An experimental power plant at the Natural Energy Laboratory of Hawaii generates electricity from the temperature gradient of the ocean. The surface and deep-water temperatures are 28 ∘C and 5 ∘C, respectively.

When L = L₁, the circuit is a series RL circuit. The voltage across the inductor is given by Vₗ = XL₁i, where XL₁ = ωL₁ is the inductive reactance and i is the current flowing through the circuit. The voltage across the resistor is given by VR = Ri. The voltage across the capacitor is zero since it is connected in series with the inductor and the resistor.

The total voltage across the circuit is given by v = V√2 cos(ωt). By Kirchhoff's voltage law, we have v = Vₗ + VR,

V√2 cos(ωt) = XL₁i + Ri

The current i can be written as i = (1/Z) V√2 cos(ωt - φ), where Z = √(R² + XL₁²) is the impedance of the circuit and φ is the phase angle between the current and the voltage. Substituting i into the equation above, we get:

V√2 cos(ωt) = XL₁/Z V√2 cos(ωt - φ) + R/Z V√2 cos(ωt - φ)

Equating the coefficients of cos(ωt) and cos(ωt - φ), we get:

1 = XL₁/Z cos φ + R/Z sin φ

XL₁ sin φ = Z - R cos φ

tan φ = XL₁ / (Z - R cos φ)

The voltage across the inductor is given by:

Vₗ = XL₁ i = XL₁/Z V√2 cos(ωt - φ)

Vₗ/V = XL₁/Z cos φ

Substituting tan φ into this equation, we get:

Vₗ/V₁ = XL₁/Z₁ √[1 - (XL₁/Z₁)²] ... (1)

When L = L₂, the circuit is a series RC circuit. The voltage across the capacitor is given by VC = XC₂i, where XC₂ = 1/(ωC₂) is the capacitive reactance.  The total voltage across the circuit is given by v = V√2 cos(ωt). By Kirchhoff's voltage law, we have v = VC + VR, which gives:

V√2 cos(ωt) = XC₂i + Ri

Following the same steps as in the previous case, we can show that:

Vₗ/V₂ = XC₂/Z₂ √[1 - (XC₂/Z₂)²] ... (2)

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Field representation of a positive or  of a positive or negative charge cannot be a representation for the gravitational field around one mass. It height from the ground must be determined.

The gravitational force is a kind of force by which an object attracts other objects into its center of a mass. Earth attracts every objects in its surface in to the ground and that is why we are all standing on the ground.

Gravitational force between two objects depends on their mass and distance between them. The field representation of the charge does not represent a gravitational field but it can show an electric field between them.

The height of the mass from the surface have to be determined to represent the gravitational field. The gravitational field is not at all depending on the charge of the object.

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Answer:

the units of work and energy is joule and unit of power is Watt

For the graph of Annual cray fish caught in Lambert Bay for the last 20 years, the axes are as follows:

Y-axis: amount of fish caught

X-axis: year

A graph is a pictorial illustration of data.

There are different types of graphs such as

A graph has the vertical axis, known as the Y-axis, and the horizontal axis known as the X-axis.

For the graph of Annual cray fish caught in Lambert Bay for the last 20 years, the axis can be labelled as follows:

Y-axis: amount of fish caught

X-axis : year

In conclusion, in a graph, the vertical axis is known as the Y-axis while  the horizontal axis is known as the X-axis.

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Answer:

5070

Explanation:

add them up and then you get your answers

Answer:

30 N to the right

Explanation:

i got it on study island

Answer:

17.15 m/s.

Explanation:

The following data were obtained from the question:

Maximum height (h) = 15 m

Final velocity (v) = 0 m/s

Acceleration due to gravity (g) = –9.8 m/s² (since the ball is going again gravity)

Initial velocity (u) =.?

The initial velocity with which the baseball pitcher threw the ball can be obtained as follow:

v² = u² + 2gh

0² = u² + (2 × –9.8 × 15)

0 = u² + (–294)

0 = u² – 294

Collect like terms

0 + 294 = u²

294 = u²

Take the square root of both side

u = √294

u = 17.15 m/s

Therefore, the baseball pitcher threw the ball with an initial velocity of 17.15 m/s

The car is moving at approximately 12.5 meters per second.

The kinetic energy (KE) of an object can be calculated using the formula:

KE = 1/2 * m * \(v^2\)

where

KE = kinetic energy,

m =Mass of the object, and

v = velocity.

