The graph shows the force and displacement of an object that is being pushed. How much work is done to push the object 0.12m?

The Graph Shows The Force And Displacement Of An Object That Is Being Pushed. How Much Work Is Done To

Answers

Answer 1

Answer:

Work done, W = 0.72 J

Explanation:

From the given graph, it is clear that the force applied is 6 N if the extension is 0.12 m.

Work done is given by :

W = Fx

[tex]W=6\ N\times 0.12\ m\\\\W=0.72\ J[/tex]

So, 0.72 J of work is done to push the object 0.12 m


Related Questions

How are the elements in the same row similar

Answers

Answer:

All elements in a row have the same number of electron shells. Each next element in a period has one more proton and is less metallic than its predecessor. Arranged this way, groups of elements in the same column have similar chemical and physical properties, reflecting the periodic law.

A mutation causes a dog to be born with a tail that is shorter than normal.

Which best describes this mutation?

Answers

Answer:

A mutation causes a dog to be born with a tail that is shorter than normal. Which best describes this mutation? It is harmful because it obviously affects the dog’s survival. It is harmful because it affects the dog’s physical appearance. It is neutral because it does not obviously affect the dog’s survival. It is beneficial because it affects the dog’s physical appearance.

Explanation:

Answer:

C

Explanation:

:)))

What is the gravitational force related to the distance between two objects?

Answers

Answer:

Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases.

answer: It is inversely proportional to the square of the distance.

Explanation:

for the team

Einstein developed much of his understanding of relativity through the use of gedanken, or thought, experiments. In a gedanken experiment, Einstein would imagine an experiment that could not be performed because of technological limitations, and so he would perform the experiment in his head. By analyzing the results of these experiments, he was led to a deeper understanding of his theory. In each the following gedanken experiments, Albert is in the exact center of a glass-sided freight car speeding to the right at a very high speed vvv relative to you. Albert has a flashlight in each hand and directs them at the front and rear ends of the freight car. Albert switches the flashlights on at the same time.

In Albert's frame of reference, which beam of light travels at a greater speed, the one directed toward the front or the one toward the rear of the train, or do they travel at the same speed? Which beam travels faster in your frame of reference? Enter the answers for Albert's frame of reference and your frame of reference separated by a comma using the terms front, rear, and same. For example, if in Albert's frame of reference the beam of light directed toward the front of the train travels at a greater speed and in your frame of reference the two beams travel at the same speed, then enter front,same.

Answers

Answer:

For eintein's frame of reference, both beam travel at the same speed.

For my own frame of reference, both beams travel at the same speed.

Explanation:

According to special relativity, the speed of light is the same in all direction on all reference frame. If not for this law we will assume the from beam will have a relative speed that will be the speed of light plus the speed of the fright car. This is not so and it violates the speed limit of light which according to the first law is the highest speed possible and nothing can go beyond that.

A 200.0 g block rests on a frictionless, horizontal surface. It is pressed against a horizontal spring with spring constant 4500.0 N/m (assume that the spring is massless). The block is held in position such that the spring is compressed 4.00 cm shorter than its undisturbed length. The block is suddenly released and allowed to slide away on the frictionless surface. Find the speed the block will be traveling when it leaves the spring.

Answers

Answer:

 6 m/s

Explanation:

Given that :

mass of the block   m =  200.0 g  = 200 × 10⁻³ kg

the horizontal spring constant   k  =  4500.0 N/m

position of the block (distance x) = 4.00 cm  = 0.04 m

To determine the speed the block will be traveling when it leaves the spring; we applying the  work done on the spring as it is stretched (or compressed) with the kinetic energy.

i.e [tex]\frac{1}{2} kx^2 = \frac{1}{2} mv^2[/tex]

[tex]kx^2 = mv^2[/tex]

[tex]4500* 0.04^2 = 200*10^{-3} *v^2[/tex]

[tex]7.2 =200*10^{-3}*v^{2}[/tex]

[tex]v^{2} =\frac{7.2}{200*10^{-3}}[/tex]

[tex]v =\sqrt{\frac{7.2}{200*10^{-3}}}[/tex]

v = 6 m/s

Hence,the speed the block will be traveling when it leaves the spring is  6 m/s

10) Two students want to use a 12-meter long rope to create standing waves. They first measure the speed at which a single wave pulse moves from one end of the rope to another and find that it is 36 m/s. What frequency must they vibrate the rope at to create the second harmonic

