I need help with one through six please

Answer:

heart, lungs, blood vessels

Explanation:

Answer:

d

Explanation:

im pretty sure. just makes the best sense. If im wrong im very sorry.

Answer:

D

Explanation:

because its right

Gravitational potential energy increases as weight and height increases.

Potential energy is stored or conserved energy in a material or object. The amount of stored energy depends on the arrangement, organization, and state of the object or substance.

You might think of it as energy with the "potential" to be productive. Any changes to the object's position, configuration, or status will result in the release of any stored energy.

For instance, in order to compress a spring, energy is required, but what happens to that energy once the spring has been broken It is impossible to generate or destroy energy because it can only transform from one form to another.

The kinetic energy that was used to compress our spring has now been converted to potential energy

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The final temperature of the air is (c) 193°C.

Explanation:-

Given,

The initial pressure, P1 = 1 MPa

The initial temperature, T1 = 400 °C

The final pressure, P2 = 110 kPa

The polytropic exponent, n = 1.2

We need to determine the final temperature of the air.

Solution:

For the polytropic process with a given polytropic exponent n, the work done can be given as;

W = P1V1 (1 - n) / (n - 1) [1 - (P2 / P1) ((n - 1) / n)] -------------- [1]

Where,V1 = (mRT1) / P1 ------------- [2]

V2 = (mRT2) / P2 ------------- [3]

Combining equations [2] and [3], we get;

V2 / V1 = P1 / P2 * T2 / T1T2

= T1 (V2 / V1) * (P2 / P1) --------------- [4]

Now, substituting equation [2] into equation [1],

we get;

W = (mRT1) / P1 * (1 - n) / (n - 1) [1 - (P2 / P1) ((n - 1) / n)]

Also, substituting equation [4] into equation [1],

we get;

W = (mR T1) / P1 (1 - n) / (n - 1) [1 - (P2 / P1) ((n - 1) / n)]T2

= T1 (1 - n) / n [1 - (P2 / P1) (n - 1)] ------------ [5]

where m is the mass of the gas and R is the gas constant.

For air, the value of R is 0.287 kJ/kg.K.

Substituting the values in equation [5],

we get;T2 = 400 × (1 - 1.2) / 1.2 [1 - (110 / 1000) (1.2 - 1)]

= -220.34 K (-53.81°C)

However, the final temperature of the air cannot be negative.

Therefore, the process must be irreversible or isothermal.

Thus, the option (c) 193°C is the correct answer.

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The event that marks the end of a star's evolutionary life before becoming a white dwarf is a planetary nebula. So, option E is correct.

A planetary nebula is characterized by cosmic rays of gas and dust surrounding a dying star. This dying star becomes a white dwarf after the complete depletion of hydrogen in the core.

It gets its name from a scientist who described the gases to be looking like two planets around a dying star. Thus, it's called planetary nebulae. This marks the final stage of a dying star that becomes a white dwarf. It consists of the outer layers of the star.

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Answer:

When fired from (1,3) the rocket will hit the target at (4,0)

When fired from (2, 2.5) the rocket will hit the target at (12,0)

When fired from (2.5, 2.4) the rocket will hit the target at \((\frac{35}{2},0)\)

When fired from (4,2.25) the rocket will hit the target at (40,0)

Explanation:

All of the parts of the problem are solved in the same way, so let's start with the first point (1,3).

Let's assume that the rocket's trajectory will be a straight line, so what we need to do here is to find the equation of the line tangent to the trajectory of the airplane and then find the x-intercept of such a line.