In this case, we are given the mass (m) of the car as 1600 kg and the kinetic energy (KE) as 125,000 J. To find the velocity .

Substituting the  values , we have:

125,000 J = 1/2 * 1600 kg *\(v^2\)

Now, we can solve for v by rearranging the equation:

\(v^2\) = (2 * 125,000 J) / 1600 kg

\(v^2\) = 156.25 \(m^2/s^2\)

Taking the square root, we find:

v = √156.25\(m^2/s^2\)

v ≈ 12.5 m/s

Therefore, the car is moving at approximately 12.5 meters per second.

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Answer:

no coaster car does not travel as high

Answer:

The net force is equal to zero.

Explanation:

Two equal forces going in opposite directions will always result in a net force of 0.

Answer:

The net force would be 0.

Explanation:

If you have equal amounts of force on each side it is 0. If you had different amounts of force on each side the side you would subtract the greater one form the smaller one equalling your net force.

Answer:

below

Explanation:

Newton's law of universal gravitation states that two bodies in space pull on each other with a force proportional to their masses and the distance between them.

Newton's First Law of Motion (Law of Inertia)

Newton's Second Law of Motion (Law of Mass and Acceleration)

Newton's Third Law of Motion

The Law of Orbits: All planets move in elliptical orbits, with the sun at one focus.

2. The Law of Areas: A line that connects a planet to the sun sweeps out equal areas in equal times.

3. The Law of Periods: The square of the period of any planet is proportional to the cube of the semimajor axis of its orbit

Thus, Kepler's laws and Newton's laws taken together imply that the force that holds the planets in their orbits by continuously changing the planet's velocity so that it follows an elliptical path is (1) directed toward the Sun from the planet, (2) is proportional to the product of masses for the Sun and planet, and (3) is inversely proportional to the square of the planet-Sun separation. This is precisely the form of the gravitational force, with the universal gravitational constant G as the constant of proportionality. Thus, Newton's laws of motion, with a gravitational force used in the 2nd Law, imply Kepler's Laws, and the planets obey the same laws of motion as objects on the surface of the Earth!

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The displacement of the runner after four seconds is 78.4 m.

We know that in this case, we are dealing with a case of an object that has a motion under gravity. We are told that the road runner an fell straight down off a cliff. The fact that we have been told that the runner just fell down the cliff means that the initial velocity of the runner would have to be taken in this context as zero since the runner was dropped from a height as shown.

Acceleration of the runner (g) = 9.8 m/s^2

Initial velocity of the runner (u) = 0 m/s

Time take (t) = 4 seconds

We then have;

h = ut + 1/2gt^2

If we then know that the initial velocity of the person is zero, then we have;

h = 1/2gt^2

h = 0.5 * 9.8 * (4)^2

h = 78.4 m

The height is 78.4 m.

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Answer:

its just built different

I could give you a big, fancy scientific answer, but it often comes down to the fact that the way that most instructors teach math is not the way the student(s) can most easily understand it.

Answer:

163n

Explanation:

Weight is force due to gravity, weight of object is 980 N

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

The gravitational constant, denoted by the capital letter G, is an empirical physical constant involved in the calculation of gravitational effects in Sir Isaac Newton's law of universal gravitation and in Albert Einstein's theory of general relativity.

Weight = mass.gravity

Weight = 100*9.8

Weight = 980 N

Weight is force due to gravity, weight of object is 980 N

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Answer:

A reference point is a place or object used for comparison to determine if something is in motion. An object is in motion if it changes position relative to a reference point.

The maximum height to which the ball attain before falling back down is 1147.96 m

The following data were obtained from the question:

The maximum height reached by the ball can be obtained as illustrated below:

v² = u² – 2gh (since the ball is going against gravity)

0² = 150² – (2 × 9.8 × h)

0 = 22500 – 19.6h

Collect like terms

0 – 22500 = –19.6h

–22500 = –19.6h

Divide both side by –19.6

h = –22500 / –19.6

h = 1147.96 m

Thus, the maximum height reached by the ball is 1147.96 m

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In a reference frame fixed to Earth, a point on the equator moves zero distance in 1 hour.

A frame of reference in physics is made up of an abstract coordinate system and the collection of physical reference points that fix the coordinate system specifically and regulate measurements inside it.

When we take a reference frame fixed to Earth,  the displacement due to earth's rotation can not be measured. So, any point on earth remains in rest through out the day with respect to the reference frame fixed to Earth. Hence, a point on the equator moves no distance in 1 hour in a reference frame fixed to Earth.

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The compressive force that would be enough to produce a fracture is 4000 N.

We know that the femur is one of the most important bones that we have in the human body. In this case, we have been told that In the human femur, bone tissue is strongest in resisting compressive force, approximately half as strong in resisting tensile force, and only about one- fifth as strong in resisting shear force.

Then we know that;

Tensile force = 8000

The compressive force would be half of this magnitude as such;

Compressive force = 4000 N

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Answer:

Yes, air can indeed make shadows. A shadow occurs when an object in a light beam prevents some of the light from continuing on in the forward direction. When the light beam hits a wall or the ground, a darker shape is visible where less light is hitting the surface. Refraction is responsible.

Explanation:

Here is a clip of an example of air making a shadow

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Answer:

i think its stress (c) hope this helpa

Answer:

C. the ability to withstand stress

Explanation:

The strength of a material refers to its ability to withstand stress. Stress is a measure of the force applied to a material, and it can be either tensile (stretching) or compressive (squeezing). A material's strength is typically measured by its ability to resist deformation or failure under a given amount of stress.

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Answer:

Energy is:

A property that must be transferred to perform work.

Answer:

Explanation:

Answer:

The forward motion of a body in space, such as a planet or moon, and the pull of gravity on it from another body in space.

Explanation:

Earth and many other bodies—including asteroids, comets, and the other planets—move around the sun in curved paths called orbits. Generally, the orbits are elliptical, or oval, in shape. Because of the sun’s relatively strong gravity, Earth and the other bodies constantly fall toward the sun, but they stay far enough away from the sun because of their forward velocity to fall around the sun instead of into it. As a result, they keep orbiting the sun and never crash to its surface. The motion of Earth and the other bodies around the sun is called orbital motion. Orbital motion occurs whenever an object is moving forward and at the same time is pulled by gravity toward another object.

(1/2)mv^2 = (1/2) * 100 * (2*3)^2 = 1800 [J]

The formula for potential difference (also known as voltage) is, V = ΔE/q, where V is the potential difference in volts (V), ΔE is the change in electric potential energy in joules (J), and q is the charge in coulombs (C).

Electric potential difference, also known as voltage, is a measure of the electric potential energy per unit of charge required to move a charge from one point to another in an electric circuit. It is the difference in electric potential between two points in an electric circuit.

The formula for potential difference, V = ΔE/q, reflects this relationship. The numerator, ΔE, represents the change in electric potential energy between the two points, while the denominator, q, represents the charge that moves between the two points.

For example, if a charge of +1 C moves from a point A to a point B in an electric circuit, and the electric potential energy at point B is greater than at point A by 1 J, then the potential difference between points A and B is 1 V. This means that it takes 1 J of energy to move a unit of charge from point A to point B in the circuit.

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Answer:

B) They remain in a fixed position.

Explanation:

They remain in a fixed position. The amplitude of the resultant wave is greater than the amplitude of either of the two individual waves.

Answer: Hope this helps:

(a) The sum of two vectors a and b in unit-vector notation can be calculated using vector addition:

a + b = (4.3 m)î + (3.4 m)ĵ + (-13.0 m)î + (7.6 m)ĵ

= (4.3 m - 13.0 m)î + (3.4 m + 7.6 m)ĵ

= (-8.7 m)î + (11.0 m)ĵ

(b) The magnitude of a vector can be calculated using the Pythagorean theorem:

|a + b| = √((-8.7 m)² + (11.0 m)²) = 12.8 m

(c) The direction of a vector relative to î can be calculated using the tangent of the angle between the vector and the x-axis (î):

tan(θ) = |(11.0 m) / (-8.7 m)| = 1.26

θ = arctan(1.26) = 57.3°

So, the direction of the vector a + b is 57.3° relative to the x-axis (î).

According to Gauss' equation, the total flux of an electric field in a confined surface is directly proportional to the charge enclosed.

1)To determine the outward electric flux through the rounded "side" of the cylinder, we can use Gauss's law. We choose a cylindrical Gaussian surface with radius r and length L, centered at the origin (where the charged sphere is located). The electric field due to the sphere is spherically symmetric, so the electric field lines are parallel to the cylinder's axis and perpendicular to its sides.

E = (1/4πϵ0) (Q/r^2)

where r is the distance from the origin (center of the sphere) to the point on the Gaussian surface.