Answers

Answer:

To create a second harmonic the rope must vibrate at the frequency of 3 Hz

Explanation:

First we find the fundamental frequency of the rope. The fundamental frequency is the frequency of the rope when it vibrates in only 1 loop. Therefore,

f₁ = v/2L

where,

v = speed of wave = 36 m/s

L = Length of rope = 12 m

f₁ = fundamental frequency

Therefore,

f₁ = (36 m/s)/2(12 m)

f₁ = 1.5 Hz

Now the frequency of nth harmonic is given in general, as:

fn = nf₁

where,

fn = frequency of nth harmonic

n = No. of Harmonic = 2

f₁ = fundamental frequency = 1.5 Hz

Therefore,

f₂ = (2)(1.5 Hz)

f₂ = 3 Hz

What do you think will be different about cars in the future? Describe a change that is already being developed or that you think should be invented.

Answers

Answer:

Flying cars.

Explanation:

The shaft of a motor has an angular displacement θ that is a function of time given by the equation: θ(t) = 4.40 t 3 rad/s3 + 1.40 t2 rad/s2 . At time t = 0.00 s the wheel is at rest and is oriented at θ = 0.00 rad. a) Derive the equation that specifies the angular velocity of the shaft as a function of time. b) Derive the equation that specifies the angular acceleration as a function of time.

Answers

Answer:

a) [tex]\omega = 13.2t^2\frac{rad}{s^3}+2.80t\frac{rad}{s^2}[/tex]

b) [tex]\alpha=26.4t\frac{rad}{s^3}+2.80\frac{rad}{s^2}[/tex]

Explanation:

You have that the angular displacement is given by:

[tex]\theta=4.40t^3\frac{rad}{s^3}+1.40t^2\frac{rad}{s^2}[/tex]

a) the angular velocity is given by the derivative in time, of the angular displacement, that is:

[tex]\omega=\frac{d\theta}{dt}=\frac{d}{dt}[4.40 t^3 rad/s^3 + 1.40 t^2 rad/s^2]\\\\\omega=\frac{d\theta}{dt}=13.2t^2\frac{rad}{s^3}+2.80t\frac{rad}{s^2}[/tex]

b) the angular acceleration is the derivative, in time, of the angular velocity:

[tex]\alpha=\frac{d\omega}{dt}=\frac{d}{dt}[13.2t^2\frac{rad}{s^3}+2.80t\frac{rad}{s^2}]\\\\\alpha=26.4t\frac{rad}{s^3}+2.80\frac{rad}{s^2}[/tex]

Yellow light with wavelength 600 nm is travelling to the left (in the negative x direction) in vacuum. The light is polarized along the z direction. (a) Draw a neat snapshot mode labeled vector picture of the wave. (b) Draw a neat movie mode labeled vector picture of the wave. (c) If the wave were to represent blue light instead of yellow light, how would your pictures in parts a and b change? If there is no change, say so explicitly.

Answers

Answer: (a) and (b) => check attached file.

(c). Picture (a) and (b) will both remain the same.

Explanation:

IMPORTANT: The solution to the question (a) and (b) that is  (a) Draw a neat snapshot mode labeled vector picture of the wave. (b) Draw a neat movie mode labeled vector picture of the wave is there in the ATTACHED FILE/PICTURE.

It is also worthy of note to know that in anything Electromagnetic wave, the magnetic field, the Electric Field and their direction of propagation are perpendicular to each other.

Therefore, knowing the fact above we can say that in yellow light, the magnetic field is in the y-direction and the Electric Field is in the z-direction.

Hence, the solution to option C is given below;

(C).If the wave were to represent blue light instead of yellow light, picture (a) will remain the same because both light are Electromagnetic wave, although the wavelength will have to change. Picture (b) will also remain the same because they are both Electromagnetic waves and possess similar properties.

A radiator rests snugly on the floor of a room when the temperature is 10 oC. The radiator is connected to the furnace in the basement by a pipe that is 15 m long. How far off the floor will the radiator be lifted when it is filled with steam at 102 oC? The iron expands 1.0 * 10-5 / oC.

Answers

Answer: 10




Explanation subtract

Help ill give you brainliest !!!

