A baseball pitcher throws a baseball horizontally at a linear speed of 42.5 m/s (about 95 mi/h). Before being caught, the baseball travels a horizontal distance of 16.0 m and rotates through an angle of 44.5 rad. The baseball has a radius of 3.67 cm and is rotating about an axis as it travels, much like the earth does. What is the tangential speed of a point on the "equator" of the baseball?

Answer:

Explanation:

Given

acceleration of a particle is given by [tex]a(t)=pt^2-qt^3[/tex]

[tex]a=\frac{\mathrm{d} v}{\mathrm{d} t}=pt^2-qt^3[/tex]

[tex]dv=(pt^2-qt^3) dt[/tex]

Integrating we get

[tex]\int dv=\int \left ( pt^2-qt^3\right )dt[/tex]

[tex]v=\frac{pt^3}{3}-\frac{qt^4}{4}+c_1[/tex]

at [tex]t=0\ v=0[/tex]

therefore [tex]c_1=0[/tex]

We know velocity is given by

[tex]v=\frac{\mathrm{d} x}{\mathrm{d} t}[/tex]

[tex]vdt=dx[/tex]

integrating

[tex]\int dx=\int \left ( \frac{pt^3}{3}-\frac{qt^4}{4}+\right )dt[/tex]

[tex]x=\frac{pt^4}{12}-\frac{qt^5}{20}+c_2[/tex]

using conditions

at [tex]t=0\ x=0[/tex]

[tex]c_2=0[/tex]

[tex]x=\frac{pt^4}{12}-\frac{qt^5}{20}[/tex]                          

(a) The velocity of the particle as function of time is pt³/3 - qt⁴/4 + C.

(b) The position of the particle as function of time is pt⁴/12 - qt⁵/20 + Ct + C.

The velocity of the particle is the integral of the acceleration of the particle.

The velocity of the particle is calculated as follows;

v = ∫a(t)

v = ∫(pt² - qt³)dt

v = pt³/3 - qt⁴/4 + C

The position of the particle as function of time is the integral of velocity of the particle.

x = ∫v

x = ∫(pt³/3 - qt⁴/4 + C)dt

x = pt⁴/12 - qt⁵/20 + Ct + C

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Answer:

qU = Ek = ½mv²

6.00 * 10^-7  * 3000 = ½ * 0.0026 * v²

v² = 0.7222167

v = 0.8498 m/s

Answer:

kinetic frictional force opposes the relative motion between the surfaces in contact of the book and the table.

Explanation:

Answer: Symbol A

Explanation:

The four symbols described here represent:

- Symbol A shows two dots and a line draw from one not connected to the other.  --> this is an open switch. A switch is component of a circuit that is used to open/close the circuit in order to interrupt/allow the flow of current through the circuit. In this case, the switch is open, since the line does not connect the second dot.

- Symbol B shows two dots and a line draw from one connected to the other.  --> this is the symbol used to represent the switch when it is closed, so it is a closed switch.

- Symbol C shows vertical lines in the pattern long, short, long, and short with a plus and minus symbol on it. --> this symbol represents a battery, which consists of two or more cells and provides the electromotive force that pushes the electrons along the circuit.

Therefore, the correct symbol representing the open switch is

Symbol A

For this case we have that by definition, the switch is a control component that opens or closes a circuit.

It is necessary to emphasize that if the switch is open then the flow of electrons from one point to another is not allowed. Thus, it is said that we are in the presence of an open circuit.

The correct option is symbol A, an open switch is shown graphically in the attached image.

Answer:

Option B

Answer:

The correct answer is

35.01 m/s

Explanation:

To solve the question, we note the given variables thus

Height of platform S = 1.6 m

Distance from the platform the ball landed = 20 m

S = ut + 0.5gt² therefore 1.6 = 0.5 × 9.81 × t² or t = 0.57 s

Distance = 20 m  = velocity × time

Therefore velocity = Distance / time = 20 m/0.57 s = 35 m/s

Answer:

Explanation:

Frequency division multiplexing operates by dividing the signal into different frequencies

Explanation:

The technique that is used in the networking is the Frequency Division Multiplexing. using this technique, the existing bandwidths can be partitioned into different frequency bandwidths. These are not interrupting with each other. Each bandwidth can be used for carrying signals individually.

