In 1 to 2 sentences, Explain how a decrease in the resistance caused by friction change the amount of force needed to move an object? When you answer the question please put a explanation for a younger child would be greatly appreciated!!

Answer:

Explanation:

according to the graph at B the potential energy of the particle is 2J

therefore we can use the kinetic energy equation to calculate the particle's velocity or speed.

\(E_{k} =1/2mv^{2}\)

2J= 1/2*1/2kg*v^2

8=v^2

v= 2√2 ms-1

Answer:

The velocity time graph which shows a negative constant velocity (horizontal line below the horizontal axis) Labeled as Graph #2

Explanation:

Notice that the position-time graph shows the position decreasing in the form of a line with negative slope. Recall that the velocity graph is associated with the change in position with time, and this change in position is in the form of a line of constant negative slope. The slope of the position-time graph is in fact the velocity of the object, which therefore is negative and constant.

Such behavior is represented by the graph labeled as Graph 2 of a constant (horizontal) line below the horizontal axis.  

Answer:

evaporation?

sorry if im wrong

Explanation:

Temperature is inversely propotional to the density of liquid

The nature of the image formed by the convex lens is virtual, the position of the image is 30 cm away from the lens on the same side as the object, and the size of the image is twice the size of the object. The magnification is 2, meaning the image is magnified.

Given:

Object height (h) = 5 cm

Focal length of the convex lens (f) = 10 cm

Object distance (u) = 15 cm (positive since it's on the same side as the incident light)

To determine the nature, position, and size of the image, we can use the lens formula:

1/f = 1/v - 1/u

Substituting the given values:

1/10 = 1/v - 1/15

To simplify the equation, we find the common denominator:

3v - 2v = 2v/3

Simplifying further:

v = 30 cm

The image distance (v) is 30 cm. Since the image distance is positive, the image is formed on the opposite side of the lens from the object.

To find the magnification (M), we use the formula:

M = -v/u

Substituting the values:

M = -30 / 15 = -2

The magnification is -2, indicating that the image is inverted and twice the size of the object.

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The gravitational potential energy of an object depends on  mass of the object, gravitational field and height of the object.

The gravitational potential energy of an object is the energy possessed by the object due to the object's position above the ground level.

P.E = mgh

where;

The factors affecting gravitational potential energy of an object include;

Thus, the gravitational potential energy of an object depends on  mass of the object, gravitational field and height of the object.

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Answer:

Key points that can be found in the realist philosophical position​ are as follows:

Answer:

Solids

Is this what your looking for, It might tell you the answer?

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Answer:

Therefore, the orbital speed of a satellite in a geosynchronous circular orbit 3.58 x 10^7 m above the surface of the Earth is approximately 3.07 x 10^3 m/s.

Explanation:

A geosynchronous circular orbit is an orbit in which a satellite revolves around the Earth once every 24 hours so that it appears to be stationary in the sky relative to an observer on the ground. The radius of such an orbit is known as the geostationary radius and is approximately 42,164 km or 3.58 x 10^7 m above the surface of the Earth.

To find the orbital speed of a satellite in a geosynchronous circular orbit, we can use the formula:

v = (GM / r)^0.5

Where v is the orbital speed, G is the gravitational constant (6.674 x 10^-11 N m^2/kg^2), M is the mass of the Earth (5.97 x 10^24 kg), and r is the distance from the center of the Earth to the satellite (in this case, r = 3.58 x 10^7 m + radius of the Earth).

The radius of the Earth is approximately 6,371 km or 6.371 x 10^6 m. Therefore, the distance from the center of the Earth to the satellite is:

r = 3.58 x 10^7 m + 6.371 x 10^6 m

r = 4.217 x 10^7 m

Now we can plug in the values for G, M, and r into the formula and solve for v:

v = (GM / r)^0.5

v = [(6.674 x 10^-11 N m^2/kg^2) x (5.97 x 10^24 kg) / (4.217 x 10^7 m)]^0.5

v = 3.07 x 10^3 m/s

Therefore, the orbital speed of a satellite in a geosynchronous circular orbit 3.58 x 10^7 m above the surface of the Earth is approximately 3.07 x 10^3 m/s.

Explanation:

The orbital speed of a satellite in a circular orbit around the Earth can be found using the following formula:

v = sqrt(G*M/R)

where G is the gravitational constant, M is the mass of the Earth, and R is the distance from the center of the Earth to the satellite's orbit.

For a geosynchronous orbit, the satellite has a period of 24 hours, which means it completes one orbit in 24 hours. This corresponds to an orbital radius of:

R = 3.58 x 10^7 m + 6.38 x 10^6 m = 4.22 x 10^7 m

where 6.38 x 10^6 m is the radius of the Earth.

The mass of the Earth is approximately 5.97 x 10^24 kg, and the gravitational constant is approximately 6.6743 x 10^-11 N*(m/kg)^2. Substituting these values into the equation above, we get:

v = sqrt(GM/R) = sqrt(6.6743 x 10^-11 N(m/kg)^2 * 5.97 x 10^24 kg / 4.22 x 10^7 m) = 3074 m/s

Therefore, the orbital speed of a satellite in a geosynchronous circular orbit 3.58 x 10^7 m above the surface of the Earth is approximately 3074 m/s.

Answer:

B

Explanation:

transferred

The pressure after the temperature has risen to 40.0 °C, assuming that there are no appreciable leaks or changes in volume is 4.35×10⁵ Pa

From the question given above, the following data were obtained:

The pressure the temperature has risen to 40 °C can be obtained as follow:

P₁V₁ / T₁ = P₂V₂ / T₂

Volume = contant

P₁ / T₁ = P₂ / T₂

4.00×10⁵ / 288 = P₂ / 313

Cross multiply

P₂ × 288 = 4.00×10⁵ × 313

Divide both sides by 288

P₂ = (4.00×10⁵ × 313) / 288

P₂ = 4.35×10⁵ Pa

Thus, the pressure at 40 °C is 4.35×10⁵ Pa

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The average density of the material from which the coin is made is 9.67 g/cm³.

The volume of the coin at the given diameter is calculated as follows;

V = Ah

where;

V = πd²/4 x h

V = π(2.8)²/4  x (0.21 cm)

V = 1.293 cm³

The average density of the material from which the coin is made is calculated as follows;

density = mass/volume

density = 12.5 g / (1.293 cm³)

density = 9.67 g/cm³

Thus, the average density of the material from which the coin is made is 9.67 g/cm³.

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Answer:

  B. When Ellen started pedaling her bike, she began moving along the street

Explanation:

Kinetic energy is energy of motion. Ellen's kinetic energy changed when she went from rest to being in motion (began moving).

Answer:

control

Explanation:

you always need a control in experiment

Answer:

If he's 500 kg and 1 kg = 10 N then it would be 5,000 N

Explanation:

The surface area is unimportant and they'll use it to trick you.

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:

What Is Moral Subjectivism? Moral subjectivism is based on an individual person's perspective of what is right or wrong. An individual can decide for themselves that they approve or disapprove of a certain behavior, and that is what determines if the behavior is right or wrong.

The CMB radiation's current temperature of 2.7 Kelvin is a result of the cooling and redshifting caused by the expansion of the Universe, which has been slowing down over time due to gravity.

The Cosmic Microwave Background (CMB) radiation is a remnant of the early hot and dense Universe. About 380,000 years after the Big Bang, the Universe cooled enough to allow the formation of neutral atoms, which made it transparent to radiation. This released the CMB radiation, which has been traveling through the Universe ever since, and its temperature has been cooling due to the expansion of the Universe. The CMB radiation is currently observed at a temperature of about 2.7 Kelvin, much cooler than the early Universe's temperature of 3000 Kelvin. This cooling is due to the expansion of the Universe, which causes the radiation to redshift to lower energies and longer wavelengths. This effect is known as the cosmological redshift. Moreover, the expansion of the Universe is not constant, but rather it is slowing down due to the gravitational pull of matter. This means that the early Universe's rate of expansion was faster than it is today, causing the CMB radiation to cool more rapidly in the past.

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(1) + 1) (w(1) - w(2) + 1) ... (w(r - 1) - w(r) + 1) (5) Proof: We will prove this statement by induction. First, for n = 1, the statement is true: wc) - w(1) + 1 = w(1). Now, assume that the statement holds for n = r.

Induction is a type of reasoning that involves making inferences based on observations and facts. It is a form of logical reasoning that enables us to use observations to make generalizations about a certain phenomenon. It is used to form hypotheses and draw conclusions. Induction can be used to develop theories, evaluate evidence, and make predictions. Induction is often used in scientific research as a way to test theories and make predictions about the behavior of a system or phenomenon. It is a powerful tool for making sense of data and providing insights into how things work.

Then, for n = r + 1, we have: w(r+1) = w(r) - x(w(r) - w(r+1)) + 1 = (w(r) - w(r+1) + 1) + (w(r-1) - w(r) + 1) + ... + (w(1) - w(2) + 1) + w(1) which confirms the statement.

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a) Yes, the force of thrust of the engine does work as it is causing the plane to accelerate and lift off the ground.

b) Yes, the kinetic energy of the plane increases as it takes off and gains speed.

c) Yes, the gravitational potential energy of the plane increases as it gains height.

d) Yes, the force of thrust of the engine changes the mechanical energy of the plane.

The kinetic energy of an object is the form of energy that it possesses due to its motion which is also defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.

The work done by the engine's thrust increases the plane's kinetic energy and gravitational potential energy, thereby changing its mechanical energy.

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The tax rate you will pay is displayed in tax brackets for each category of taxable income.

Thus, For instance, in 2022, the first $10,275 of your taxable income is subject to the lowest tax rate of 10% if you are single.

Up until the maximum amount of your taxable income, the following portion of your income is taxed at a rate of 12%.

As taxable income rises, the tax rate rises under the progressive tax system. Overall, this has the result that taxpayers with higher incomes often pay a greater rate of income tax than taxpayers with lower incomes.

Thus, The tax rate you will pay is displayed in tax brackets for each category of taxable income.

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A program that could do an instantaneous analysis of video motion will be useful it in a sportscast to analyze events as they occur.

A motion video is defined as the display of video images at a rate (such as thirty frames per second) that causes objects to appear to move smoothly and continuously.

Sports inherently involve fast and accurate motion, which can be difficult for competitors to master but also for coaches and trainers to analyze and audiences to follow. Because of the nature of most sports, monitoring with sensors or other devices attached to players or equipment is generally not possible. This opens up a plethora of opportunities for the use of computer vision techniques to assist competitors, coaches, and the audience

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Answer:

(a) 22 kN

(b) 36 kN, 29 kN

(c) left will decrease, right will increase

(d) 43 kN

Explanation:

(a) When the truck is off the bridge, there are 3 forces on the bridge.

Reaction force F₁ pushing up at the first support,

reaction force F₂ pushing up at the second support,

and weight force Mg pulling down at the middle of the bridge.

Sum the torques about the second support.  (Remember that the magnitude of torque is force times the perpendicular distance.  Take counterclockwise to be positive.)

∑τ = Iα

(Mg) (0.3 L) − F₁ (0.6 L) = 0

F₁ (0.6 L) = (Mg) (0.3 L)

F₁ = ½ Mg

F₁ = ½ (44.0 kN)

F₁ = 22.0 kN

(b) This time, we have the added force of the truck's weight.

Using the same logic as part (a), we sum the torques about the second support:

∑τ = Iα

(Mg) (0.3 L) + (mg) (0.4 L) − F₁ (0.6 L) = 0

F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.4 L)

F₁ = ½ Mg + ⅔ mg

F₁ = ½ (44.0 kN) + ⅔ (21.0 kN)

F₁ = 36.0 kN

Now sum the torques about the first support:

∑τ = Iα

-(Mg) (0.3 L) − (mg) (0.2 L) + F₂ (0.6 L) = 0

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.2 L)

F₂ = ½ Mg + ⅓ mg

F₂ = ½ (44.0 kN) + ⅓ (21.0 kN)

F₂ = 29.0 kN

Alternatively, sum the forces in the y direction.

∑F = ma

F₁ + F₂ − Mg − mg = 0

F₂ = Mg + mg − F₁

F₂ = 44.0 kN + 21.0 kN − 36.0 kN

F₂ = 29.0 kN

(c) If we say x is the distance between the truck and the first support, then using our equations from part (b):

F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L − x)

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (x)

As x increases, F₁ decreases and F₂ increases.

(d) Using our equation from part (c), when x = 0.6 L, F₂ is:

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L)

F₂ = ½ Mg + mg

F₂ = ½ (44.0 kN) + 21.0 kN

F₂ = 43.0 kN

Answer:

Explanation:

given:

weight of bridge = 44 kN

weight of truck = 21 kN

a) truck is off the bridge

since the bridge is symmetrical, left support is equal to right support.

Left support = Right support = 44/2

Left support = Right support = 22 kN

b) truck is positioned  as shown.

to get the reaction at left support, take moment from right support = 0

∑M at Right support = 0

Left support (0.6) - weight of bridge (0.3) - weight of truck (0.4) = 0

Left support = 44 (0.3) + 21 (0.4)  

                                  0.6

Left support = 36 kN

Right support = weight of bridge + weight of truck - Left support

Right support = 44 + 21 - 36

Right support = 29 kN

c)

as the truck continues to drive to the right, Left support will decrease

as the truck get closer to the right support,  Right support will increase.

d) truck is directly under the right support, find reaction at Right support?

∑M at Left support = 0

Right support (0.6) - weight of bridge (0.3) - weight of truck (0.6) = 0

Right support = 44 (0.3) + 21 (0.6)  

                                  0.6

Right support = 43 kN

Answer:

The can was dropped from a height of 0.74 meters.

Explanation:

We can use the Conservation of Energy to evaluate the height.

The formula for the Conservation of Energy is

\(\frac{1}{2} mv^2=mgh\)

Lets solve for \(h\).

Cancel the common factor of \(m\).

\(\frac{1}{2} v^2=gh\)

Rewrite \(\frac{1}{2} v^2\) as \(\frac{v^2}{2}\) .

\(\frac{v^2}{2}=gh\)

Multiply both sides by 2.

\(v^2=2gh\)

Divide both sides by \(2g\).

\(\frac{v^2}{2g}=h\)

Numerical Evaluation

We are given

\(v=3.8\\g=9.81\)

Substituting our values into our equation for height gives us

\(h=\frac{3.8^2}{2*9.81}\)

Evaluate \(3.8^2\).

\(h=\frac{14.44}{2*9.81}\)

Multiply \(2\) and \(9.81\).

\(h=\frac{14.44}{19.62}\)

\(h=0.736235\)

Note: We also could have solved this using the kinematics equations. Both methods give you the same answer.

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Answer:

What are igneous rocks?

Igneous rocks (from the Latin word for fire) form when hot, molten rock crystallizes and solidifies. The melt originates deep within the Earth near active plate boundaries or hot spots, then rises toward the surface. Igneous rocks are divided into two groups, intrusive or extrusive, depending upon where the molten rock solidifies

Answer:

The Anatomy of a Lens

Refraction by Lenses

Image Formation Revisited

Converging Lenses - Ray Diagrams

Converging Lenses - Object-Image Relations

Diverging Lenses - Ray Diagrams

Diverging Lenses - Object-Image Relations

The Mathematics of Lenses

Ray diagrams can be used to determine the image location, size, orientation and type of image formed of objects when placed at a given location in front of a lens. The use of these diagrams was demonstrated earlier in Lesson 5 for both converging and diverging lenses. Ray diagrams provide useful information about object-image relationships, yet fail to provide the information in a quantitative form. While a ray diagram may help one determine the approximate location and size of the image, it will not provide numerical information about image distance and image size. To obtain this type of numerical information, it is necessary to use the Lens Equation and the Magnification Equation. The lens equation expresses the quantitative relationship between the object distance (do), the image distance (di), and the focal length (f)

The near point of a human eye is about a distance of 25 cm.

The closest distance that an object may be viewed clearly without straining is known as the near point of the eye.

This distance (the shortest at which a distinct image may be seen) is 25 cm for a typical human eye.

The closest point within the accommodation range of the eye at which an object may be positioned while still forming a focused picture on the retina is also referred to as the near point.

In order to focus on an item at the average near point distance, a person with hyperopia must have a near point that is further away than the typical near point for someone of their age.

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Answer:

You need to be patient with them. Also, a positive attitude is the single most important quality for anyone who works with children with special needs. I’ve seen highly trained specialists unable to interact with Louie because of their negative attitude and assumptions