Answers

Answer:

1. B

2. A

3. C

4. B

5. A

6. Muscular strength is different than muscular endurance because of the fact that muscular strength is the amount of force that can be exerted in one instance. Muscular endurance is how long that you can exert that force without being completely exhausted.

7. Some benefits to strength training is the increase in muscular endurance. There is also the benefit of better muscular strength.

Explanation:

A long solid conducting cylinder with radius a = 12 cm carries current I1 = 5 A going into the page. This current is distributed uniformly over the cross section of the cylinder. A cylindrical shell with radius b = 21 cm is concentric with the solid cylinder and carries a current I2 = 3 A coming out of the page. 1)Calculate the y component of the magnetic field By at point P, which lies on the x axis a distance r = 41 cm from the center of the cylinders.

Answers

Answer:

Explanation:

We shall use Ampere's circuital law to find magnetic field at required point.

The point is outside the circumference of two given wires so whole current will be accounted for .

Ampere's circuital law

B = ∫ Bdl = μ₀ I

line integral will be over circular path of radius r = 41 cm .

Total current  I  = 5A -3A = 2A .

∫ Bdl = μ₀ I

2π r B = μ₀ I

2π x .41  B = 4π x 10⁻⁷ x 2

B = 2 x 10⁻⁷ x 2 / .41

= 9.75 x 10⁻⁷ T . It will be along - ve Y - direction.

5.00 kg of liquid water is heated to 100.0 °C in a closed system. At this temperature, the density of liquid water is 958 kg/m3 . The pressure is maintained at atmospheric pressure of 1.01 x 105 Pa. A moveable piston of negligible weight rests on the surface of the water. The water is then converted to steam by adding an additional amount of heat to the system. When all of the water is converted, the final volume of the steam is 8.50 m3 . The latent heat of vaporization of water is 2.26 x 106 J/kg. Calculate how much work is done and the change in the internal energy during this isothermal process.

Answers

Answer:

1.04 x 107 J.

Explanation:

We can use the following method to do the calculation

Total energy given to water to convert intosteam

dQ = m* l

dQ = 5.00* 2.26 * 106

= 1.13* 107 J

Work done at constantpressure dW = P* dV

Initialvolume V1 = 5.00kg / 958

= 5.22* 10-3 m3

Finalvolume = 8.50 m3

=> dW = 1.01* 105 * ( 8.50 - 5.22 * 10-3)

= 8.58* 105 J

First law of thermodynamicsis dQ = ΔU + dW

Change in internalenergy ΔU = 1.13* 107 - 8.58 *105

= 1.04 x 107 J as our answer

A piston with stops containing water goes through an expansion process through the addition of heat. State 1 the pressure is 200 kPa and the volume is 2 m3. After half of the heat has been delivered the piston hits the stops corresponding to a volume of 5 m3. After all the heat has been delivered, state 2, the pressure is 1000 kPa with the piston resting on the stops. What is the work?

Answers

Answer:

The work will be "600 kJ/kg".

Explanation:

(1-a) ⇒ Constant Pressure

(a-2) ⇒ Constant Volume

The given values are:

In state 1,

Pressure, P₁ = 200 kPa

Volume, V₁ = 2m³

In state 2,

Pressure, P₂ = 1000 kPa

Volume, V₁ = 5m³

Now,

In process (1-a), work will be:

W₁₋ₐ = P₁(Vₐ - V₁)

On putting the values, we get

⇒ W₁₋ₐ = 200(5-2)

⇒         = 200(3)

⇒         = 600 kJ/kg

In process (a-2), work will be:

Wₐ₋₂ = 0

∴ (The change in the volume will be zero.)

So,

Total work = (W₁₋ₐ) + (Wₐ₋₂)

⇒                    = 600 + 0

⇒                    = 600 kJ/kg

How the musculoskeletal and nervous system develop as a human grows

Answers

Answer:

Explanation:

A fan pushes hot air out of a vent and into a room. The hot air displaces cold air in the room, causing the cold air to move closer to the floor.

The hot air displacing the cold air is an example of  transfer by

A ball with a mass of 4 kg is initially traveling at 2 m/s and has a 5 N force applied for 3 s. What is the initial momentum of the ball?

Answers

Answer:

The initial momentum of the ball is 8 kg-m/s.