Using this technique many users can share a particular communication medium and they will not be interrupted with each other's communication.Hence this technique can also be termed as Frequency Division Multiple Access.

Frequency Division Multiplexing operates by dividing a signal into different frequencies to enable multiple transmissions at the same time. This method is used in FM radio and television broadcasts, as well as cell phone conversations and computer data transmissions.

Frequency Division Multiplexing (FDM) operates by dividing the signal into different frequencies. This can be seen in how FM (Frequency Modulation) radio signals are used. In FM radio transmission, the information is carried by varying the frequency of the carrier wave and keeping its amplitude constant. This forms different channels on which multiple signals can be transmitted simultaneously without interfering with each other.

Another example is how television broadcasts. Since a vast amount of visual and audio information needs to be carried, each channel requires a larger range of frequencies, these fall under VHF and UHF (high to ultra-high frequencies).

Cell phone conversations and computer data are also transmitted using a similar approach by converting the signal into a sequence of binary ones and zeros. This binary sequence can be transmitted via different frequencies, allowing multiple data transmissions to take place at the same time.

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Answer: a) 12.9 m b) 0.40 rad/s² c) 335.4 J d) 1349.06 J

Explanation:

a)

Distance traveled= R∅

where ∅= [tex]\frac{(2\pi )^2 }{(2.8)^2}[/tex] / 2*α

To find α we have

Sum of torque= Iα

F x r x number of children = m*r²/2 * α

26 x [tex]\frac{4.1}{2}[/tex] * 4 = (255) * (4.10/2)² /2 *α

Solving we get,

α= [tex]\frac{213.2}{535.82}[/tex]

α= 0.40

So,

∅= [tex]\frac{(2\pi )^2 }{(2.8)^2}[/tex] / 2*α

Becomes

∅= 6.29

Now,

Distance traveled= 4.1/2 * 6.29

Distance traveled= 12.9 m

b)

Angular acceleration= 0.40

c)

Work done= F x distance traveled

Work done= 26 x 12.9

Work done= 335.4 J

d)

Kinetic energy= [tex]\frac{1}{2}* I * (\frac{2\pi }{2.8})^2[/tex]

Putting values we get,

Kinetic energy= 267.91 * 5.03

Kinetic energy= 1349.06 J

The merry-go-round problem involves calculating the distance each child runs, the angular acceleration, the work done by each child, and the kinetic energy of the system, using principles of rotational dynamics and energy.

We are given a merry-go-round scenario with children exerting force to set it in motion. Let's break down the problem step by step.

(a) Distance run by each child:

(b) Angular acceleration:

(c) Work done by each child:

(d) Kinetic energy of the merry-go-round:

Final answer:

To find the Earth's average density, we divide its mass (5.94×10²⁴ kg) by its volume converted to cubic meters (1.08×10²⁴ m³), resulting in a density of approximately 5.5 kg/m³.

Explanation:

An object's average density is defined as the ratio of its mass to its volume (rho = M/V). To calculate the Earth's average density, we use the given mass of the Earth (5.94×10²⁴ kg) and the given volume of the Earth (1.08×10¹² km³).

It's important to convert the volume from cubic kilometers to cubic meters to ensure consistency in units since density is typically expressed in kilograms per cubic meter (kg/m³). There are 1,000,000,000,000 (1×10¹²) cubic meters in one cubic kilometer, so the volume in cubic meters is 1.08×10¹² km³ × 1×1012 m³/km³ = 1.08×1024 m³.

Now, dividing the mass by the volume in consistent units gives:

Density = Mass / Volume = 5.94×10²⁴ kg / 1.08×10²⁴ m³ = 5.5 kg/m³

Therefore, The average density of Earth is roughly 5.5 kg/m³.

Answer:

A) 0.20Joules

Explanation:

The type mechanical energy acting on the body is potential energy since the body is covering a particular height.

