If you move 2 newtons of mess for 5 meters how much work did you do?

The acceleration on a body that approached the earth and comes within 6 earth radii of the earth’s surface is 0.2 m/s².

The acceleration of the body can be calculated using the equation-

g = g0 R/(R + H)².

In the question, we have, h = 6R

g = 9.8 x R/(R + 6R)²

By putting in the values and calculating, we get,

g = 0.2 m/s².

Where g' = Gravitational Acceleration at Height, R = Earth's Radius, and g = Gravitational Acceleration at Earth's Surface. As a result, at a distance of 2R from the earth's surface, the acceleration of a body caused by the earth's attraction is equal to g/9.

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Answer:  NEWTONS FIRST LAW

One's body moves to the side when a car makes a sharp turn.

Tightening of seat belts in a car when it stops quickly.

A ball rolling down a hill will continue to roll unless friction or another force stops it.

If pulled quickly, a tablecloth can be removed from underneath dishes.

                                NEWTONS SECOND LAW

1· Try to move an object.

2· Pushing a car and a truck.

3· Racing Cars.

4· Rocket launch.

5· Kick the ball.

6· Car crash.

7· Two people walking.

8· Object thrown from a height

   Newton's Second Law of Motion says that acceleration (gaining speed) happens when a force acts on a mass (object). Riding your bicycle is a good example of this law of motion at work. Your bicycle is the mass. Your leg muscles pushing on the pedals of your bicycle is the force.

                                     NEWTONS THIRD LAW

                  Examples

The recoil of a Gun. ...

Swimming. ...

Pushing the Wall. ...

Diving off a Raft. ...

Space Shuttle. ...

Throwing a Ball. ...

Walking. ...

Hammering a Nail.

More items...

Common examples of newton's third law of motion are: A horse pulls a cart, a person walks on the ground, a hammer pushes a nail, and magnets attract paper clips. In all these examples a force is exerted on one object and that force is exerted by another object.

The density of the rod is 3.0 kg/m at the origin,then the center of mass of the rod is located 2.15 m from the origin.

The center of mass is the point in an object or system of objects where the mass of the object or system can be considered to be concentrated. In other words, it is the point in which an object or system can be balanced and behaves as if all the mass is concentrated at that point.

To find the center of mass of the rod, we need to integrate the product of the density and position of each infinitesimal element of the rod, divided by the total mass of the rod.

We can express the position of each element as x, where x=0 at the origin and x=4.5 m at the other end of the rod.

Let's start by finding the total mass of the rod. We can divide the rod into two parts: one from the origin to the midpoint, and the other from the midpoint to the other end. The mass of each part can be found by integrating the density over its length:

m₁ = ∫(0 to 2.25) 3x² dx = 15.19 kg

m₂ = ∫(2.25 to 4.5) (22.5x - 45.375) dx

                                = 95.63 kg

The total mass of the rod is then derived as follows:

                 m = m₁ + m₂ = 110.82 kg

We can now evaluate the location of the centre of mass. We can express the position of the center of mass as xcm, where x=0 at the origin and x=4.5 m at the other end of the rod.

We can use the formula:

xcm = (1/m) ∫(0 to 4.5) ρ(x) x dx

where (x) represents the density at position x.

We can express the density as a quadratic function of x:

                          ρ(x) = ax² + bx + c

Using the given values of density at the origin, midpoint, and other end, we can solve for the coefficients a, b, and c:

                               c = 3

                         a + b + c = 11/4.5

                           = 2.44

4a + 2b + c = 3

Solving these equations gives:

a = -1.96 kg/m³

b = 7.33 kg/m²

Now we can plug in the expression for ρ(x) and evaluate the integral:

xcm = (1/m) ∫(0 to 4.5) (-1.96x⁴ + 7.33x³ + 3x) dx

Evaluating this integral gives:

                                       xcm = 2.15 m

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In a genetic cross Rr x rr the percentage of offspring will be round will be 50% since this trait is associated with the recessive genotype (rr).

The recessive genotype in a genetic cross is the proportion of offspring that will be homozygous recessive and therefore will express the recessive trait such as in this case round plants that are produced by the combination of recessive and heterozygous parents.

Therefore, with this data, we can see that recessive genotypes in a genetic cross are generated by the combination of two recessive gametes.

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The perpendicular force exerted by the incline on the car is Fw cosθ.

Normal reaction force is defined as a contact force that is exerted by the surface on an object which is in contact with it. It prevents the object from passing through the surface.

Here,

Weight of the car is Fw

The incline makes an angle θ with the horizontal.