In order to find the line tangent to the graph of the trajectory of the airplane, we need to start by finding the derivative of such a function:

\(y=2+\frac{1}{x}\)

\(y=2+x^{-1}\)

\(y'=-x^{-2}\)

\(y'=-\frac{1}{x^{2}}\)

so, we can substitute the x-value of the given point into the derivative, in this case x=1, so:

\(y'=-\frac{1}{x^{2}}\)

\(y'=-\frac{1}{(1)^{2}}\)

m=y'=-1

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

\(y-y_{1}=m(x-x_{1})\)

\(y-3=-1(x-1})\)

\(y-3=-1x+1\)

\(y=-x+1+3\)

\(y=-x+4\)

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

\(-x+4=0\)

and solve for x

x=4

so, when fired from (1,3) the rocket will hit the target at (4,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2, 2.5)

so, we can substitute the x-value of the given point into the derivative, in this case x=2, so:

\(y'=-\frac{1}{x^{2}}\)

\(y'=-\frac{1}{(2)^{2}}\)

\(m=y'=-\frac{1}{4}\)

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

\(y-y_{1}=m(x-x_{1})\)

\(y-2.5=-\frac{1}{4}(x-2})\)

\(y-2.5=-\frac{1}{4}x+\frac{1}{2}\)

\(y=-\frac{1}{4}x+\frac{1}{2}+\frac{5}{2}\)

\(y=-\frac{1}{4}x+3\)

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

\(-\frac{1}{4}x+3=0\)

and solve for x

x=12

so, when fired from (2, 2.5) the rocket will hit the target at (12,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2.5, 2.4)

so, we can substitute the x-value of the given point into the derivative, in this case x=2.5, so:

\(y'=-\frac{1}{x^{2}}\)

\(y'=-\frac{1}{(2.5)^{2}}\)

\(m=y'=-\frac{4}{25}\)

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

\(y-y_{1}=m(x-x_{1})\)

\(y-2.4=-\frac{4}{25}(x-2.5})\)

\(y-2.4=-\frac{4}{25}x+\frac{2}{5}\)

\(y=-\frac{4}{25}x+\frac{2}{5}+2.4\)

\(y=-\frac{4}{25}x+\frac{14}{5}\)

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

\(-\frac{4}{25}x+\frac{14}{5}=0\)

and solve for x

\(x=\frac{35}{20}\)

so, when fired from (2.5, 2.4) the rocket will hit the target at \((\frac{35}{2},0)\)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (4, 2.25)

so, we can substitute the x-value of the given point into the derivative, in this case x=4, so:

\(y'=-\frac{1}{x^{2}}\)

\(y'=-\frac{1}{(4)^{2}}\)

\(m=y'=-\frac{1}{16}\)

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

\(y-y_{1}=m(x-x_{1})\)

\(y-2.25=-\frac{1}{16}(x-4})\)

\(y-2.25=-\frac{1}{16}x+\frac{1}{4}\)

\(y=-\frac{1}{16}x+\frac{1}{4}+2.25\)

\(y=-\frac{1}{16}x+\frac{5}{2}\)

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

\(-\frac{1}{16}x+\frac{5}{2}=0\)

and solve for x

\(x=40\)

so, when fired from (4,2.25) the rocket will hit the target at (40,0)

I uploaded a graph that represents each case.

its c                                     tjrxjxfjzjzrjrzjrzjkxkrzizikxikskiskskioxi

Answer:

it's C velocity

Explanation:

The total kinetic energy of the system before the collisions was equal to the total kinetic energy of the system after the collisions.

It appears that both conservation of momentum and conservation of energy hold across all trials regardless of initial conditions. This can be inferred from the fact that the elastic collisions were perfectly elastic, meaning that there was no loss of kinetic energy during the collisions. As a result, the system's total kinetic energy before the collisions was equal to the system's total kinetic energy after the collisions.

As for the conservation of momentum, this can be confirmed by calculating the momentum of the system before and after each collision and comparing the results. In a perfectly elastic collision, the total momentum of the system is conserved, which means that the momentum before the collision is equal to the momentum after the collision.

There do not appear to be any significant patterns based on the information provided regarding whether higher mass or faster velocities did a better job of showing momentum or kinetic energy conservation. However, it is important to note that in a perfectly elastic collision, both momentum and kinetic energy are conserved regardless of the initial conditions of the system.

