Light travels at 2.4 x 10^8 m/s in a medium. What is the index of refraction for that medium? (speed of light in a vacuum= 3.0 x 10^8 m/s

.8

1.25

2.4

not enough information

1. The maximum height above its launch level is 1.45 m

2. The time of flight of the ball is 1.1 s

1. How do I determine the maximum height?

From the question given above, the following data were obtained:

The maximum height can be obatianed as follow:

H = u²Sine²θ / 2g

H = [5.5² × (Sine 76)²] / (2 × 9.8)

Maximum height = 1.45 m

How do I determine the time of flight?

The time of flight of the ball can be obtained as follow:

T = 2uSineθ / g

T = [2 × 5.5 × Sine 76] / 9.8

Time of flight = 1.1 s

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The average acceleration of the motorcycle over the given distance is approximately 9.39 m/s².

To calculate the average acceleration of the motorcycle, we can use the formula:

Average acceleration = (final velocity - initial velocity) / time

First, let's convert the final velocity from km/h to m/s since the distance is given in meters. We know that 1 km/h is equal to 0.2778 m/s.

Converting the final velocity:

Final velocity = 78 km/h * 0.2778 m/s = 21.67 m/s

Since the motorcycle starts from rest (initial velocity is zero), the formula becomes:

Average acceleration = (21.67 m/s - 0 m/s) / time

To find the time taken to reach this velocity, we need to use the formula for average speed:

Average speed = total distance/time

Rearranging the formula:

time = total distance / average speed

Plugging in the values:

time = 50 m / 21.67 m/s ≈ 2.31 seconds

Now we can calculate the average acceleration:

Average acceleration = (21.67 m/s - 0 m/s) / 2.31 s ≈ 9.39 m/s²

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Answer:

Ratio of kinetic energy of object A to B = 1:5

Explanation:

Given:

Mass of object A = 200 kg

Mass of object B = 1,000 kg

Find:

Ratio of kinetic energy of object A to B

Computation:

Kinetic energy = (1/2)(m)(v²)

Kinetic energy of object A = (1/2)(200)(v²)

Kinetic energy of object A = (100)(v²)

Kinetic energy of object B = (1/2)(1,000)(v²)

Kinetic energy of object B = (500)(v²)

Ratio of kinetic energy of object A to B = Kinetic energy of object A / Kinetic energy of object B

Ratio of kinetic energy of object A to B = (100)(v²) / (500)(v²)

Ratio of kinetic energy of object A to B = 100 / 500

Ratio of kinetic energy of object A to B = 1/5

Ratio of kinetic energy of object A to B = 1:5

Answer: 0 m/s^2

Explanation: acceleration is the change in velocity per time (slope of the tangent line to the velocity-time graph),  a=Δv/Δt. If the speed(velocity) is constant in any direction, there is no change to the velocity. if Δv=0, then a=0

Answer:

a)40 meters

b)2 m/s

c)2 m/s

d)0 m/s

e)45 degrees northeast

Explanation:

a) The displacement from C to A is the distance directly across the river, which is 40 meters.

b) The speed of the boat as seen by people standing at A is the magnitude of the boat's velocity vector, which is equal to the displacement divided by the time taken:

Speed = displacement / time = 40 m / 20 s = 2 m/s.

c) Let v be the speed of the water in the river. The boat is moving at right angles to the flow of the river, so the water exerts a perpendicular force on the boat. The time taken for the boat to travel from A to C is 20 seconds, during which time the boat will have been carried downstream by the river by a distance equal to v times the time taken.

Distance carried downstream = v × time = v × 20 m.

Since the boat landed at C, which is directly across the river from A, the distance it traveled horizontally is 40 meters. Therefore:

40 m = (boat speed) × (time taken) = (boat speed) × 20 s.

Hence, the speed of the boat is:

Boat speed = 40 m / 20 s = 2 m/s.

