{1}})^{2} = 4*[2m(\frac{d}{T_{1}})^{2}] = 4W_{1}\)    

Therefore, you would have to do four times the work W₁ to make the box go the same distance in half the time.

I hope it helps you!    

Answer:

30 newtons

explanations

data given

mass=6kg

acceleration=5

f=m×a

6×5=30

Waves can travel through the air, granite rock, water, sandstone, and mudstone. Waves can also travel through molten magma, but only in certain conditions.

In physics, a wave is a disturbance that travels through space and time, usually accompanied by the transfer of energy. Waves can be characterized by their amplitude, wavelength, frequency, and speed.

Here,

Waves can travel through the air, granite rock, water, sandstone, and mudstone. Waves can also travel through molten magma, but only in certain conditions.

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The distance formula used for a falling object is y = 1/2gt^2 where g is gravity which is 9.80m/s^2 and t is the time it falls ( reaction time).

Y = 1/2(9.8)0.2^2

Y = 0.196

It travels 0.196 meters

Explanation:

The most established layers are on the base, and the most youthful layers are on the top. Since dregs here and there incorporate once-living creatures, sedimentary stone regularly contains a great deal of fossils. Fossils are once living beings that have been transformed into rock, fit as a fiddle or type of the creature can in any case be seen.

Answer:

A

Explanation:

A constant velocity means the position graph has a constant slope. It's a straight line sloping up.

Answer:Force is to the right

Explanation: because the right side has 75N compared to the 25N on the left.

Answer:

Explanation:

As temperature is constant , we shall apply Boyle's law

P₁V₁ = P₂V₂

P₁ = pressure at depth of 10 m

= P + hdg , h = 10 , d = 10³ , g = 10

P is atmospheric pressure which is 10⁵ Pa

P₁ = 10⁵ + 10 x 10³ x 10

= 2 x 10⁵

applying the formula

2 x 10⁵ x 6 = 10⁵ x v

v = 2 x 6 = 12 L

volume will be doubled at the surface .

B )

warming of air at the surface will increase the volume of air in her lungs so so she will need more lung capacity .

C )

The rms value of a gas depends upon the temperature of the gas . As temperature of the gas is constant , the rms value of the gas particles will remain constant when she goes to the surface .

The lungs will expand 12 L when she reaches the surface of the water, and the warming of the air results in more lung capacity, and \(\rm V_{rms}\) the value remains the same.

According to the law, the pressure of the gas is inversely proportional to the volume of the gas. In other words when the pressure of the gas increases the volume of the gas decreases.

We know the pressure at the 10 meters depth:

\(\rm P_1 = P+h\times \rho\times g\)

Where P = Atmospheric pressure

            h = Depth

            ρ =Density of the water

We have: \(\rm P = 10^5 \ Pa\), h = 10 meters, and \(\rm \rho = 1000 \ kg/m^3\), and \(\rm g = 10 \ m/s^2\)

Putting the values in the above equation, we get:

\(\rm P_1 = 10^5+ 10\times 1000\times 10\)

\(\rm P_1 = 2\times 10^5\)

From the Boyle's law:

\(\rm P_1\times V_1 = P_2\times V_2\)

\(\rm 2\times10^5\times 6 = 10^5\times V_2\)

\(\rm V_2 = 12 \ L\)

We know that as the air at the surface warms, the volume of air in her lungs expands, requiring more lung capacity.

The temperature of the gas is constant and \(\rm V_{rms}\) values for gas depend on the temperature of the gas, but here the temperature of the gas is constant thus, the  \(\rm V_{rms}\) will remains constant.

Thus, the lungs will expand 12 L when she reaches the surface of the water, and the warming of the air results in more lung capacity, and \(\rm V_{rms}\) the value remains the same.

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Answer:

Earth rotates on its axis from west to east approximately every 24 hours

Explanation:

hope this helps :)

Answer:

The velocity of the box is related to the angular velocity of the spool, which is given by the equation:

v = r * ω

where r is the radius of the spool and ω is the angular velocity of the spool. The angular velocity of the spool, in turn, is related to the torque applied to the spool by the tension in the string, which is given by the equation:

τ = I * α

where τ is the torque, I is the moment of inertia of the spool, and α is the angular acceleration of the spool.