From the figure,

The perpendicular force exerted by the incline, N = Fw cosθ

Hence,

The perpendicular force exerted by the incline on the car is Fw cosθ.

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E. Vital Functions of a Team. Developing camaraderie is an important part of creating a team atmosphere, as it helps team members to learn to trust and rely on each other.

Atmosphere is the layer of gases that surrounds the Earth. It is composed of roughly 78% nitrogen, 21% oxygen, and 1% other gases, such as argon, carbon dioxide and water vapor. The atmosphere protects the Earth from harmful ultraviolet radiation from the sun, shelters us from extreme temperatures, and creates the air pressure that allows us to breathe. The air we breathe is a mixture of gases, and the atmosphere also contains dust, smoke and other particles.

Paying attention during training helps to ensure that team members have all the necessary information to be able to perform their duties, while taking part in team activities and training helps to strengthen team bonds and create a sense of unity.

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Answer:131s

Explanation:

time = 5.15*10^5 J / 3940

Answer:

3 second is the answer of your question

The final velocity of block A is -2 m/s to the left ( in the opposite direction of its initial velocity).

Quantity that designates how fast and in what direction a point is moving is known as velocity.

As we know, m_A * v_Ai + m_B * v_Bi = m_A * v_Af + m_B * v_Bf

m_A and m_B are the masses of blocks A and B, v_Ai and v_Bi are their initial velocities, and v_Af and v_Bf are their final velocities.

5 kg * 10 m/s + 10 kg * 0 m/s = 5 kg * v_Af + 10 kg * 6 m/s

50 kgm/s = 5 kg * v_Af + 60 kgm/s

-10 kg*m/s = 5 kg * v_Af

So, v_Af = -2 m/s

Therefore, the final velocity of block A is -2 m/s to the left (i.e., in the opposite direction of its initial velocity).

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Answer:

Explanation:

Force = mass*acceleration

15=mass*5

mass=3kg

When a town puts in new power lines during the height of summer, it needs to be able to handle the increased electrical load on the power grid that will probably come from more people using air conditioners.

The power grid needs to be able to handle the increased demand for electricity that will result from the use of air conditioning when new power lines are installed. In order to guarantee that there is sufficient capacity to meet the increased demand, this may necessitate improvements to the existing infrastructure, which may include transformers, distribution lines, and substations.

In addition, it may be necessary to coordinate the installation of new power lines with the local utility company to avoid disrupting the existing power supply to homes and businesses. This may necessitate temporary power outages or other measures to guarantee that the installation goes off without a hitch and does not pose any safety risks.

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Explanation:

The work done by a force F over a displacement d is given by the dot product of the force and the displacement:

W = F · d

where · denotes the dot product.

We are given that the work done is 33 J, so

F · d = 33 J

We can find the dot product by taking the sum of the products of the corresponding components:

F · d = (2i + 5j) · (-i + 7j) = 2(-1) + 5(7) = 33

Now, we can find the magnitude of the force and displacement vectors:

|F| = √(2² + 5²) = √29

|d| = √((-1)² + 7²) = √50

The angle between the force and displacement vectors can be found using the dot product:

F · d = |F| |d| cos θ

cos θ = (F · d) / (|F| |d|)

cos θ = 33 / (√29 √50)

cos θ ≈ 0.603

θ ≈ 53.2°

However, we are looking for the angle between the force and the displacement, which is the supplement of θ, so:

angle = 180° - 53.2° ≈ 126.8°

None of the answer choices match this result, so we must have made a mistake.

Let's check our work. We made an error in the calculation of the cosine of θ. It should be:

cos θ = (F · d) / (|F| |d|)

cos θ = 33 / (√29 √50)

cos θ ≈ 0.433

θ ≈ 64.2°

The angle between the force and displacement vectors is approximately 64.2°.

Now we can check the answer choices:

A. 17.4° - Too small

B. 21.6° - Too small

C. 29.9° - Too small

D. 32.7° - Too small

E. 39.8° - Too small

None of the answer choices match the calculated angle of 64.2°. Therefore, the correct answer is not among the choices given.

The final kinetic energy of the student is 352 J. We can use the formula for kinetic energy, $KE=\frac{1}{2}mv^2$, to solve for the final velocity, $v$,

Kinetic energy is the energy of motion. It is the energy that an object has due to its motion. It is one of the forms of energy that exist in nature. Kinetic energy is related to the mass of an object and its velocity. The greater the mass of an object, or the faster its velocity, the more kinetic energy the object possesses. Kinetic energy can be converted into other forms of energy such as heat or electricity, and can also be transferred from one object to another.