So, we have two equations:

Distance carried downstream = v × 20 m

Boat speed = 2 m/s

From the first equation, we get:

v × 20 m = 40 m

Therefore, the speed of the water in the river is:

v = 40 m / 20 m = 2 m/s.

d) The speed of the boat as seen by a fish drifting with the river is the difference between the speed of the boat and the speed of the water in the river:

Boat speed - Water speed = 2 m/s - 2 m/s = 0 m/s.

So, the speed of the boat as seen by a fish drifting with the river is zero.

e) The boat should head in a direction that makes its velocity vector point directly from A to B. Since A and B are directly opposite each other, this means the velocity vector should be perpendicular to the line connecting A and B.

We know the boat's velocity vector has a magnitude of 2 m/s and is at right angles to the velocity vector of the water in the river, which has a magnitude of 2 m/s. So, we can draw a vector diagram with the velocity vector of the boat pointing straight up and the velocity vector of the water pointing straight to the right. The vector connecting the tail of the water velocity vector to the head of the boat velocity vector will then point directly from A to B.

The angle between the boat's velocity vector and the line connecting A and B can be found using trigonometry. Let θ be this angle. Then:

tan(θ) = (boat speed) / (water speed) = 2 m/s / 2 m/s = 1.

Taking the inverse tangent of both sides gives:

θ = tan^(-1)(1) = 45°.

So, the boat should head in a direction 45 degrees to the right of straight up, or northeast.

Fan made of plastic bottle is an innovative way to reuse plastic bottles and can be easily made at home. This can be done by following simple steps, and a few materials are required to make it. The fan can be created by using a plastic bottle, scissors, string, a ruler, and a marker.

First, the bottle needs to be cut into half, and the upper part needs to be cut into three equal sections, then fold each section to make a blade. With the help of a ruler and marker, make a mark on each section, then make a hole in the center of each blade. Insert a string through the holes and tie the ends of the strings. The fan is ready to use by holding the string and swinging it back and forth.

The use of plastic bottle fans can significantly reduce the number of plastic waste and provide a practical solution to avoid environmental pollution. Besides, it is easy to make, and the materials are readily available, which can be used for various occasions, such as picnics, camping, or any outdoor activities.

In conclusion, the creation of a fan made of plastic bottle with string is an excellent way to reuse plastic bottles and can be made with simple steps. This project encourages everyone to contribute to environmental protection by utilizing what is available at home and reducing the number of plastic wastes.

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Taking into account the Newton's second law, the acceleration of the object is 5 m/s².

Newton's second law states that this force will change the speed of an object because the speed and/or direction will change and these are called acceleration.

Newton's second law states that: "The acceleration of an object is directly proportional to the force acting on it and inversely proportional to the mass."

Mathematically, Newton's second law is expressed as:

F= m×a

where:

In this case, you know:

Replacing in the definition of Newton's second law:

200 N= 40 kg× a

Solving:

a= 200 N÷40 kg

a= 5 m/s²

Finally, the acceleration is 5 m/s².

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As temperature increases, molecules of liquid become more active and they move more rapidly.

Answer:Hope it will help you.

Explanation:

Answer:

\(\boxed {\boxed {\sf a= 1.5 \ m/s^2}}\)

Explanation:

Acceleration is the change in velocity over time, so it is calculated using the following formula.

\(a= \frac{v_f-v_i}{t}\)

The object is initially moving at 10.5 meters per second. After 2 seconds, it is moving at 13.5 meters per second.

Substitute the values into the formula.

\(a= \frac{13.5 \ m/s - 10.5 \ m/s}{2 \ s}\)

Solve the numerator.

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

Divide.

\(a=1.5 \ m/s/s \\a= 1.5 \ m/s^2\)

The acceleration of the object is 1.5 meters per second squared.

Answer:

Cold fronts and Warm fronts

Explanation:

the ground state of O2 is characterized by its MO configuration with two unpaired electrons in the π* orbitals. Excited states occur when electrons are promoted to higher energy orbitals, and the order of energy levels is essential for understanding the transitions between these states.