The tension in the string is equal to the weight of the box, which is given by:

T = m * g

Putting all of these equations together, we can solve for the time it takes for the box to reach the floor. Here's how:

First, we can find the angular acceleration of the spool using the torque equation:

τ = I * α

T = m * g = τ

m * g = I * α

α = (m * g) / I

α = (35.30 kg * 9.81 m/s^2) / 4.00 kg*m^2

α = 86.53 rad/s^2

Next, we can find the angular velocity of the spool using the kinematic equation:

ω^2 = ω_0^2 + 2 * α * θ

where ω_0 is the initial angular velocity (which is zero), θ is the angle through which the spool has turned (which is equal to the distance the box has fallen divided by the radius of the spool), and ω is the final angular velocity (which is what we want to find). Solving for ω, we get:

ω^2 = 2 * α * θ

ω = sqrt(2 * α * θ)

ω = sqrt(2 * 86.53 rad/s^2 * (3.50 m / 0.10 m))

ω = 166.6 rad/s

Finally, we can find the time it takes for the box to reach the floor using the equation:

v = r * ω

v = 0.10 m * 166.6 rad/s

v = 16.66 m/s

t = d / v

t = 3.50 m / 16.66 m/s

t = 0.21 s

The skater's final angular velocity is approximately 9.86 rad/s.

The skater's final angular velocity can be calculated using the principle of conservation of angular momentum. The equation for angular momentum is given by:

L = Iω

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Initially, the skater has an angular momentum of:

L_initial = I_initial * ω_initial

Substituting the given values:

L_initial = 2.12 kg m² * 3.25 rad/s

The skater's final angular momentum remains the same, as angular momentum is conserved:

L_final = L_initial

The final moment of inertia is given as 0.699 kg m². Therefore, the final angular velocity can be calculated as:

L_final = I_final * ω_final

0.699 kg m² * ω_final = 2.12 kg m² * 3.25 rad/s

Solving for ω_final:

ω_final = (2.12 kg m² * 3.25 rad/s) / 0.699 kg m²

Hence, the skater's final angular velocity is approximately 9.86 rad/s.

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The time elapsed between the first splash and the second splash is approximately 0.69 seconds.

To calculate this, we consider the motion of two rocks thrown simultaneously from a bridge. Heather throws a rock straight down with a speed of 14 m/s, while Jerry throws a rock straight up with the same speed.

We use the equation for displacement in uniformly accelerated motion: s = ut + (1/2)at^2.

For Heather's rock, which is thrown downwards, the initial velocity (u) is positive and the acceleration (a) due to gravity is negative (-9.8 m/s^2). The displacement (s) is the height of the bridge (46 m).

Solving the equation, we find two possible values for the time (t): t ≈ -4.91 s and t ≈ 1.91 s.

Since time cannot be negative in this context, we discard the negative value and consider t ≈ 1.91 s as the time it takes for Heather's rock to hit the water.

For Jerry's rock, thrown upwards, we use the same equation with the same initial velocity and acceleration. The displacement is also the height of the bridge, but negative.

Solving the equation, we find t ≈ -5.68 s and t ≈ 1.22 s. Again, we discard the negative value and consider t ≈ 1.22 s as the time it takes for Jerry's rock to reach its maximum height before falling back down.

To find the time difference between the first and second splash, we subtract t ≈ 1.91 s (Heather's rock) from t ≈ 1.22 s (Jerry's rock). This gives us a time difference of approximately 0.69 seconds.

Therefore, the time elapsed between the first splash and the second splash is approximately 0.69 seconds.

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Answer:

Approximately \(0.10\), assuming that \(g = 9.81\; {\rm N\cdot kg^{-1}}\) and a level surface.

Explanation:

Under the assumption, the normal force between the ice and the hockey puck is equal to the weight of the puck:

\(\begin{aligned}m\, g &= (0.16\; {\rm kg})\, (9.81\; {\rm N\cdot kg^{-1}}) \\ &\approx 1.57\; {\rm N}\end{aligned}\).

The friction on the puck is considered "static" when as long as the puck is not moving relative to that surface. In this question, the maximum value of this static friction is \(0.157\; {\rm N}\). When the external horizontal force exceeds \(0.157\; {\rm N}\!\), the puck would start moving relative to the ice.

Divide maximum static friction by the normal force to find the coefficient of static friction:

\(\begin{aligned}\mu_{s} &= \frac{(\text{maximum static friction})}{(\text{normal force})} \\ &\approx \frac{0.157\; {\rm N}}{1.57\; {\rm N}} \\ &\approx 0.10\end{aligned}\).

The speed of the artillery shell when it hits its target is 100 m/s.