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The equations mentioned by Emily in the valuation analysis for Aspeon are as follows:

1) Equation (1): This equation represents the value of equity (S) and calculates it based on the EBIT (earnings before interest and taxes), the tax shield provided by debt (D), and the required return on debt (kd) and equity (ks). It implies that the value of equity is equal to the adjusted EBIT after deducting the tax shield from debt.

2) Equation (2): This equation calculates the total enterprise value (V) by adding the value of equity (S) and debt (D). It represents the total worth of the company, considering both equity and debt.

3) Equation (3): This equation calculates the price per share (P) by dividing the total enterprise value (V) minus the initial debt (D0) by the number of shares (n0). It represents the price per share based on the valuation of the company.

4) Equation (4): This equation calculates the new number of shares (n1) by subtracting the dividend (D) from the initial number of shares (n0) divided by the price per share (P). It represents the adjusted number of shares after the payment of dividends.

5) Equation (5): This equation calculates the levered value (VL) by adding the unlevered value (VU) with the tax shield value (TD). It represents the value of the company after considering the tax advantages of debt.

These equations provide a framework for valuation analysis, considering factors such as earnings, taxes, debt, and equity. They help assess the value and financial implications of Aspeon's growth prospects.

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At 0.0804 km altitude the 1% of the mass of the atmosphere lies above and 99% of the mass lies below. Assumptions made are the global mean surface pressure of 1000 hPa is a representative value and the scale height is assumed to be constant throughout the entire atmosphere.

First, we need to calculate the pressure at the desired percentiles (1% and 99%) relative to the surface pressure.

For 1% of the mass lying above, we consider the pressure to be 1% of the surface pressure:

1% of 1000 hPa = 0.01 × 1000 hPa

= 10 hPa.

For 99% of the mass lying below, we consider the pressure to be 99% of the surface pressure:

99% of 1000 hPa = 0.99 × 1000 hPa

= 990 hPa.

Next, we use the exponential relationship between pressure and altitude:

P = P0 × exp(-z/H),

where P is the pressure at a given altitude, P0 is the surface pressure, z is the altitude, and H is the scale height.

To find the altitude z at which the pressure is equal to 10 hPa (1% of the surface pressure), we rearrange the equation:

10 hPa = 1000 hPa × exp(-z/H).

Taking the natural logarithm (ln) of both sides, we have:

ln(10 hPa / 1000 hPa) = -z / H.

ln(0.01) = -z / 8 km.

z = -8 km × ln(0.01).

Evaluating the expression:

z = -8 km × (-4.605)

= 36.84 km.

Therefore, 1% of the mass of the atmosphere lies above an altitude of approximately 36.84 km.

Similarly, to find the altitude z at which the pressure is equal to 990 hPa (99% of the surface pressure), we follow the same procedure:

990 hPa = 1000 hPa × exp(-z/H).

ln(990 hPa / 1000 hPa) = -z / H.

ln(0.99) = -z / 8 km.

z = -8 km × ln(0.99).

Evaluating the expression:

z = -8 km × (-0.01005)

= 0.0804 km.

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Answer:

3040 m

Explanation:

The formula to apply here is ;

d=∅*r  where

d is the diameter of the spot , ?

∅ is angle of divergence in radians, = 0.8 * 10⁻⁵rad

r is distance of Moon from Earth, =380,000 km

Applying the formula as;

d=∅*r

d= {0.8 *10⁻⁵} * 380,000 * 1000

d= 3040 m

Observing galaxies at different distances provides valuable information about how galaxies existed in different periods of time. When we look at galaxies that are farther away, we are also looking back in time because the light from those galaxies took a long time to reach us. This allows us to study galaxies as they existed in the past.

By analyzing the properties of these distant galaxies, such as their size, shape, and composition, we can gain insights into the early stages of galaxy formation and evolution. For example, studying galaxies at different distances helps us understand how galaxies have grown and changed over billions of years.

Observations of galaxies at various distances also allow us to study the expansion of the universe. The light from distant galaxies can reveal information about the rate at which the universe has been expanding over time.