In its ground state, O2 has a MO configuration of σ2(1s), σ2*(1s), σ2(2s), σ2*(2s), σ2(2pz), π4(2px, 2py), and π2*(2px, 2py). This configuration results in O2 being a diradical with two unpaired electrons in its π* orbitals.

Excited states occur when electrons are promoted from lower energy MOs to higher energy MOs. In O2, possible excited states involve promoting an electron from the π(2px, 2py) or π*(2px, 2py) MOs to higher energy orbitals, such as the σ(3s) or σ(3pz) MOs.

The order of energy levels, from lowest to highest, is as follows: σ(1s), σ*(1s), σ(2s), σ*(2s), σ(2pz), π(2px, 2py), π*(2px, 2py), σ(3s), σ(3pz). In this order, the energy gap between π*(2px, 2py) and σ(3s) or σ(3pz) determines the energy required for electronic transitions to excited states. These transitions can be induced by processes like  the absorption of light or energy transfer from other excited molecules.

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A sound wave will suffer reflection, absorption, and transmission when it comes into contact with a concrete wall. The characteristics of the wave and the substance determine how much of each of these interactions occur.

When a mechanical wave encounters a particular material, its behavior depends on the properties of both the wave and the material. Let's take an example of a sound wave interacting with a wall made of concrete.

Reflection: When the sound wave meets the concrete wall, some of the wave energy gets reflected back toward the source. The amount of reflection depends on the angle of incidence and the acoustic impedance of the materials. In this case, a significant amount of sound will be reflected since the acoustic impedance of air and concrete is very different.

Absorption: Some of the sound energy may be absorbed by the concrete wall as it passes through it. The amount of absorption depends on the thickness of the wall, its density, and its acoustic absorption coefficient. Concrete walls are known to have low absorption coefficients, which means that they are not very effective at absorbing sound.

Transmission: The sound wave can also transmit through the concrete wall, and the amount of transmission depends on the wall's thickness and the frequency of the sound. Lower-frequency sounds tend to transmit better through concrete walls than higher-frequency sounds.

Therefore, when a sound wave encounters a concrete wall, it will experience reflection, absorption, and transmission. The amount of each of these interactions depends on the properties of the wave and the material.

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Making progress by conducting studies for research, project or article and presenting this progress to scientists represent a category that belongs to Science (Option A).

Sciences can be defined as a body of empirical knowledge whose development is based on the presentation of results from experiments in peer-reviewed publications and the generation of projects to investigate different issues of the real world.

Therefore, we can conclude that Science is a body of info based on the transference of knowledge which involves developing projects or writing scientific articles (Option A).

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The quantity of heat energy that must be transferred to 2.25 kg of brass to raise its temperature from 20 °C to 240 °C is 195030 J

First, we shall list out the given parameters from the question. This is shown below:

The quantity of heat energy that must be transferred can be obtained as follow:

Q = MCΔT

= 2.25 × 394 × 220

= 195030 J

Thus, we can conclude quantity of heat energy that must be transferred is 195030 J

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A simplified version of Rutherford’s calculation of the size of the gold nucleus. A piece of gold foil that is 0.010 cm thick and whose area is 1 cm x 1 cm is used in the experiment. It is observed that 99.93% of all particles go through undeflected. The density of gold is 19,300 kg/m3.

Rutherford's experiment involved firing alpha particles (helium nuclei) at a thin sheet of gold foil to study the structure of atoms. Based on the results of this experiment, Rutherford was able to deduce that atoms have a small, dense nucleus at their center.

In this problem, we will go through a simplified version of Rutherford's calculation of the size of the gold nucleus.

First, we need to calculate the total number of gold atoms in the foil. We know that the foil is 0.010 cm thick and has an area of 1 cm x 1 cm, so its volume is

V = thickness x area = 0.010 cm x (1 cm x 1 cm) = 0.010 \(cm^{3}\)

The density of gold is 19,300 kg/\(m^{3}\), which is equivalent to 19.3 g/\(cm^{3}\)Therefore, the mass of the gold foil is

m = density x volume =  19.3 g/\(cm^{3}\) x 0.010 \(cm^{3}\) = 0.193 g.