Potential Energy = mass×acceleration due to gravity × height

Mass of the body = 0.1kg

acceleration due to gravity = 10m/s²

h is the height of the object when dropping = 1.0meters

Substituting this values in the formula to get the energy of the body on dropping, we have;

PE = 0.1×10×1.0

PE = 1.0Joules

On bouncing back to height of 0.8m, the potential energy becomes

PE = 0.1×10×0.8

PE = 0.8Joules

The mechanical energy lost by the ball as it bounces will approximately be the difference in its potential energy when dropping and when it bounces back i.e 1.0Joules - 0.8Joules = 0.20Joules

0.2 J

mgh (before) = (0.1)(9.8)(1) = 0.98

mgh (after) = (0.1)(9.8)(0.8) = 0.784

0.98-0.784=0.196

0.196 is approximately 0.2.  

Answer: 0.5m/s

This value may vary depending on the initial velocity value.

Explanation:

The question is incomplete due to absence of the initial velocity (U) of the rock. Taking the initial velocity as any value say 10m/s at 30° above the horizontal.

Time it will take to reach maximum height (Tmax) will be Usin(theta)/g

Where U is the initial velocity = 10m/s theta = 30° g is the acceleration due to gravity = 10m/s²

Substituting the values in the formula we have;

Tmax = 10sin30°/10

Tmax = sin 30°

Tmax = 0.5second

Therefore, it will take for the rock 0.5s to reach the maximum height of its trajectory if the velocity is 10m/s.

The time varies though depending on the value of the initial velocity.

Answer:

8.64 seconds

Explanation:

According to the question

[tex]10^{10}\ years=1\ day[/tex]

[tex]1\ year=\dfrac{1}{10^{10}}\ day[/tex]

Humans have lived for [tex]10^6[/tex] years

In one universe day

[tex]10^6\ years=10^6\times \dfrac{1}{10^{10}}=10^{-4}\ day[/tex]

In seconds

[tex]10^{-4}\times 24\times 60\times 60=8.64\ seconds[/tex]

In one universe day humans have lived for 8.64 seconds

Humans have been around for 8.64 'universe seconds'.

The question asks for a conversion of human existence into 'universe seconds' given that humans have been around for approximately 106 years and the universe is approximately 10^10 years old. To answer this, one must conceptualise a 'universe day' as representing the entire 10^10 years of the universe's existence and break it down into 'universe seconds' in the same way that a normal day is broken down into normal seconds.

A regular day has 86,400 seconds (24 hours × 60 minutes per hour × 60 seconds per minute). As humans have existed for 10^6 years in the context of the universe's 10^10 years, we can calculate the fraction of the universe's existence that humans have been present for:

Human existence fraction = (Human years) / (Universe years) = 10^6 / 10^10 = 1 / 10^4

Now, we multiply this fraction by the total number of seconds in a day to find the equivalent 'universe seconds' of human existence:

Universe seconds of human existence = 1 / 10^4 × 86,400

Universe seconds of human existence = 8.64

Therefore, humans have existed for 8.64 'universe seconds' when defining the age of the universe as 1 'universe day'.

Answer:

348.57 kg

Explanation:

From the law of conservation of momentum,

Total momentum before the small child was picked = Total momentum after the small child was picked

mu + m'u' = V(m+m')................... Equation 1

Where m = mass of the skater, u = initial velocity of the skater, m' = mass of the small child, u' = initial velocity of the small child, V = common velocity after the child was picked.

Note: Assuming the small child was stationary before he was picked, the u' = 0 m/s.

making m' the subject of the equation

m' = (mu-mV)/V........................... Equation 2

Given: m = 80 kg, u = 6 m/s, V = 1.12 m/s.

Substitute into equation 2

m' = [(80×6)-(80×1.12)]/1.12

m' = (480-89.6)/1.12

m' = 390.4/1.12

m' = 348.57 kg.

Thus the mass of the small child = 348.57 kg

Answer:

d. interaction atmosphere and biosphere interaction

Explanation:

hydrosphere, lithosphere, and biosphere interaction.

Answer:

Explanation:

Given

Charge [tex]Q_1=+5.5\mu C[/tex]

[tex]Q_2=-17\ mu C[/tex]

When the two charges are touch then the net charge on the two sphere will be same after some time

[tex]Q=\frac{Q_1+Q_2}{2}[/tex]

[tex]Q=\frac{5.5-17}{2}=-5.75\ \mu C[/tex]

Now the force between this tow charges are when they are separated with [tex]d=60\ mm[/tex]

[tex]F=\frac{kq_1q_2}{r^2}[/tex]

[tex]F=\frac{9\times 10^9\times (-5.75)^2}{(60\times 10^{-3})^2}[/tex]

[tex]F=82.65\ N[/tex]

Answer: d= 0.57* l

Explanation:

We need to check that before ladder slips the length of ladder the painter can climb.