Overall, observing galaxies at different distances provides a glimpse into the past and helps us piece together the puzzle of how galaxies have evolved and shaped the universe as we know it today.

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Answer:

James is traveling at 4 miles per hour.

Explanation:

3 miles per 45 minutes

x miles per 60 minutes(1 hour)

___________________

3*60=45*x

180=45x |:45

x=4

Answer:

The magnet produces an electric current in the wire

Explanation:

Answer: the first one is right

Explanation:

Answer:

Explanation:

3 and 4G networks, Bluetooth, and Wi-Fi technologies. my opinions

Copernicus's model of the universe differs from Ptolemy's because Copernicus believed the Sun is at the center of the universe and majority of the bodies in the universe revolves around the Sun.

The universe is defined as the open place that contains the planets, stars, galaxies, and all other forms of matter and energy.

There are different models of the universe including which contains the Copernicus's model and Ptolemy's model.

The Copernicus's model of the universe differs from Ptolemy's because Copernicus believed the Sun is at the center of the universe and majority of the bodies in the universe revolves around the Sun.

While the Ptolemy's model of the universe differs from the Copernicus's model because Ptolemy believed that the Earth is at the center of the universe and everything in the universe revolves around the Earth.

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Answer:

I'm pretty sure its 3m/s^2 for the acceleration but I don't know the force part sorry .

Explanation:

15m/s - 0m/s divided by 5 s = 3m/s

I'm no expert or anything so I could be wrong but this is the best I can give you. Sorry

As the ball is pulled inward, the radius of the circle decreases, which means that the tangential speed of the ball must increase in order to maintain uniform centripetal motion. Option (A)

This increase in tangential speed means that the angular velocity of the ball also increases, as angular velocity is directly proportional to tangential speed divided by the radius of the circle.

Since angular momentum is given by the product of rotational inertia and angular velocity, any change in angular velocity will result in a change in angular momentum.

As a result, the proper prediction for the angular momentum and rotational inertia of the ball about the axis of revolution as the ball is drawn inward is: A) The angular momentum of the ball rises. The rotational inertia of the ball about the axis of revolution remains constant.

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Answer:

When the string of a bow and arrow is pulled from equilibrium, the elastic potential energy in the bow is converted to kinetic energy of the arrow when the string is released.

Kinetic energy of an arrow can be found by using the formula KE=(mv²)/450,240 where m = mass of the arrow in grains and v = velocity of the arrow in fps.

Two factors determine the amount of energy a bow can hold. Its draw weight is the amount of force required to draw the bow. A bow's draw weight increases the farther back you pull the string.

Does a stretched bow and arrow have energy?

An arrow held in a stretched bow also contains this stored energy. Specifically, the bow has elastic potential energy. When the bow is released, the arrow will move. ... When work is done on an object, the work may be converted into either kinetic or potential energy.

Explanation:

The electric force between the droplets, and the horizontal acceleration this force produce on the droplets are: 5.84*10^-7 N and 0.32 m/s² respectively

In physics the electric force is the force that attracts or repels two charges (q) separated at a distance called (r), this is expressed in the international system of units in Newton.

To solve this exercise the electric force formulas and the procedures we will use are:

Where:

Given Info:

Applying the electric force formula we have:

F = (k * q1 * q2)/r²

F = [(9 *10^9 N*m²/C² * (2.9*10^-11 C) * (2.9*10^-11 C)]/ (3.6*10^-3 m)²

F = 7.569*10^-12 N*m² /1.296*10^-5 m²

F = 5.84*10^-7 N

Applying the force formula and clearing the acceleration, we get:

a= F/m

a= 5.84*10^-7 N/ 1.8*10^-6 kg

a= 0.32 m/s²

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Answer: it eventually cracks the rock and breaks pieces of the rock off

Explanation:

Answer:

emissions line spectrum

Answer:

The direction of friction force is toards north.

Explanation:

A tool box is at rest on the back of a track. The truck is accelerating towards north.

As the frame of reference is acceleratinf so it is a non inertial frame  of reference.