The molar mass of gold is 197 g/mol, so the number of gold atoms in the foil is

N = (0.193 g) / (197 g/mol) x (6.022 x \(10^{23}\) atoms/mol) = 1.86 x \(10^{21}\) atoms

Next, we need to determine the fraction of alpha particles that are deflected by the gold nucleus. We are told that 99.93% of all alpha particles go through undeflected, which means that only 0.07% of the alpha particles are deflected. This is a very small fraction, which suggests that the size of the gold nucleus must be very small compared to the size of the atom.

Assuming that the alpha particles are deflected only by the gold nucleus and not by the electrons, we can use the principle of conservation of momentum to estimate the size of the gold nucleus. When an alpha particle approaches the gold nucleus, it experiences a repulsive electrostatic force that causes it to change direction. The magnitude of this force is given by Coulomb's law

F = k\(q_{1}\)\(q_{2\) / \(r^{2}\)

Where k is Coulomb's constant, \(q_{1}\) and \(q_{2}\) are the charges of the alpha particle and gold nucleus, respectively, and r is the distance between them. Since the alpha particle has a positive charge and the gold nucleus has a positive charge, the force is repulsive.

If we assume that the alpha particle is initially moving directly toward the center of the gold nucleus, then at the point of closest approach, the alpha particle will have a velocity v that is perpendicular to the direction from the alpha particle to the gold nucleus. At this point, the force on the alpha particle will be perpendicular to its velocity, which means that it will change only the direction of the alpha particle's velocity, not its magnitude.

Using conservation of momentum, we can relate the angle of deflection θ to the distance of closest approach r.

m\(v^{2}\) / r = k\(q_{1}\)\(q_{2\) / \(r^{2}\)

Where m is the mass of the alpha particle. Solving for r, we get

r = k\(q_{1}\)\(q_{2\) / m\(v^{2}\)

To estimate the size of the gold nucleus, we assume that the alpha particles are deflected by a single, stationary gold nucleus at the center of the atom. In reality, the gold nucleus is not stationary, but this assumption gives us a rough estimate of its size.

Hence, the alpha particles are undeflected with a probability of 0.9993, we can assume that they do not interact with the gold nucleus and that their path is a straight line through the foil.

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The surface density of atoms in the (1,1,1) plane of the BCC lattice is 0.16 atoms/angstrom².

To calculate the surface density of atoms in the (1,1,1) plane, we first need to determine how many atoms are in the plane. Since the plane passes through the center of the unit cell, it intersects one-eighth of each of the corner atoms and the full central atom. Therefore, the total number of atoms in the (1,1,1) plane is:

N_atoms = 1 + 8*(1/8) = 2

Next, we need to calculate the area of the (1,1,1) plane. To do this, we can consider the lengths of the sides of the unit cell. Since the BCC lattice has a cubic unit cell, all sides have length a=5 angstroms. The (1,1,1) plane passes through the center of the unit cell and intersects the sides at a 45-degree angle. The distance from the center of the unit cell to the midpoint of one of the sides is a/sqrt(2), so the length of the diagonal of the square face of the unit cell is 2a/√(2) = a√(2). The area of the square face is then (a√(2))² = 2a².

The (1,1,1) plane intersects the square face at a 45-degree angle, so its area is 2*a²/2 = a².

The surface density of atoms is the number of atoms per unit area. Dividing the number of atoms in the (1,1,1) plane by the area of the plane gives:

surface density = N_atoms / area

= 2 / a² = 2 / (5 angstroms)² = 0.16 atoms/angstrom²

Therefore, the surface density of atoms in the (1,1,1) plane of the BCC lattice is 0.16 atoms/angstrom².

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Answer:

The PE of the mass increased by 6,972.95 J

Explanation:

Gravitational Potential Energy

It's the energy stored in an object because of its vertical position or height in a gravitational field.