So we need to satisfy the equilibrium conditions.

So for ∑Fx=0, ∑Fy=0 and ∑M=0

We have,

At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal

At the top of ladder, N₂ acting horizontal

And Between somewhere we have the weight of painter acting downward equal to= mg

So, we have N₁=mg

and also mg*d*cosФ= N₂*l*sin∅

So,

d=[tex]\frac{N2}{mg}*l[/tex] * tan∅

Also, we have f₁=N₂

As f₁= чN₁

So f₁= 0.357 * 69.1 * 9.8

f₁= 241.75

Putting in d equation, we have

d= [tex]\frac{241.75}{69.1*9.8} *l[/tex] * tan 58

d= 0.57* l

So painter can be along the 57% of length before the ladder begins to slip

Answer:

The weight of the ladder is given as well as the length ;  l = 11.9 m long and weighing W = 46.1 N rests against a smooth vertical wall.

how far along the length of the ladder = x = 4.14m

Explanation:

The detailed step and calculation is as shown in the attached file.

Answer:

40 seconds

Explanation:

Thrusting speed = V

Where V=at

a= F/m

Therefore V = Ft/m

F = 1200kN = 1200000N

t = 20s

m = 8.0 * 104 kg

V = 1200000 * 20 / 80000

V = 300m/s

Time for the spaceship to coast a distance of 12km

Distance = Vt

t = 12000/300 = 40seconds

The time taken for taking the  spaceship to coast this distance is 40 seconds

Thrusting speed = V

Where

V=at

And,  

[tex]a= F\div m[/tex]

Therefore [tex]V = Ft\div m[/tex]

Now  

F = 1200kN

= 1200000N

t = 20s

[tex]m = 8.0 \times 104 kg[/tex]

Now  

[tex]V = 1200000 \times 20 \div 80000[/tex]

V = 300m/s

Time for the spaceship to coast a distance of 12km

Distance = Vt

[tex]t = 12000\div 300[/tex]

= 40seconds

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Answer:

Case A: - x axis

Case B: - x axis

Case C: no force as the direction of v and B are parallel.

Explanation:

Solution:

- The direction of Force exerted by the magnetic field B with a moving charge q with a velocity of v is given by an expression:

                                     F = q*(v x B)

- Where F: force vector exerted by the field

             v: velocity vector of charged particle q.

             B: magnetic field direction

- The cross product of v with B:

       We will use right hand rule in which our fingers curl from v to B, the direction of the thumb denotes the direction of applied force. Hence,

                                Case A: - x axis

                                Case B: - x axis

                                Case C: no force as the direction of v and B are parallel.

Final answer:

The direction of the magnetic force on a positively charged particle moving through a uniform magnetic field is determined by the right-hand rule 1 (RHR-1). The force is perpendicular to the plane formed by the particle's velocity and the magnetic field. The direction of the force depends on the charge of the particle and is opposite for positive and negative charges.

Explanation:

The direction of the magnetic force on a positively charged particle moving through a uniform magnetic field is determined by the right-hand rule 1 (RHR-1). According to RHR-1, if you point your right thumb in the direction of the particle's velocity (v) and your right fingers in the direction of the magnetic field (B), then your right palm will point in the direction of the magnetic force (F) acting on the particle.

For example, if the particle is moving perpendicular to the magnetic field (B), as shown in the figure, and the particle has a positive charge, then the force (F) will be directed outward from the plane formed by v and B.

It's important to note that the direction of the magnetic force on a negatively charged particle will be in the opposite direction to that on a positively charged particle.

Final answer:

After deploying the parachute, a skydiver does experience nonzero acceleration as the increased drag causes rapid deceleration until reaching a new lower terminal velocity.

Explanation:

Immediately after deploying her parachute, a skydiver does indeed have a nonzero acceleration. When a parachute opens, it rapidly increases the area exposed to air resistance, causing a significant increase in drag. As a result, the force of air resistance acting on the skydiver greatly exceeds the force of gravity, which causes a rapid deceleration of the skydiver. The magnitude of this deceleration is determined by the net force acting on the skydiver, which now includes the strong opposing force from the parachute's drag. This deceleration continues until the skydiver reaches a new, much lower terminal velocity with the parachute open.