Explanation:

Given that,

Mass of the ball is 4 kg

Initial speed of the ball is 2 m/s

Force applied to the ball is 5 N for 3 seconds

It is required to find the initial momentum of the ball. Initial momentum means that the product of mass and initial velocity of the ball. It is given as :

[tex]p_i=mu\\\\p_i=4\ kg\times 2\ m/s\\\\p_i=8\ kg-m/s[/tex]

So, the initial momentum of the ball is 8 kg-m/s.

A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes and the car comes to a rest uniformly in a distance of 160 m. What are the magnitude and direction of the net force applied to the car to bring it to rest?

Answers

Answer:

Force applied to stop the car = 1,250 N

Explanation:

Given:

Mass of car (M) = 1,000 kg

Initial velocity (U) = 20 m/s

Final velocity (V) = 0 m/s

Distance (S) = 160 m

Find:

Force applied to stop the car.

Computation:

[tex]v^2 = u^2 + 2as\\\\0^2=20^2+2(a)(160)\\\\0=400+320(a)\\\\Acceleration = a = -1.25m/s^2\\\\Force = ma \\\\Force= 1,000(1.25)\\\\Force = 1,250 N[/tex]

Force applied to stop the car = 1,250 N

A fisherman fishing from a pier observes that the float on his line bobs up and down, taking 2.4 s to move from its highest point to its lowest point. He also estimates that the distance between adjacent wave crests is 48 m. What is the speed of the waves going past the pier?Immersive Reader

Answers

Answer:

  v = 10 m / s

Explanation:

For this exercise we will use the relationship between the speed of a wave and its frequency and wavelength

         v = λ f

the wavelength is the distance at which the wave repeats, in this case the distance between the two ridges λ = 48 m.

the frequency is the number of oscillations per unit of time, it is also the inverse of the period which is the time in a complete oscillation, in this case they give us the time of half a period, ½ T = 2.4 s

   T = 4.8 s

the frequency is

        f = 1 / T

        f = 1 / 4.8

        f = 0.2083 Hz

let's calculate

        v = 0.2083 48

        v = 10 m / s

Based on the simple blackbody radiation model described in class, answer the following question. The planets Mars and Venus have albedo values of 0.15 and 0.75, and observed surface temperatures of approximately 220 K and 700 K, respectively. The average distance of Mars from the sun is 2.28 x 108 km, and the average distance of Venus is 1.08 x 108 km. Given that the radius of the sun is 7 x 108 m, and the energy flux at the surface of the sun is 6.28 x 107 W/m2 , what is the extent of the greenhouse effect for Mars

Answers

Answer:

The extent of greenhouse effect on mars is [tex]G_m = 87 K[/tex]  

Explanation:

From the question we are told that

     The albedo value of Mars is  [tex]A_1 = 0.15[/tex]

      The albedo value of Mars is  [tex]A_2 = 0.15[/tex]

       The surface temperature of Mars is  [tex]T_1 = 220 K[/tex]

        The surface temperature of Venus is  [tex]T_2 = 700 K[/tex]

          The distance of Mars from the sun is [tex]d_m = 2.28*10^8 \ km = 2.28*10^8* 1000 = 2.28*10^{11} \ m[/tex]

          The distance of Venus from sun is  [tex]d_v = 1.08 *10^{8} \ km = 1.08 *10^{8} * 1000 = 1.08 *10^{11} \ m[/tex]

       The radius of the sun is [tex]R = 7*10^{8} \ m[/tex]

        The energy flux is   [tex]E = 6.28 * 10^{7} W/m^2[/tex]

The solar constant for Mars is mathematically represented as

 

          [tex]T = [\frac{E R^2 (1- A_1)}{\sigma d_m} ][/tex]

Where [tex]\sigma[/tex] is the Stefan's constant with a value  [tex]\sigma = 5.6*10^{-8} \ Wm^{-2} K^{-4}[/tex]

So substituting values

         [tex]T = \frac{6.28 *10^{7} * (7*10^8)^2 * (1-0.15)}{(5.67 *10^{-8}) * (2.28 *10^{11})^2)}[/tex]

          [tex]T = 307K[/tex]

So the greenhouse effect on Mars is  

           [tex]G_m = T - T_1[/tex]

           [tex]G_m = 307 - 220[/tex]

          [tex]G_m = 87 K[/tex]

   

Electric fields are MOST associated with ________.

Answers

With each point in space

Forests have been completely wiped out in some parts the world. In areas that are economically depressed, people use the cut wood to fuel fires for cooking and for making shelters or homes. How would deforestation most likely impact Earth's biogeochemical cycles?

Group of answer choices


Less carbon dioxide will be present in the atmosphere because the people living in the area still breath in oxygen.


An increase in carbon dioxide will occur in the atmosphere because there are less trees to use the gas in the process of photosynthesis.


Carbon dioxide levels will not change, since plants do not have a role in the carbon cycle.


More water will be released into the atmosphere because transpiration and condensation will produce more precipitation.

Answers

Answer: An increase in carbon dioxide will occur in the atmosphere because there are less trees to use the gas in the process of photosynthesis.

Explanation: Photosynthesis can be simply defined as a process where green plants uses minerals from the soil, water and carbon dioxide to produce their own food (energy-rich organic compound/starch) while oxygen is released into the environment.

On the other hand, Human beings and other animals inhales the oxygen released into the environment by green plants and breath out carbon dioxide.

Deforestation is the permanent removal trees either for commercial or household use or in some cases, to make room or space for other activities or development.

Now when these trees (green plants) are permanently cut down without replacement, humans and other animals continuously exhale carbon dioxide and there will be little or no green plants (trees) available to convert this carbon dioxide into oxygen.

Hence, there will be an increased amount of carbon dioxide in the atmosphere as their are little or no tress to use the gas (carbon dioxide) in the process of photosynthesis.

A steam engine takes in superheated steam at 270 °C and discharges condensed steam from its cylinder at 50 °C. The engine has an efficiency of 30%, and taken in 50 kJ from the hot steam per cycle. If a Carnot engine takes in the same amount of heat per cycle and operates at these temperatures, the work it can turn into is most likely to be:a) 15 kJ. b) 20 kJ. c) 10 kJ. d) 50 kJ.

Answers

Answer:

b) 20 kJ

Explanation:

Efficiency of carnot engine = (T₁ - T₂ ) / T₁  Where T₁ is temperature of hot source  and T₂ is temperature of sink .

T₁ = 270 + 273 = 543K

T₂ = 50 + 273 = 323 K

Putting the given values of temperatures

efficiency = (543 - 323) / 543

= .405

heat input = 50 KJ

efficiency = output work / input heat energy

.405 = output work / 50

output work = 20.25 KJ.

= 20 KJ .

Which is the correct representation of the right-hand rule for a current flowing to the right?

Answers

Answer:

The third image

Explanation:

The one with the thumb pointing to the right

Answer:

3, correct on Edge 2020

To understand thermal linear expansion in solid materials. Most materials expand when their temperatures increase. Such thermal expansion, which is explained by the increase in the average distance between the constituent molecules, plays an important role in engineering. In fact, as the temperature increases or decreases, the changes in the dimensions of various parts of bridges, machines, etc., may be significant enough to cause trouble if not taken into account. That is why power lines are always sagging and parts of metal bridges fit loosely together, allowing for some movement. It turns out that for relatively small changes in temperature, the linear dimensions change in direct proportion to the temperature.
For instance, if a rod has length L0 at a certain temperature T0 and length L at a higher temperature T, then the change in length of the rod is proportional to the change in temperature and to the initial length of the rod: L - L0 = αL0(T - T0),
or
ΔL = αL0ΔT.
Here, α is a constant called the coefficient of linear expansion; its value depends on the material. A large value of α means that the material expands substantially as the temperature increases; smaller values of α indicate that the material tends to retain its dimensions. For instance, quartz does not expand much; aluminum expands a lot. The value of α for aluminum is about 60 times that of quartz!
Questions:
A) Compared to its length in the spring, by what amount ΔLwinter does the length of the bridge decrease during the Teharian winter when the temperature hovers around -150°C?
B) Compared to its length in the spring, by what amount ΔLsummer does the length of the bridge increase during the Teharian summer when the temperature hovers around 700°C?

Answers

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

What types of mediums are involved in the energy transfer

Answers

Answer:

In electromagnetic waves, energy is transferred through vibrations of electric and magnetic fields. In sound waves, energy is transferred through vibration of air particles or particles of a solid through which the sound travels. In water waves, energy is transferred through the vibration of the water particles.