It can be calculated with the equation:

U=m.g.h

Where m is the mass of the object, h is the height with respect to a fixed reference, and g is the acceleration of gravity or 9.8 m/s^2.

We are given the mass of m=16 slug raised by a height h=10 ft. Both units will be converted to SI standard:

1 slug = 14.59 Kg, thus

16 slug = 16*14.59 Kg=233.44 Kg

1 ft = 0.3048 m, thus:

10 ft = 10*0.3048 m = 3.048 m

Thus, the PE of the mass increased by:

U = 233.44 * 9.8 * 3.048 = 6,972.95 J

the PE of the mass increased by 6,972.95 J

The potential energy will be "5120 ft.lb".

According to the question,

Mass,

Height,

As we know the formula,

→ \(U = mgh\)

By substituting the values, we get

→     \(= 16\times 32\times 10\)

→     \(= 5120 \ ft.lb\)

Thus the response above is correct.

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1. The force between the two charges is 1.78 × 10⁻⁵ pounds, with opposite signs indicating attraction between the charges.

2. The resistance of 10 gauge aluminum wire over a 40 ft distance would be 0.506 ohms.

3. The cost of electricity from the automobile battery is 38.6 cents per kWh.

4. The current that would be carried through the body is 0.8 mA if dry.

1. The force between two point charges can be calculated using Coulomb's law, which states that the force is proportional to the product of the charges and inversely proportional to the square of the distance between them.

Using the values given, the force can be calculated as F = (k * q1 * q2) / r², where k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between them. Plugging in the values, the force can be calculated as 1.78 × 10⁻⁵ pounds, with opposite signs indicating attraction between the charges.

2. The resistance of a wire is determined by its length, cross-sectional area, and resistivity. The resistivity of aluminum is higher than that of copper, so a larger cross-sectional area is required to achieve the same resistance. Using the gauge size conversion chart, 10 gauge aluminum wire has a cross-sectional area of 5.26 mm², which is approximately 83% of the cross-sectional area of 12 gauge copper wire.

Thus, the resistance of 10 gauge aluminum wire over a 40 ft distance can be calculated as R = (rho * L) / A, where rho is the resistivity of aluminum, L is the length, and A is the cross-sectional area. Plugging in the values, the resistance can be calculated as 0.506 ohms.

3. To calculate the cost of electricity per kWh, the total cost and the total amount of energy supplied must be known. Since the battery supplies 12 V and 51 A for one hour, the total energy supplied can be calculated as E = V * I * t, where V is the voltage, I is the current, and t is the time.

Plugging in the values, the total energy supplied can be calculated as 612 watt-hours (Wh). Since one kWh is equal to 1000 Wh, the total energy supplied can be converted to 0.612 kWh. Dividing the total cost by the total energy supplied gives the cost per kWh, which is 38.6 cents.

4. The current through the body can be calculated using Ohm's law, which states that current is equal to voltage divided by resistance. Using the values given, the resistance can be either 100,000 ohms or 300 ohms depending on whether the skin is dry or wet.

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\( \sf \red{what \: is \: propagation}\)

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Propagation is an asèxual method of reproduction. In propagation, the DNA from the parents is taken to form another organism.

\(\green\star\)There are two types of propagation:

1. Vegetative propagation: In vegetative propagation, new plants are obtained from the parts of old plants (like stem, roots and leaves), without the help of any reproductive organs.

2. Artificial propagation: Human grow by some man made methods are called artificial propagation.

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Answer:

removal of heat by cooling towers

ANSWER

9.8 m/s² up

EXPLANATION

Given:

• The elevator's weight, Fg = 50,000 N

• The upward force acting on the elevator, F = 100,000 N

Find:

• The elevator's acceleration, a

By Newton's second law of motion, the net force acting on an object of mass m is,

In this case, there are two forces acting on the elevator: the upward force of the brakes and the downward force of the elevator's weight, so the net force is,

We can find the mass of the elevator from its weight, knowing that g = 9.8 m/s²,

Solving the net-force equation for a,

Replace the known values and solve,

Hence, the elevator has an upward acceleration of 9.8 m/s² while braking.