Answer:

44.1613858478 m/s

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity

s = Displacement = 99.4

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 99.4+0^2}\\\Rightarrow v=44.1613858478\ m/s[/tex]

If air resistance was absent Dan Koko would strike the airbag at 44.1613858478 m/s

Answer : The velocity of an electron is, [tex]6.1\times 10^{6}m/s[/tex]

Explanation :

According to de-Broglie, the expression for wavelength is,

[tex]\lambda=\frac{h}{p}[/tex]

and,

[tex]p=mv[/tex]

where,  

p = momentum, m = mass, v = velocity

So, the formula will be:

[tex]\lambda=\frac{h}{mv}[/tex]       .............(1)

where,

h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]

[tex]\lambda[/tex] = wavelength  = [tex]1.2\times 10^{-10}m[/tex]

m = mass  of electron = [tex]9.11\times 10^{-31}kg[/tex]

v = velocity of electron = ?

Now put all the given values in formula 1, we get:

[tex]1.2\times 10^{-10}m=\frac{6.626\times 10^{-34}Js}{(9.11\times 10^{-31}kg)\times v}[/tex]

[tex]v=6.1\times 10^{6}m/s[/tex]

Thus, the velocity of an electron is, [tex]6.1\times 10^{6}m/s[/tex]

Answer: B. Sensation

Explanation:

Sensation is input about the physical world obtained by our sensory receptors, and perception is the process by which the brain selects, organizes, and interprets these sensations. In other words, senses are the physiological basis of perception. Perception of the same senses may vary from one person to another because each person’s brain interprets stimuli differently based on that individual’s learning, memory, emotions, and expectations.

The sensitivity of a given sensory system to the relevant stimuli can be expressed as an absolute threshold. Absolute threshold refers to the minimum amount of stimulus energy that must be present for the stimulus to be detected 50% of the time.

Sometimes, we are more interested in how much difference in stimuli is required to detect a difference between them. This is known as the just noticeable difference (jnd) or difference threshold.

Final answer:

The initial detection of physical stimuli such as odors, light, and sound is known as sensation, which is distinct from perception, the interpretation of these stimuli.

Explanation:

When we initially detect physical stimuli such as odors, light, and sound, the process is referred to as sensation. Sensation occurs when sensory receptors detect sensory stimuli, involving the conversion of physical energy like light or sound waves into a form of energy that the brain can understand, which is electrical stimulation. On the other hand, perception involves the organization, interpretation, and conscious experience of those sensations. It is during the perception process that we can identify specific objects, sounds, or smells and comprehend what they mean or represent in our environment. While absolute threshold refers to the minimum amount of stimulus energy required to be detected about 50% of the time, it is not the initial detection of stimuli itself.

When a bungee cord is stretched a distance x, it exerts an opposing force given by the formula F = bx2. Integrated over the distance of stretch, this results in a work done of 2 Joules for a stretch distance of 1 meter and force constant b = 6 N/m^2.

The work done in stretching an elastic object like a bungee cord is given by the integral of the force over the distance of stretch. Given the force exerted by the bungee cord is F = bx^2, where b = 6 N/m^2 and the work W is defined as the integral from 0 to x of F dx, the formula to calculate work done becomes W = ∫from 0 to 1 (bx^2) dx. Solving this integral gives W = (b/3)x^3 evaluated from 0 to 1, which simplifies to W = (6/3) (1) = 2 Joules. So, the correct answer is 2 Joules.

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Impulse delivered

or

Change in momentum.

Final answer:

The area under the net external force vs time graph is equal to impulse or change in momentum, representing the impulse-momentum theorem.

Explanation:

The area under the curve of the net external force vs time graph is equal to impulse or change in momentum. This relationship arises from Newton's second law of motion, which states that the force acting on an object is equal to the rate of change of its momentum. In graphical terms, when force is plotted on the y-axis and time on the x-axis, the area under the curve represents the impulse, which is the product of the force and the time interval over which it acts. This impulse is equivalent to the change in momentum of the object according to the impulse-momentum theorem.