$352 = \frac{1}{2}mv^2$

$v = \sqrt{\frac{352}{m}}$

where $m$ is the mass of the student.

Now, we can use the equation for work, $W=Fd$, to calculate the distance the student must be pushed to reach the final velocity:

$W = Fd = mv^2$

$d = \frac{mv^2}{F} = \frac{m\sqrt{\frac{352}{m}}}{45} = \frac{\sqrt{352m}}{45}$

Therefore, the student must be pushed a distance of $\frac{\sqrt{352m}}{45}$ to reach a final kinetic energy of 352 J.

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It is not possible for a tree to survive if the inner part of the tree is missing. This is because, the inner part of the tree is vital for survival and health.

Trees are the plants which carry out the life processes that all plants share on the planet. It is not possible for a tree to survive without the inner part. If the entire inner part is missing, including the heartwood and sapwood, is missing. The inner part of a tree is vital to the survival as well as the health, as it is responsible for the transportation of water, nutrients, and sugars between the root and the leaves.

However, it is possible for a tree to survive if only some part of the inner part is missing. Trees have a remarkable ability to compartmentalize the damage and decay, which means that they can wall off or can isolate the injured or infected parts of the tree for the prevention of the spread of damage to the rest of the tree.

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The work done in raising an object is equal to the force applied to the object multiplied by the distance it is moved.

Objects in programming are entities that contain data and can be manipulated by code. They are commonly used to represent real-world things, such as a customer or an employee, or even abstract concepts such as a bank account. Objects are made up of properties, which are the data associated with the object, and methods, which are functions that objects can perform. Objects interact with one another through messages, which are sent from one object to another, allowing them to work together.

Therefore, we can calculate the distance the apple is lifted using the formula:

Work = Force × Distance

2.0 J = (0.180 kg)(9.8 m/s2) × d

d = 2.0 J / (0.180 kg)(9.8 m/s2)

d = 12.2 m

The apple is lifted 12.2 meters.

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The second mass will have a velocity of v² = 4√F of the force, F. Since the displacement and the force are the same in both cases

Frictional force is a force that opposes motion between two surfaces in contact. When two surfaces are in contact, there are microscopic irregularities that prevent them from sliding over each other easily. As a result, a force is required to overcome this resistance and move one surface over the other.

The frictional force between two surfaces depends on several factors, including the type of surfaces in contact, the force pressing the surfaces together (normal force), and the presence or absence of lubrication.

According to the work-energy principle, the work done by a force on an object is equal to the change in the object's kinetic energy. This principle can be used to solve the problem.

For the first case, the work done by the force F is:

                           W = Fd

where d is the displacement of the mass. This work is equal to the change in kinetic energy of the mass:

                                W = ΔK

where ΔK = (1/2)mv² - 0 is the change in kinetic energy, since the mass starts from rest.

Combining these equations, we get:

                                 Fd = (1/2)mv²

For the second case, the work done by the force 2F is:

                            W = 2Fd

This work is equal to the change in kinetic energy of the mass:

                                 W = ΔK

where ΔK = (1/2)(1/2)m v₂² - 0 is the change in kinetic energy, since the mass starts from rest.

Combining these equations, we get:

2Fd = (1/4)mv₂²

Solving for v², we get:

v² = √(8Fd/m)

Since the displacement and the force are the same in both cases, we can substitute the first equation into the expression for v²:

                             v² = √(8F(1 m)/(1/2m))

                                   v² = √(16F)

                                      v² = 4√F

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Answer:

speed

Explanation:

speed

Answer:

Equal to

Explanation:

work output is equal to the work input, so the efficiency is equal to 1 ( or 100% ) and the mechanical advantage is numerically equal to the velocity ratio. So, M.A. = V.R.

The final momentum of the artist-clay system immediately after collision is -82 kgm/s.

Assuming that the initial momentum of the clay was 3.0 kgm/s and the initial momentum of the potter was -85 kgm/s, we can use the law of conservation of momentum to find the final momentum of the artist-clay system immediately after the collision:

Initial momentum = Final momentum

(3.0 kgm/s) + (-85 kgm/s) = Final momentum

-82 kg×m/s = Final momentum

Therefore, the final momentum of the artist-clay system immediately after the collision is -82 kgm/s.

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Answer:

31,500 joules

Explanation:

Work = force * distance * cos(theta)

Force = 4500N (weight of the barbell)

Distance = 7m (the height lifted)

Theta = 0 (cosine of 0 degrees is 1, since the force and distance are in the same direction)

So, Work = 4500 N * 7m * cos(0)

Work = 31,500 joules

The object's kinetic energy, in contrast to potential energy, is relative to other fixed and moving things that are present inside its immediate vicinity.For instance, if the item is positioned at a higher height, its kinetic energy will be higher.