Because the weight of the helium gas is less than the weight of the air it has displaced from the volume of the balloon, the helium-filled balloon rises in the atmosphere.

Chemical element helium has the atomic number 2 and the symbol He. It is the first member of the noble gas group in the periodic table and is a colorless, odorless, tasteless, non-toxic, inert, monatomic gas. Its melting point at ordinary pressure is zero, and its boiling point is the lowest of all the elements.

Helium is most commonly used to fill balloons for celebrations and parades since it is a secure, non-flammable gas. Helium, however, is an essential component in numerous industries, including as national security, high-tech manufacturing, medical technology, and scientific research.

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Answer:

D.reflected waves of the same wavelength and less intensity than the original sound source

a. Va-Vb= 0 volts

The two are perpendicular and therefore would have a difference of zero

b. Vc-Vb= -Ed

We already know V=Ed and in this case d is negative

c. -2.736 volts

(-7.2)*(.38)= -2.736

Plug in what was given for E and d into the equation for b

The problem is about the relation between electric field and potential difference. They both are related according to:

Vb -Va = - ∫ E× dl

The solution is:

a) V(B) - V(A)  = 0

c) V(C) - V(B) =  2.74 (V)

a) The segment AB is perpendicular to the lines of the electric field, then the point product is zero and V(B) - V(A) = 0

b) The segment CB is parallel to the lines of the electric field, and it is uniform, the:

V(C) - V(B) = - ∫₀⁰°³⁸ E × dl = E× d |₀⁰°³⁸ = - (7.2)×(0.38) = - 2.74 (V)

So the expression for the potential difference between point C and point B in terms of E and d is:

V(C) - V(B) = - ( E×d)

And finally     V(C) - V(B) =  2.74 (V)

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The size of the angle θ of a point object moving from point A to point B along a circular path is 2πR / L.

To understand this, consider a simple example. Suppose that a point object that moves from point A to point B along a circular path with a radius of 1 meter. The distance between points A and B is also 1 meter. Therefore, the size of the angle θ is equal to 2π × 1 / 1 = 2π radians.

In general, the size of the angle θ = ratio of the circumference of the circle to the distance between points A and B. The circumference of the circle is equal to 2πR, where R = radius of the circle. Therefore, the size of the angle θ is equal to 2πR / L.

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Complete question:

A point object moves from point A to point B along a circular path with a radius R. What is the size of the angle θ?

The angular velocity is 3.75 m/s and the centripetal force is 225 N respectively.

The angular velocity of an object with respect to some extent is a degree of the way rapid that item actions through the point's view, within the feel of the way speedy the angular function of the item modifications. An instance of angular pace is a ceiling fan. One blade will whole a complete round in a certain amount of time T, so its angular speed with respect to the middle of the ceiling fan is twoπ/T.

Calculation:-

A. angular velocity ω = v/r

                                     = 30 /8

                                      = 3.75 m/s

B. Centripetal force = mv²/r

                             = 2×30²/8

                              = 225 N

There are 3 formulations we will use to find the angular velocity. the primary comes instantly from the definition. The angular pace is the rate of alternate of the position attitude of an object with respect to time, so w = theta / t, in which w = angular pace, theta = position attitude, and t = time.

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Explanation:

The energy of the system before the collision must equal the energy after the collision.

After the collision the bullet and the block have a total mass of 18.41 kg and they move at a speed of 8.8 m/s. The kinetic energy after the collision is

\(\frac{18.41 kg (8.8 m/s)^2}{2} = 713 J\)

Before the collision only the bullet has kinetic energy.

So we can now determine the speed of the bullet using

\(\frac{0.11kg (v^2)}{2} = 713 J\\v = 114 m/s\)

Answer:B

Explanation:

the circuit can have the same amount of power with the amount of flow of electricity, or current

Answer:

Moving vehicles have a lot of kinetic energy, and when brakes are applied to slow a vehicle, all of that kinetic energy has to go somewhere.

Explanation: