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slamgirl [31]
3 years ago
8

Describe the mechanical energy of a roller coaster car immediately before it begins traveling down a long track

Physics
1 answer:
GrogVix [38]3 years ago
5 0
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

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A 100 Ω resistor is connected in series with a 47 µF capacitor and a source whose maximum voltage is 5 V, operating at 100.0 Hz.
Pani-rosa [81]

Answer:

X_c=-33.86275385\Omega

|Z|=105.5778675\Omega

I=0.04735841062A

\phi=20.78612878\°

Explanation:

The electrical reactance is defined as:

X_c=-\frac{1}{2\pi fC}

Where:

f=Frequency\\C=Capacitance

So, replacing the data provided by the problem:

X_c=\frac{1}{2\pi *100*(47*10^{-6} )} =-33.86275385\Omega

Now, the impedance can be calculated as:

Z=R+jX_c

Where:

R=Resistance\\X_c= Capacitive\hspace{3}reactance

Replacing the data:

Z=100-j33.86275385

In order to find the magnitude of the impedance we can use the next equation:

|Z|=\sqrt{(R^2)+(X_c^2)}=\sqrt{(100)^2+(-33.86275385)^2} =105.5778675\Omega

We can use Ohm's law to find the current:

V=I*Z\\I=\frac{V}{Z}

Therefore the current is:

I=\frac{5}{100-j33.86275385}=0.04485638113+0.01518960593j

And its magnitude is:

|I|=\sqrt{(0.04485638113)^2+(0.01518960593)^2} =0.04735841062\Omega

Finally the phase angle of the current is given by:

\phi=arctan(\frac{0.01518960593}{0.04485638113})=20.78612878\°

5 0
3 years ago
Why gravitational force at the center og earth is zero
KengaRu [80]
F = G Mm/r²
mg = G Mm/r²
g = GM/r²
At centre of earth, r=0
g = GM/0
g =0
8 0
3 years ago
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A ball is dropped and bounces off the floor. Its speed is the same immediately before and immediately after the collision. Which
Korolek [52]

Answer:

option B.

Explanation:

The correct answer is option B.

when the ball drops, the velocity of the ball before the collision is v

After the collision, the velocity of the ball is the same but in the opposite direction.

Impulse delivered to the ball and the floor, in this case, is not zero.

The magnitude of the momentum remains the same but the direction of the ball changes.

7 0
3 years ago
Suppose that the magnetic field in some region has the form B = kz ˆx (where k is a constant). Find the force on a square loop (
AleksAgata [21]

Answer:

F = ika^2

Explanation:

As we know that loop is placed in YZ plane and magnetic field is along x direction

So here net force on the side of the loop which lies along Y axis is given as

F_1 = i (\vec L \times \vec B)

here we know that on Y axis z = 0

so B = 0

so we have

F_1 = 0

now on the opposite side we have z = a

so magnetic field is given as

B = ka

so force on that side is given as

F = i(\vec L \times \vec B)

F = i(a)(ka) sin90

F_2 = ika^2

so net force on the loop is given as

F = F_1 + F_2

F = ika^2

5 0
3 years ago
Struggling on this, really need help
Bad White [126]

Answer:

Guessing you just need help with the definition but if it's the question I can still help you.

7 0
2 years ago
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