Thus, teh toolkit experineces a pseudo force towards the south direction.

According to the question, the toolbox is at rest so the fiction force is balances by teh psheudoforce, and thus teh friction force is acting toards north.

f = 0.72 n

where f = magnitude of the friction force and n = mag. of the normal force. So we have

n = (14 N) / 0.72 ≈ 19.4 N

By Newton's second law, the object's weight and the normal force cancel, so that

n - m g = 0

where m = mass of the object and g = 9.8 m/s². Then the weight of the object is

m g = n ≈ 19 N

Encontramos que la distancia recorrida en los primeros 9 segundos es 288 metros.

Los datos dados son:

El cuerpo tiene una rapidez inicial de 5 m/s

El cuerpo tiene una aceleración de 6 m/s^2

Queremos calcular la distancia recorrida durante los primeros 9 segundos de movimiento.

Lamentablemente no contamos con los gráficos ni las tablas, así que se procede a obtener las ecuaciones de movimiento.

La aceleración será:

a(t) =  6m/s^2

Para la ecuación de la velocidad tenemos que integrar la ecuación de arriba, obteniendo:

v(t) = (6m/s^2)*t  + v0

Donde v0 es la rapidez inicial, que conocemos es igual a 5 m/s, así tenemos:

v(t) = (6m/s^2)*t  + 5m/s

Para la ecuación de la posición debemos integrar nuevamente, así obtendremos:

p(t) = (1/2)*(6m/s^2)*t^2  + (5m/s)*t + p0

Donde la p0 es la posición inicial, la cual podemos definir como p0 = 0m

p(t) = (3m/s^2)*t^2  + (5m/s)*t

Para encontrar la distancia recorrida en los primeros 9 segundos, simplemente debemos remplazar t por 9s en la ecuación de posición:

p(9s) =  (3m/s^2)*(9s)^2  + (5m/s)*9s = 288 m

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The energy associated with an infrared photon with a wavelength of 7.93 micrometers (μm) is approximately 2.49 ×  \(10^{-20}\)joules (J).

In order to calculate the energy associated with an infrared photon, we can use the equation E = (hc) ÷ λ, where E represents energy, h is Planck's constant (approximately 6.62607015 × \(10^{-34}\) J·s), c is the speed of light (approximately 299,792,458 m/s), and λ is the wavelength of the photon.

Given that the wavelength is 7.93 μm (7.93 × \(10^{-6}\) m), we can substitute these values into the equation to find the energy:

E = (6.62607015 × \(10^{-34}\) J·s × 299,792,458 m/s) ÷ (7.93 × \(10^{-6}\) m)

Evaluating this expression using a calculator, we find:

E ≈ 2.49 × \(10^{-20}\) J

Therefore, an infrared photon with a wavelength of 7.93 μm is associated with an energy of approximately 2.49 × 10^(-20) joules (J).

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For weightless object, gravity is equal to centripetal acceleration. The radius of the loop is calculated as 8 km.

The property of the motion of an object traversing a circular path is known as centripetal acceleration. Any object that is moving in circle and has an acceleration vector pointed towards center of that circle is Centripetal acceleration.

As Centripetal Acceleration formula is;

a = v²/r

Given, V-  velocity =  280 m/s

And we know that for weightless object,

Gravity =  centripetal acceleration

So, a =  g

given, g =9.8 m/s²

So,9.8 = 280²/r

r= 8000 m

Radius of the loop = 8 km

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Answer: Independent variable

Explanation: i think

Answer:

of course is clear

Explanation:

very clear

Answer:

Always on the move

Nothing stays the same

The sun comes and goes

The night you can never escape

Nothing can last for ever

Everything we know rotates

I thought this would be more clever

But it's late and I don't know what else to say

Those were eight lines only

Counting lines makes you lonely

Sixteen is a big number

I feel like I'm going under

It's always daytime somewhere

But it will never be forever

I move in circles

This poem is over

Explanation:

Sorry I thought it would end up better than this