As height is reduced, potential energy is reduced as well.If height lowers, the kinetic energy for a freefall similarly reduces.That is as a result of the shorter distance to be traveled.

A doubling in height will lead to a doubling in gravitational potential energy since an object's gravitational potential energy is directly proportionate to its height above the zero point.The potential energy of gravity will triple with a tripling of height.

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To solve this problem, we can use conservation of energy:

Initial mechanical energy (at the top of the hill) = final mechanical energy (at the bottom of the hill)

Initial mechanical energy = mgh, where m is the total mass of the child and sled, g is the acceleration due to gravity (9.8 m/s^2), and h is the vertical height of the hill. Since the hill makes an angle of 16.9° with respect to the horizontal, we can use trigonometry to find the height:

h = (length of slope) * sin(16.9°)

We don't know the length of the slope, but we can solve for it by using the final speed of the child and sled:

Final mechanical energy = (1/2)mv^2, where v is the final speed of the child and sled.

Setting these equal, we get:

mgh = (1/2)mv^2

Substituting for h and solving for the length of the slope:

(length of slope) = v^2 / (2g*sin(16.9°))

Plugging in the given values:

(length of slope) = (6.8 m/s)^2 / (29.8 m/s^2sin(16.9°)) ≈ 12.5 meters

Therefore, the child and sled slide about 12.5 meters along the slope.

Answer:

Given:

R1 = 20 ohms

R2 = 20 ohms

R3 = 30 ohms

Total resistance = 10 ohms

Power = 0.36 kW

1/Req = 1/20 + 1/20 + 1/30

1/Req=2/15

Req 15/2

Req = 7.5 ohms

Hence, 10 - 7.5 = 2.5 ohms

PIR, hence 0.36 x 103

= I²(10)

I = √360/10 = √36 = 6 A

The potential energy of the mass is 0.0452 J when it is at a distance of

0.0600 m from equilibrium.

How to calculate potential energy?

The potential energy stored in a spring when it is stretched or compressed by a displacement x is given by:

[tex]$PE = \frac{1}{2}kx^2$[/tex]

where k is the spring constant.

In this problem, the spring constant is given as 26.6 N/m, and the displacement of the mass from equilibrium is

[tex](0.120 m - 0.0600 m) = 0.0600 m.[/tex]

Therefore, the potential energy stored in the spring when the mass is at a distance of 0.0600 m from equilibrium is:

[tex]$PE = \frac{1}{2}kx^2 = \frac{1}{2}(26.6 N/m)(0.0600 m)^2 = 0.0452 J$[/tex]

So the potential energy of the mass is 0.0452 J when it is at a distance of 0.0600 m from equilibrium.

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Answer: 0.0479

Explanation: This answer works for Acellus! :D

hope this helps!

The magnitudes of the vectors (A+1B) and (2A - B) would be 6.71 and 10.25 respectively

a) The vector A + B can be found by adding the corresponding components of A and B:

A + B = (3i - 1j + 4k) + (2i + 3j - 0k) = 5i + 2j + 4k

The magnitude of A + B is:

|A + B| = sqrt((5)^2 + (2)^2 + (4)^2) = sqrt(45) ≈ 6.71

b) The vector 2A - B can be found by multiplying the components of A and B by the appropriate scalar values and then subtracting:

2A - B = 2(3i - 1j + 4k) - (2i + 3j - 0k) = 6i - 2j + 8k - 2i - 3j = 4i - 5j + 8k

The magnitude of 2A - B is:

|2A - B| = sqrt((4)^2 + (-5)^2 + (8)^2) = sqrt(105) ≈ 10.25

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The acceleration of the moving vehicle when velocity and time is given would be = 15.2 km/hr²

Acceleration is defined as the quantity that shows the change on velocity of a moving object with respect to change in time.

That is;

acceleration = ∆velocity/∆time.

Velocity is defined as the change in the distance of a moving object with respect to time.

The velocity of the moving vehicle = 38km/hr

The time given = 2.5 hrs

Acceleration = velocity/time

= 38/2.5

= 15.2 km/hr²

Therefore, the acceleration of the vehicle when velocity and time is given = 15.2 km/hr².

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Answer:

F = G M1 M2 / R^2      attraction between 2 masses

M1 = F R^2 / (G M2)

M1 = 2.20E4 * (4.2E6)^2 / (6.67E-11 * 4.95E3) kg

M1 = 2.2 * 4.2^2 / (6.67 * 4.95) E24

M1 = 1.18E24 kg