The area element of the Gaussian surface is dA = 2πRdz, where dz is an element of length along the cylinder's axis. The electric flux through the top and bottom surfaces of the Gaussian surface is then given by:

Φ = ∫E⋅dA = E ∫dA = E(2πR)L

Substituting the expression for the electric field, we have:

Φ = (Q/2ϵ0r^2)(2πRL)

Therefore, the outward electric flux through the rounded "side" of the cylinder is:

Φ = (Q/ϵ0)(R/Lr^2)

2)To determine the electric flux upward through the circular cap at the top of the cylinder, we use a flat Gaussian surface with radius R and height r, centered at the top of the cylinder. The electric field due to the charged sphere is perpendicular to the Gaussian surface, so the electric flux through the top cap is simply the flux through the flat Gaussian surface. The electric field at any point on the Gaussian surface is given by Coulomb's law as:

E = (1/4πϵ0) (Q/R^2)

The area element of the Gaussian surface is dA = πR^2, so the electric flux through the top cap is given by:

Φ = ∫E⋅dA = E ∫dA = EπR^2

Substituting the expression for the electric field, we have:

Φ = (Q/ϵ0)(R/r^2)

3)To determine the electric flux downward through the circular cap at the bottom of the cylinder, we use a similar flat Gaussian surface with radius R and height r, centered at the bottom of the cylinder. The electric flux through the bottom cap is also given by:

Φ = (Q/ϵ0)(R/r^2)

4)Adding the results from parts 1-3, we have the total outward electric flux through the closed cylinder as:

Φ_total = Φ_side + Φ_top + Φ_bottom

= (Q/ϵ0)(R/Lr^2) + 2(Q/ϵ0)(R/r^2)

Simplifying this expression, we have:

Φ_total = (Q/ϵ0) [(2R/r^2) + (R/Lr^2)]

5)According to Gauss's law, the total outward electric flux through a closed surface is proportional to the total charge enclosed within that surface. In this case, the closed surface is the cylindrical Gaussian surface with radius r and length L, centered at the origin (where the charged sphere is located). The charge enclosed within this surface is simply the charge of the sphere, which is +Q. Therefore, we expect the total outward electric flux through the closed cylinder to be:

Φ_total = Q/

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At 30°C, the rms speed of a sample of oxygen with a molar mass of 16 g/mol is approximately 482.34 m/s.

The root mean square (rms) speed of a gas molecule is a measure of the average speed of the gas particles in a sample. It can be calculated using the formula:

vrms = √(3kT/m)

Where:

vrms is the rms speed

k is the Boltzmann constant (1.38 x 10^-23 J/K)

T is the temperature in Kelvin

m is the molar mass of the gas in kilograms

To calculate the rms speed of oxygen at 30°C (303 Kelvin) with a molar mass of 16 g/mol, we need to convert the molar mass to kilograms by dividing it by 1000:

m = 16 g/mol = 0.016 kg/mol

Substituting the values into the formula, we have:

vrms = √((3 * 1.38 x 10^-23 J/K * 303 K) / (0.016 kg/mol))

Calculating this expression yields the rms speed of the oxygen sample:

vrms ≈ 482.34 m/s

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A VfL (Vertical Field LED) backlighting system is commonly used in LCD (Liquid Crystal Display) televisions.

LCD TVs rely on a backlight to illuminate the liquid crystal layer, which controls the passage of light to create the visual image. The VfL technology is a specific type of LED backlighting arrangement used in certain LCD TV models. In a VfL backlighting system, the LEDs (Light-Emitting Diodes) are positioned vertically along the edges of the LCD panel.

The light emitted by these LEDs is directed across the panel using light guides or optical films, illuminating the liquid crystal layer uniformly. One advantage of VfL backlighting is its ability to provide consistent illumination across the LCD panel, reducing any potential inconsistencies in brightness or color uniformity. The vertical orientation of the LEDs allows for more precise control over light distribution, improving overall image quality.

Additionally, VfL backlighting offers potential advantages in terms of power efficiency. By selectively dimming or turning off specific zones of LEDs, local dimming techniques can be employed to enhance contrast and black levels, resulting in improved picture quality while conserving energy. It's important to note that VfL backlighting is just one of several backlighting technologies available for LCD TVs.

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Answer:

I think you are trying to ask 'what happens to the average speed (of humans or animals.) if the distance decreased and time stayed the same (the time such as what hour it is or what second?)' If that is the case then the average speed will be more since the distance decreased. And the time staying the same might affect it I am not fully sure so I will not say anything about that. I hope this helps!