A student performs an experiment that involves the motion of a pendulum. The student attaches one end of a string to an object of mass M and secures the other end of the string so that the object is at rest as it hangs from the string. When the student raises the object to a height above its lowest point and releases it from rest, the object undergoes simple harmonic motion. As the student collects data about the time it takes for the pendulum to undergo one oscillation, the student observes that the time for one swing significantly changes after each oscillation. The student wants to conduct the experiment a second time. Which two of the following procedures should the student consider when conducting the second experiment?
a) Make sure that the length of the string is not too long.
b) Make sure that the mass of the pendulum is not too large.
c) Make sure that the difference in height between the pendulum's release position and rest position is not too large.
d) Make sure that the experiment is conducted in an environment that has minimal wind resistance.

Answers

Answer:

the answers the correct one is cη

Explanation:

In this simple pendulum experiment the student observes a significant change in time between each period. This occurs since an approximation used is that the sine of the angle is small, so

              sin θ = θ

 

with this approach the equation will be surveyed

     d² θ / dt² = - g / L sin θ

It is reduced to

      d² θ / dt² = - g / L θ

in which the time for each oscillation is constant, for this approximation the angle must be less than 10º so that the difference between the sine and the angles is less than 1%

The angle is related to the height of the pendulum

         sin θ = h / L

         h = L sin θ.

Therefore the student must be careful that the height is small.

When reviewing the answers the correct one is cη

Considering the approximation of simple harmonic motion, the correct option is:

(c) Make sure that the difference in height between the pendulum's release position and rest position is not too large.

Simple Harmonic Motion

According to Newton's second law in case of rotational motion, we have;

[tex]\tau = I \alpha[/tex]

Applying this, in the case of a simple pendulum, we get;

[tex]-mg\,sin\,\theta =mL^2 \,\frac{d^2 \theta}{dt^2}[/tex]

On, rearranging the above equation, we get;

[tex]mL^2 \,\frac{d^2 \theta}{dt^2} + mg\,sin\,\theta=0\\\\\implies \frac{d^2 \theta}{dt^2} +\frac{g}{L} sin \,\theta=0[/tex]

Now, if angular displacement is very small, i.e.; the bob of the pendulum is only raised slightly.

Then, [tex]sin\, \theta \approx \theta[/tex]

[tex]\implies \frac{d^2 \theta}{dt^2} +\frac{g}{L} \,\theta=0[/tex]

This is now in the form of the equation of a simple harmonic motion.

[tex]\frac{d^2 \theta}{dt^2} +\omega^2 \,\theta=0[/tex]

Comparing both these equations, we can say that;

[tex]\omega = \sqrt{\frac{g}{L}}[/tex]

[tex]T=2\pi\sqrt{\frac{L}{g}}[/tex]

This relation for the time period can only be obtained if the angular displacement is very less.

So, the correct option is;

Option (c): Make sure that the difference in height between the pendulum's release position and rest position is not too large.

Learn more about simple harmonic motion here:

https://brainly.com/question/26114128

the speed of sound is 343m/s. dezeirey is positioned 5m behind her. how many seconds will it take for the echo from the wall to reach her

Answers

Answer:

t = 0.029 s

Explanation:

We have,

Speed of sound is 343 m/s.

Dezeirey is positioned 5 m behind her.

It is required to find the time taken for the echo from the wall to reach her. The total distance covered by the echo when it reaches her is 2d or 10 m.

Time taken,

[tex]t=\dfrac{d}{v}\\\\t=\dfrac{10\ m}{343\ m/s}\\\\t=0.029\ s[/tex]

So, it will take 0.029 seconds for the echo from the wall to reach her.

A solid cylinder of mass m and radius R rolls down a ramp, starting from rest at a height h above a nearby horizontal surface. The coefficients of kinetic and static friction and are non-zero, and sufficiently large that the cylinder rolls down the ramp without slipping. Assume that the coefficient of rolling friction is zero. As the cylinder leaves the ramp, it continues along a horizontal surface (with the same frictional coefficients as the ramp).

Required:
What is the speed V of the cylinder after it has traveled a distance D along the horizontal surface?

Answers

Answer:

the volocity is 50

Explanation:

By which process does the heat from the Sun reach the Earth? (AKS 4b DOK 1) *

Answers

The earth radiation

does work done by the electric force depends on the path taken.

Answers

Answer:

yes but not sure

Explanation:

i don't know so much but I just guessed I'm not sure

Other Questions
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