Explanation:

First, we will calculate the electric potential energy of two charges at a distance R as follows.

                   R = 2r

                      = [tex]2 \times 0.1 m[/tex]

                      = 0.2 m

where,    R = separation between center's of both Q's. Hence, the potential energy will be calculated as follows.

                U = [tex]\frac{k \times Q \times Q}{R}[/tex]

                    = [tex]\frac{8.98 \times 10^{9} \times (3 \times 10^{-6} C)^{2}}{0.1}[/tex]

                    = 0.081 J

As, both the charges are coming towards each other with the same energy so there will occur equal sharing of electric potential energy between these two charges.

Therefore, when these charges touch each other then they used to posses maximum kinetic energy, that is, [tex]\frac{U}{2}[/tex].

Hence,   K.E = [tex]\frac{U}{2}[/tex]

                     = [tex]\frac{0.081}{2}[/tex]

                     = 0.0405 J

Now, we will calculate the speed of balls as follows.

                 V = [tex]\sqrt{\sqrt{\frac{2 \times K.E}{m}}[/tex]

                     = [tex]\sqrt{\sqrt{\frac{2 \times 0.0405}{4 kg}}[/tex]

                     = 0.142 m/s

Therefore, we can conclude that final speed of one of the balls is 0.142 m/s.

Answer:

A

Explanation:

Why is that they truly evaluate both of the theories in the same way.

4. KE increases by a factor of 16 is the answer

Explanation:

Kinetic energy  = (1/2)mv² = 0.5 mv²

where  

m  = mass, and  v = velocity.

So at 15 mph,  

K E =  0.5 m  (15) ² = 112.5 m

And at 60 mph,  

K E = 0.5 m (60)² = 1800 m

m  is the mass, and not meters.

So, 1800  m/112. 5 m = 16

16 times the Kinetic Energy.

Answer:

Explanation:

This question is quite tricky. Not all parameters were provided.

λ = 515nm(nm = namo meters), 515 nm becomes 515 x 10^-9 which in turn becomes 5.15 x 10^-7.

The velocity for the green light wasn't provided but the velocity of greenlight is a constant; velocity of green light(v) = 3.00 × 108 m/s.

frequency of the light =[tex]\frac{velocity of green light}{wavelength of light}\\[/tex]

f =[tex]\frac{3.00 X 108 m/s}{5.15 X 10^{-7} }[/tex]

f=20.971 Hertz

The frequency of light will be "20.971 Hz".

According to the question,

Wavelength,

or,

          [tex]= 5.15\times 10^{-7} \ m[/tex]

Velocity of green,

The frequency of light (f) will be:

= [tex]\frac{Velocity}{Wavelength}[/tex]

= [tex]\frac{3.00\times 108}{5.15\times 10^{-7}}[/tex]

= [tex]20.971 \ Hz[/tex]

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Answer: Acoustic or sound wave

Explanation:

Acoustic wave is a type of wave energy that travels through a medium by adiabatic compression and decompression, they have acoustic velocity which is determined by the type and nature of medium they travel through. They are mechanical and longitudinal waves with characteristic features such as amplitude, period, frequency and wavelength.

Answer:

B. 10 days

Explanation:

Answer: perpendicular to it oscillations.

Explanation: A transverse wave is a wave whose oscillations is perpendicular to the direction of the wave.

By perpendicular, we mean that the wave is oscillating on the vertical axis (y) of a Cartesian plane and the vibration is along the horizontal axis (x) of the plane.

Examples of transverse waves includes wave in a string, water wave and light.

Let us take a wave in a string for example, you tie one end of a string to a fixed point and the other end is free with you holding it.

If you move the rope vertically ( that's up and down) you will notice a kind of wave traveling away from you ( horizontally) to the fixed point.

Since the oscillations is perpendicular to the direction of wave, it is a transverse wave

In a transverse wave, the vibrations move in a direction perpendicular to the direction of wave travel. This distinguishes them from other types of waves. An example is the movement of water waves.

In a transverse wave, the vibrations or disturbances move in a direction perpendicular (at right angles) to the direction of the wave's travel. This is the key characteristic that separates transverse waves from other types of waves. A common example of transverse waves would be waves in a body of water where the water particles move up and down causing the wave to move forward.

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