Once a scientific theory is formed, there is no further investigation needed.
True
False

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.  

Standard atmospheric pressure is the average pressure exerted by the Earth's atmosphere at sea level under normal conditions. It is defined as 101.325 kPa or 1 atm and serves as a reference point for scientific measurements and comparisons related to atmospheric pressure.

Standard atmospheric pressure is the average pressure exerted by the Earth's atmosphere at sea level under normal conditions. It is defined as 101.325 kilopascals (kPa) or 1 atmosphere (atm). This pressure is used as a reference point for various scientific measurements and is crucial in fields such as meteorology, physics, and engineering.

Atmospheric pressure is caused by the weight of the air above a given point on the Earth's surface. The air is composed of molecules, mainly nitrogen (approximately 78%) and oxygen (approximately 21%), along with other trace gases. These molecules are in constant motion and exert a force on the surfaces they come into contact with, including the Earth's surface.

The standard atmospheric pressure of 101.325 kPa is equivalent to the pressure exerted by a column of mercury 760 millimeters (mm) in height in a barometer at sea level. This measurement was established as a reference point for atmospheric pressure, providing a consistent value for scientific calculations and comparisons.

It is important to note that atmospheric pressure can vary with altitude and weather conditions. As altitude increases, the atmospheric pressure decreases because the column of air above becomes thinner. In areas of high or low pressure systems associated with weather patterns, the atmospheric pressure deviates from the standard value.

The standard atmospheric pressure is a valuable reference in many applications. For instance, it is used as a standard for measuring gas pressure in laboratories and industrial processes. It is also used in meteorology to calculate and compare pressure systems and to study atmospheric phenomena such as wind patterns and weather changes.

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Final Answer:

If there is lower air pressure being "pushed" onto the surface of the water, then the water molecules do not need as much energy to overcome the force of atmospheric pressure pushing on the surface, and they can more easily escape and become a vapor. When the vapor pressure of the liquid is equal to the air pressure surrounding the liquid, it will boil. If you lower the air pressure, you will lower the temperature that water will boil at.

Explanation:

The boiling point of a liquid depends on the balance between the vapor pressure of the liquid and the external pressure, which is typically atmospheric pressure. When the external pressure is reduced, such as at higher altitudes or in a vacuum, the liquid molecules require less energy to escape the surface and become vapor. This is because there is less force pushing down on the liquid's surface, allowing for easier vaporization.

At lower air pressure, the water molecules can overcome the diminished force of atmospheric pressure more readily, requiring less heat energy to transition from liquid to vapor. When the vapor pressure of the liquid equals the surrounding air pressure, the liquid will start to boil.

Boiling point and vapor pressure are fundamental concepts in thermodynamics and phase transitions. Understanding how pressure affects the boiling point of liquids is crucial in various scientific and engineering applications, such as cooking, chemical processes, and high-altitude cooking.

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Answer:

C that's 10N I think. not sure though

Answer:

by giving a person to person

Answer:

b i hope this is correct answee

The horizontal component of the force pushing him forward is 37 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.

Frictional force due to ice, F =   μN

Normal force N = mg = 75*9.8 = 735 N

Frictional force = 0.05*735 = 36.75 N

Coefficient of friction, μ for ice is 0.05

The horizontal component of the force pushing him forward is 37 N.

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Answer:

wha is this .....................? same confusion

Answer:

D. A lower Voltage into a higher

The time interval between the man clapping and hearing his echo is approximately 1.75 seconds.

An echo is a repetition or reflection of a sound or signal. It can be caused by sound waves bouncing off a surface, signal interference, or the repetition of a message in communication.

The speed of sound in air at room temperature is approximately 343 meters per second. When a person claps, the sound waves propagate outward in all directions and reach the school building, where they bounce off and return to the person as an echo. The time it takes for the sound to travel the distance to the building and back to the person is the time interval between the clap and the echo.

To calculate the time interval, we can use the following formula:

time = distance / speed

where distance is the total distance traveled by the sound (twice the distance from the person to the school building), and speed is the speed of sound in air.

distance = 2 x 300m = 600m

speed = 343 m/s

time = 600m / 343 m/s = 1.75 seconds (rounded to two decimal places)

Therefore, the time interval between the man clapping and hearing his echo is approximately 1.75 seconds.

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A galvanometer is an instrument that measures small electric currents using the interaction between a coil and a magnet. It finds applications in current measurement and in various electrical and electronic systems.

A galvanometer is a device that detects and measures small electric currents. The instrument works on the principle of the magnetic field, which is generated by an electric current. The coil is suspended on a torsion-free suspension thread between the poles of a permanent magnet. The needle is fixed on the axis and moves along a graduated scale on a metal plate.

Galvanometer working principle

Applications of galvanometer :

Galvanometers are widely used in various electrical and electronic circuits for detecting small current flows. They are used for the following purposes:

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The potential energy of the roller coaster is 176,400 J (joules).

The potential energy of an object is given by the formula PE = mgh, where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height or vertical position of the object.

In this case, the roller coaster is at a peak of 20m and has a mass of 900kg. The acceleration due to gravity, g, is approximately 9.8 \(m/s^2\).

Using the formula, we can calculate the potential energy:

PE = mgh

= (900 kg)(9.8 \(m/s^2\))(20 m)

= 176,400 J

Therefore, the potential energy of the roller coaster is 176,400 J (joules).

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Answer:

(a) Approximately \(0.335\; \rm m\).

(b) Approximately \(1.86\; \rm m\cdot s^{-1}\).

(c) Approximately \(0.707\; \rm m\).

(d) Approximately \(0.228\; \rm m\).

Explanation:

Consider the conversion of energy in this object-spring system.

First diagram: Right before the object came into contact with the spring, the object carries kinetic energy \(\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^2\).

Second diagram: As the object moves towards the position in the third diagram, the spring gains elastic potential energy. At the same time, the object loses energy due to friction.

Third diagram: After the velocity of the object becomes zero, it has moved a distance of \(D\) and compressed the spring by the same distance.

The sum of these two energies should match the initial kinetic energy of the object (before it comes into contact with the spring.) That is:

\(\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^{2} = (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2\).

Assume that \(g = 9.81\; \rm m \cdot s^{-2}\). In the equation above, all symbols other than \(D\) have known values:

Substitute in the known values to obtain an equation for \(D\) (where the unit of \(D\!\) is \(m\).)

\(3.178 = 2.69775\, D + 25\, D^2\).

\(2.69775\, D + 25\, D^2 + 3.178 = 0\).

Simplify and solve for \(D\). Note that \(D > 0\) because the energy lost to friction should be greater than zero.

\(D \approx 0.335\; \rm m\).

The energy of the object-spring system in the third diagram is the same as the elastic potential energy of the spring:

\(\displaystyle \frac{1}{2}\,k\, D^2 \approx 2.81\; \rm J\).

As the object moves to the left, part of that energy will be lost to friction:

\((\mu \cdot m \cdot g) \, D \approx 0.905\; \rm J\).

The rest will become the kinetic energy of that block by the time the block reaches the position in the fourth diagram:

\(2.81\; \rm J - 0.905\; \rm J \approx 1.91\; \rm J\).

Calculate the velocity corresponding to that kinetic energy:

\(\displaystyle v =\sqrt{\frac{2\, (\text{Kinetic Energy})}{m}} \approx 1.86\; \rm m \cdot s^{-1}\).

As the object moves from the position in the fourth diagram to the position in the fifth, all its kinetic energy (\(1.91\; \rm J\)) would be lost to friction.

How far would the object need to move on the surface to lose that much energy to friction? Again, the size of the friction force is \(\mu \cdot m \cdot g\).

\(\displaystyle (\text{Distance Travelled}) = \frac{\text{(Work Done by friction)}}{\text{(Size of the Friction Force)}} \approx0.707\; \rm m\).

Similar to (a), solving (d) involves another quadratic equation about \(D\).

Left-hand side of the equation: kinetic energy of the object (as in the fourth diagram,) \(1.91\; \rm J\).

Right-hand side of the equation: energy lost to friction, plus the gain in the elastic potential energy of the spring.

\(\displaystyle {1.91\; \rm J} \approx (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2\).

\(25\, D^2 + 2.69775\, D - 1.90811\approx 0\).

Again, \(D > 0\) because the energy lost to friction is greater than zero.

\(D \approx 0.228\; \rm m\).

The energy transferred between the object and the spring as a closed system, therefore, conserved are;

(a) The distance of compression, d ≈ 0.3354 meters

(b) The speed in the un-stretched position wen the object is sliding to the left, v ≈ 1.8623 m/s

(c) The distance where the object comes to rest, D ≈ 0.7071 m

(d) The distance the object will come to rest attached to the spring, D ≈ 0.2278 m

The reason the above values are correct are as follows;

The known parameters are;

Mass of the object, m₁ = 1.10 kg

Coefficient of friction, μ = 0.250

The initial speed of the object, \(v_i\) = 2.60 m/s

Force constant of the spring, K = 50.0 N/m

Distance the spring is compressed by the object = d

(a) Conservation of energy principle

\(Kinetic \ energy = \dfrac{1}{2} \cdot m\cdot v^2\)

Work done = Force × Distance

Friction force, \(F_f\) = W × μ

Weight, W = m·g

Weight = Mass × Acceleration

Energy transferred by object = Work done by spring + Work done by friction

\(Energy \ transferred \ by \ object = Kinetic \ energy = \dfrac{1}{2} \times 1.10\times 2.60^2 = 3.718\)

Energy transferred by object = 3.718 J

\(Work \ done \ by \ spring = \dfrac{1}{2} \cdot k\cdot x^2\)

\(Work \ by \ spring \ to \ bring \ object \ to \ rest, \ W_{spring} = \dfrac{1}{2} \times 50\times d^2\)

\(W_{spring}\) = 25·d²

Work done by friction, \(W_{friction}\) = 1.10×9.81×0.250×d = 2.69775·d

Therefore;

3.718 = 25·d² + 2.69775·d

25·d² + 2.69775·d - 3.718 = 0

Solving gives

The distance of the compression d ≈ 0.3354 m

(b) The energy given by the spring = 25·d²

The work done by friction, \(W_{friction}\) = 2.69775·d

Kinetic energy given to object = 0.55·v²

0.55·v² = 25·d² - 2.69775·d

0.55·v² = 25×0.3354² - 2.69775×0.3354

∴ v = √(3.4682) = 1.8623

The velocity of the object at the un stretched position, v ≈ 1.8623 m/s

(c) The kinetic energy, K.E. of the object on the way left is given as follows;

K.E. = 0.5 × 1.10 kg × 3.4682 m²/s² = 1.90751 J

The work done by friction before object comes to rest = 2.69775·D

\(D = \dfrac{1.90751 \, J}{2.69775 \, N} \approx 0.7071 \, m\)

The distance where the object comes to rest, D ≈ 0.7071 m

(d) The work done on spring, \(W_{spring}\) = 25·D'²

Work done on friction, \(W_{friction}\) = 2.69775·D'

Kinetic energy of object, K.E. ≈ 1.90751 J

K.E. = \(W_{spring}\) + \(W_{friction}\)

1.90751 ≈ 25·D'² + 2.6775·D'

25·D'² + 2.6775·D' - 1.90751 = 0

Solving with a graphing calculator gives;

D' ≈ 0.2278 m

The new value of the distance D = 0.2278 m

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By attaching a mass of 100 g too spring 1. Play the simulator. Measure the duration of the stopwatch for 10 full cycles. The measured period by dividing the total time by 10 is 2 seconds.

The period of oscillation of the mass on spring 1, you would need to measure the time it takes for the mass to complete 10 complete cycles. You can do this by starting a stopwatch when the mass is in the down position and stopping the stopwatch when the mass returns to the down position after completing 10 cycles. Once you have measured the total time for 10 complete cycles, you can calculate the period by dividing the total time by 10. For example, if the total time for 10 complete cycles is 20 seconds, then the period would be 20 seconds / 10 = 2 seconds. We will call this value Tz.

After measuring the period of oscillation of the mass on spring 1, you should pause the simulator and remove the mass from the hook of the spring. This will allow you to perform any additional measurements or calculations without interference from the mass.

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Answer:

Explanation:

First last of thermodynamics, just discusses the changes that a system is undergoing and the processes involved in it. It explains conservation of energy for a system undergoing changes or processes.

Second law of thermodynamics helps in defining the process and also the direction of the processes. It tells about the possibility of a process or the restriction of a process. It states that the entropy of a system always increases.

For this to occur the energy contained by a body has to diminish without converting to work or internal energy. So imagine a machine which works with less than efficiency, this means there are losses but they don’t show up anywhere. But the energy is obtained from a higher energy source to lower.

The easy way to do this is with an imaginary device that extracts zero-point energy to heat a quantity of gas. Energy is being created, so the first law is violated, and the entropy of the system is increasing as the gas heats up.

First law is violated since the energy conversion don't apply but the direction of work is applied so second law is satisfied.

Answer:

Speesd

Explanation:

Answer:

negative plus positive with energy = battery

Explanation:

Answer:

Its more likely to be wave D.

Explanation:

Light waves for violet are closer together than all other colors.

The car that has the greater acceleration magnitude is car B

Torricelli's theorem can be used to calculate the magnitude of an object's acceleration when the value of time is not known:

                                            \(v^{2} = v_{0}^{2} + 2a\Delta S\)

Note that \(v\\\) is the final velocity value, \(v_{0}\) is the initial velocity value, \(a\) is the acceleration value, and \(\Delta S\) is the displacement.

Substituting the values ​​given by the question in the formula, we have:

                                           \(10^{2}= 0^{2}+2\times a \times 50\)

                                                    \(100 = 100a\)

                                                     \(a = 1m/s^{2}\)

    2. CAR B

                                               \(20^{2} = 10^{2} + 2\times a \times 50\)

                                                   \(400 = 100 + 100a\)

                                                        \(100a=300\)

                                                         \(a = 3m/s^{2}\)

So, comparing the two acceleration calculated above,  it is assumed that the car has the greater acceleration magnitude is car B with \(3m/s^{2}\)  comparing to \(1m/s^{2}\)  of the car A.

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Two forces acting on the rocket immediately after leaving the launching pad are the gravitational force and the thrust force.

1. Gravitational Force: The gravitational force is the force exerted by the Earth on the rocket due to their mutual gravitational attraction. It acts downward and is responsible for the rocket's weight.

This force can be represented by the equation Fg = mg, where Fg is the gravitational force, m is the mass of the rocket, and g is the acceleration due to gravity. The gravitational force acts to pull the rocket downward, opposing its upward motion.

2. Thrust Force: The thrust force is the force generated by the rocket's engines as they expel exhaust gases in the opposite direction. It acts upward and propels the rocket forward.

The magnitude of the thrust force depends on factors such as the design of the rocket engines, the amount of fuel burned, and the rate of exhaust gas expulsion. The thrust force must be greater than or equal to the gravitational force for the rocket to overcome Earth's gravity and achieve upward acceleration.

Initially, when the rocket is launched, the thrust force is at its maximum while the gravitational force remains constant. As the rocket gains altitude, the gravitational force decreases slightly due to the increasing distance from the Earth's center.

However, the thrust force continues to be the dominant force propelling the rocket upward.

It's important to note that other forces such as air resistance and wind may also act on the rocket, but immediately after leaving the launching pad, these forces are typically negligible compared to the gravitational force and thrust force.

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Answer:

A

Explanation:

Any sort of honk can frighten the animal, drive slow enough that the person would know you're coming/you're there and make sure you are taking a lot of care.

The thing that should be done when driving is A. Use reasonable care, and don't make any loud noises when approaching them, to avoid frightening the animal.

It should be noted that the person is driving on a rural road and see a person on the shoulder ahead, riding or leading an animal.

In this case, the person should reasonable care, and don't make any loud noises when approaching them, to avoid frightening the animal.

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Answer:

Different organelles play different roles in the cell — for instance, mitochondria generate energy from food molecules; lysosomes break down and recycle organelles and macromolecules; and the endoplasmic reticulum helps build membranes and transport proteins throughout the cell.

Explanation:

Hopefully this helped!

Answer:

\(d=11\sqrt2\ km\)

Explanation:

Given that,

Displacement in North = 11 km

Displacement in East = 11 km

We need to find the approximate displacement of the hiker from the camp. It can be calculated by resultant of two displacement as follows :

\(d=\sqrt{11^2+11^2}\\\\= 11\sqrt2\ km\)

Hence, the approximate displacement of the hiker from the camp is \(11\sqrt2\ km\).

The average velocity of the particle between t = 2.0 s and t = 4.0 s is 4 m/2.0 s = 2.0 m/s.

Average velocity is the rate of change of an object's position, expressed as a vector quantity that tells both the speed and direction of the object's motion.

The average velocity of the particle between t = 2.0 s and t = 4.0 s can be calculated by taking the difference in the x-position of the particle at t = 4.0 s and t = 2.0 s, and dividing it by the difference in the time.
The x-position of the particle at t = 2.0 s is 4 m and the x-position of the particle at t = 4.0 s is 8 m.
Therefore, the difference in the x-position is 8 m - 4 m = 4 m.
The difference in time is 4.0 s - 2.0 s = 2.0 s. Therefore, the average velocity of the particle between t = 2.0 s and t = 4.0 s is 4 m/2.0 s = 2.0 m/s.

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Answer:

11.76Newtons

Explanation:

Workdone = Force * distance

Given

Workdone = 180joules

Distance = 15.3 metres

Required

Force

From the formula;

Force - Work/distance

Force = 180/15.3

Force = 11.76Newtons

Hence the required force is 11.76Newtons

Explanation:

F = ma

m = 90 kg    

           a = change in v / change in time = (40-5)m/s  / 3 s =  11.67 m/s^2

F = ma = 90 kg ( 11.67 m/s^2 ) = 1050 N

Answer:

10 mls2

Explanation:

speed =distance /time

Answer:

Give me some hint please

Answer:

MA=Fo/Fi

Explanation:

Answer: Inferring,Predicting,Analyzing

if this wasn't what you were looking for, here are some more science skills to help you pass you classes

- Evaluating.

- Generating possibilities.

- Formulating hypothesis.

- Communicating

Explanation: all these science skills will help you get through your science class with no struggles