Given:

Initial vertical displacement (y) = 200 m

Vertical displacement at 100 m in the air (y') = 100 m

Final velocity in the vertical direction (vy') = 0 m/s (at the highest point of the trajectory)

Using the equation for vertical displacement in projectile motion:

y' = vy^2 / (2g),

where g is the acceleration due to gravity (approximately 9.8 m/s^2), we can solve for the initial vertical velocity (vy).

100 m = vy^2 / (2 * 9.8 m/s^2),

vy^2 = 100 m * 2 * 9.8 m/s^2,

vy^2 = 1960 m^2/s^2,

vy = sqrt(1960) m/s,

vy ≈ 44.27 m/s.

Now, since the horizontal motion is independent of the vertical motion, the horizontal speed of the shell remains constant throughout its trajectory. Therefore, the speed of the shell when it hits its target is 100 m/s.

Hence, the speed of the artillery shell when it hits its target is 100 m/s.

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Answer:

Carbohydrates are divided into four types: monosaccharides, disaccharides, oligosaccharides, and polysaccharides. Monosaccharides consist of a simple sugar; that is, they have the chemical formula C 6 H 12 O 6. Disaccharides are two simple sugars. Oligosaccharides are three to six monosaccharide units, and polysaccharides are more than six.

Answer: By dividing airflow by duct cross section.

Explanation:

In short, air velocity in the ducts is calculated by dividing airflow by duct cross-section. Airflow is expressed as a simple number. Example: Air conditioner has a max. airflow of 600 CFM.

The velocity of the ladder when the base of the ladder is 7, 15, and 24 feet from the wall is -0.875, -2.25, and -10.286 respectively.

Given, the length of the ladder= 25 ft

Let x be the distance between the wall and the ladder and h be the height of the ladder from the ground

x² + h² = (25)²

x² +h²= 625

h²= 625- x²

h= \((625- x^{2})^{1/2}\)

\(\frac{dh}{dx}\) = \(\frac{1}{2}\) \((625- x^2)^{-1/2} * (-2x)\)

\(\frac{dh}{dx}\) =  \(\frac{-x}{\sqrt{ (625-x^2)}}\)

We have dx/dt= 3 ft/sec

Therefore, by applying the chain rule, we can get the value of dh/dt as:

dh/dt= dh/dx *dx/dt

=\(\frac{-x}{\sqrt{ (625-x^2)}}*3\)

=\(\frac{-3x}{\sqrt{ (625-x^2)}}\)

Now substituting the different values of x in the acquired equation

When x= 7, dh/dt= \(\frac{-7}{8}\) ft/sec= -0.875 ft/sec

When x= 15, dh/dt= \(\frac{-9}{4}\) ft/sec= -2.25 ft/sec

When x= 24, dh/dt= \(\frac{72}{7}\) ft/sec= -10.286 ft/sec

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Answer:

The work done by friction was \(-4.5\times10^{5}\ J\)

Explanation:

Given that,

Mass of car = 1000 kg

Initial speed of car =108 km/h =30 m/s

When the car is stop by brakes.

Then, final speed of car will be zero.

We need to calculate the work done by friction

Using formula of work done

\(W=\Delta KE\)

\(W=K.E_{f}-K.E_{i}\)

\(W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{f}^2\)

Put the value of m and v

\(W=0-\dfrac{1}{2}\times1000\times(30)^2\)

\(W=-450000 \ J\)

\(W=-4.5\times10^{5}\ J\)

Hence, The work done by friction was \(-4.5\times10^{5}\ J\)

The acceleration is 0, this means there is no net force acting on the diver as he comes to rest. The net force is equal to zero.

To determine the net force on the diver, we can use the equation of motion:

(Final velocity)² = (Initial velocity)² + 2 × acceleration × distance

The initial velocity is 0 since the diver is at rest on the diving board. The final velocity is also 0 since the diver comes to rest in the water. The distance fallen is 3.0 m.

We can solve for the acceleration using the given time:

0 = 0² + 2 × acceleration × 3.0

0 = 0 + 6.0 × acceleration

acceleration = 0 m/s²

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Answer:

1,143 N

Explanation:

m = 82 kg

d = 3 m

t = 0.55 s

Vf² = Vi² + 2ad

Vf² = 0 + 2(9.8 m/s²)(3 m) = 58.8 m²/s²

Vf = √58.8 m²/s² = 7.67 m/s

Use the Impulse Force equation:

FΔt = mΔv

F = (82 kg)(7.67 m/s) / (0.55 s) ≈ 1143 N