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brilliants [131]
3 years ago
13

A very large magnet applies a repulsive force to a smaller (0.05 Kg) magnet. If the smaller magnet accelerates across a friction

less table at 2m/s2, then what is the magnitude of the repulsive force?
Physics
1 answer:
makkiz [27]3 years ago
5 0

Answer:

0.1 N

Explanation:

using F = ma yields 0.1 N.

net

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Describe the mechanical energy of a roller coaster car immediately before it begins traveling down a long track
GrogVix [38]
At the top of the hill, the cars possess a large quantity of potential energy. Potential energy - the energy of vertical position - is dependent upon the mass of the object and the height of the object. The car's large quantity of potential energy is due to the fact that they are elevated to a large height above the ground. As the cars descend the first drop they lose much of this potential energy in accord with their loss of height. The cars subsequently gain kinetic energy. Kinetic energy - the energy of motion - is dependent upon the mass of the object and the speed of the object. The train of coaster cars speeds up as they lose height. Thus, their original potential energy (due to their large height) is transformed into kinetic energy (revealed by their high speeds). As the ride continues, the train of cars are continuously losing and gaining height. Each gain in height corresponds to the loss of speed as kinetic energy (due to speed) is transformed into potential energy (due to height). Each loss in height corresponds to a gain of speed as potential energy (due to height) is transformed into kinetic energy (due to speed). A roller coaster ride also illustrates the work and energy relationship. The work done by external forces is capable of changing the total amount of mechanical energy from an initial value to some final value. The amount of work done by the external forces upon the object is equal to the amount of change in the total mechanical energy of the object. The relationship is often stated in the form of the following mathematical equation.

KEinitial + PEinitial + Wexternal = KEfinal + PEfinal

The left side of the equation includes the total mechanical energy (KEinitial + PEinitial) for the initial state of the object plus the work done on the object by external forces (Wexternal) while the right side of the equation includes the total mechanical energy (KEfinal + PEfinal) for the final state of the object.

Once a roller coaster has reached its initial summit and begins its descent through loops, turns and smaller hills, the only forces acting upon the coaster cars are the force of gravity, the normal force and dissipative forces such as air resistance. The force of gravity is an internal force and thus any work done by it does not change the total mechanical energy of the train of cars. The normal force of the track pushing up on the cars is an external force. However, it is at all times directed perpendicular to the motion of the cars and thus is incapable of doing any work upon the train of cars. Finally, the air resistance force is capable of doing work upon the cars and thus draining a small amount of energy from the total mechanical energy which the cars possess. However, due to the complexity of this force and its small contribution to the large quantity of energy possessed by the cars, it is often neglected. By neglecting the influence of air resistance, it can be said that the total mechanical energy of the train of cars is conserved during the ride. That is to say, the total amount of mechanical energy (kinetic plus potential) possessed by the cars is the same throughout the ride. Energy is neither gained nor lost, only transformed from kinetic energy to potential energy and vice versa.

The conservation of mechanical energy by the coaster car in the above animation can be studied using a calculator. At each point in the ride, the kinetic and potential energies can be calculated using the following equations.

<span> KE = 0.5 * mass * (speed)^2 PE = mass * g * height</span>

If the acceleration of gravity value of 9.8 m/s/s is used along with an estimated mass of the coaster car (say 500 kg), the kinetic energy and potential energy and total mechanical energy can be determined

5 0
3 years ago
Ben franklin showed that 1 teaspoon of oil would cover about 0.50 acre of still water. if you know that 1.0 x 104m2 = 2.47 acres
Dmitry_Shevchenko [17]

To solve for the thickness of the latter of oil, we can simply divide the volume by the total surface area:

Thickness = Volume / Surface Area

We are both given the volume and the surface area, all we have to do now is to convert the units in like terms so that we can cancel those out.

Surface Area = 0.50 acres (1.0 x 10^4 m^2 / 2.47 acres)

Surface Area = 2,024.29 m^2

Further converting this into cm:

Surface Area = 2,024.29 m^2 (100 cm / m)^2

Surface Area = 20,242,914.98 cm^2

 

Therefore the thickness is:

Thickness = 5.0 cm^3 / 20,242,914.98 cm^2

Thickness = 2.47 x 10^-7 cm

or

<span>Thickness = 2.47 x 10^-9 m = 2.47 nm = 24.7 A</span>

3 0
4 years ago
How can you determine the number of covalent bonds an element can form?
BabaBlast [244]
Yes it depends on the column on the periodic table
3 0
3 years ago
Does an object travel farther on a smooth or slippery surface or on a rough surface? Why?
Vitek1552 [10]

Answer:

There is much more friction on the rough surface than there is on the smooth surface.

Explanation:

8 0
3 years ago
Make a rough estimate of the number of quanta emitted in one second by a 100 W light bulb. Assume that the typical wavelength em
mixas84 [53]

Answer:

#_photon = 5 10²⁰ photons / s

Explanation:

For this exercise let's calculate the energy of a single quantum of energy, use Planck's law

         E = h f

         c= λ f

         E = h c / λ

          λ= 1000 nm (1 m / 109 nm) = 1000 10⁻⁹ m

Let's calculate

          E₀ = 6.6310⁻³⁴ 3 10⁸/1000 10⁻⁹

          E₀ = 19.89 10⁻²⁰ J

This is the energy emitted by a photon let's use a proportions rule to find the number emitted in P = 100 w

                #_photon = P / E₀

               #_photon = 100 / 19.89 10⁻²⁰

              #_photon = 5 10²⁰ photons / s

6 0
3 years ago
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