A transverse wave moves perpendicular to the particle vibration, while a compressional wave moves parallel to the particle vibration. Light waves are an example of a transverse wave, and sound waves are an example of a compressional wave.
A transverse wave is a type of wave in which the particles of the medium vibrate perpendicular to the direction of the wave's propagation. This means that as the wave moves forward, the particles move up and down or side to side. An example of a transverse wave is a light wave. When light travels through space, the electric and magnetic fields oscillate perpendicular to the direction of the wave.
A compressional wave is a type of wave in which the particles of the medium vibrate parallel to the direction of the wave's propagation. This means that as the wave moves forward, the particles move back and forth, creating compressions and rarefactions in the medium. An example of a compressional wave is a sound wave. When sound travels through air, the molecules of air vibrate back and forth, creating areas of high pressure (compressions) and low pressure (rarefactions).
In summary, a transverse wave moves perpendicular to the particle vibration, while a compressional wave moves parallel to the particle vibration. Light waves are an example of a transverse wave, and sound waves are an example of a compressional wave.
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Two uniform solid spheres, each with mass 0.852 kg
and radius 8.00×10−2 m are connected by a short, light rod that is along a diameter of each sphere and are at rest on a horizontal tabletop. A spring with force constant 153 N/m has one end attached to the wall and the other end attached to a frictionless ring that passes over the rod at the center of mass of the spheres, which is midway between the centers of the two spheres. The spheres are each pulled the same distance from the wall, stretching the spring, and released. There is sufficient friction between the tabletop and the spheres for the spheres to roll without slipping as they move back and forth on the end of the spring.
Assume that the motion of the center of mass of the spheres is simple harmonic. Calculate its period.
The period of the simple harmonic motion of the center of mass of the two spheres is approximately 0.770 seconds.
To find the period of the simple harmonic motion of the center of mass of the two spheres, we need to use the equation for the period of a mass-spring system:
T = 2π√(m/k)
where T is the period, m is the total mass of the system (two spheres), and k is the spring constant.
First, we need to find the total mass of the system:
m = 2m1 = 2(0.852 kg) = 1.704 kg
where m1 is the mass of one sphere.
Next, we need to find the spring constant:
k = 153 N/m
Now, we can calculate the period:
2π√(1.704 kg/153 N/m) ≈ 0.770 s
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What was the molar mass from number 1?
The question is incomplete. Could you please provide more details or clarify your question? I will explain molecular mass.
A substance's molecular mass, measured in grams per mole (g/mol), is referred to as its molar mass. It is utilized in many chemical computations and conversions and is computed by adding the atomic masses of the atoms in a molecule.
The molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all the atoms in a molecule. The atomic masses can be found on the periodic table. Molar mass is important in various applications, such as stoichiometry, where it is used to convert between the mass of a substance and the number of moles.
It is also used to determine the empirical formula and molecular formula of compounds. Molar mass plays a crucial role in chemical reactions and calculations involving the quantities of substances, providing a bridge between the microscopic realm of atoms and molecules and the macroscopic world of everyday measurements.
Therefore, Molar mass refers to the mass of one mole of a substance, expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of the atoms in a molecule and is used in various chemical calculations and conversions.
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What is the average acceleration of an
automobile moving from 0 to 60 m/s in 3.0
seconds?
a. 90 m/s²
b. -20 m/s²
c. -90 m/s²
d. 20 m/s²
Answer:
D
Explanation:
Using the formula for average acceleration, which is:
average acceleration = (final velocity - initial velocity) / time
Where the initial velocity is 0 m/s and the final velocity is 60 m/s, and the time is 3 seconds, we get:
average acceleration = (60 m/s - 0 m/s) / 3 s = 20 m/s²
Therefore, the answer is (d) 20 m/s².
A 10 KVA, 380 V, 50 Hz, 3-phas, star-connected salient pole alternator has direct axis and quadrature axis reactances of 12 ohms and 8 ohms respectively. The armature has resistance of 1 ohin per phase, The generator delivers rated load at 0.8 p,f lagging with the terminal voltage being maintained at rated value. If the load angle is 16.15, determine (i) the direct axis and quadrature axis components of armature current, (b) excitation voltage of the generator.
Direct axis and quadrature axis components of armature current are 30.28 A and 46.92 A respectively, and the excitation voltage of the generator is 765.36 V.
Given:
Apparent power (S) = 10 KVA = 10,000 VA
Line voltage (V) = 380 V
Frequency (f) = 50 Hz
Xd = 12 ohms
Xq = 8 ohms
Ra = 1 ohm
Power factor (pf) = 0.8 lagging
Load angle (δ) = 16.15 degrees
(i) Armature current's direct axis and quadrature axis components
We know that the apparent power is given by S = 3VLILcos(φ), where VL is the line voltage, IL is the line current, and φ is the angle between them. For a star-connected alternator, line voltage is equal to phase voltage, so we can write:
S = 3Vphase Iphase cos(φ)
Iphase = S / (3Vphase cos(φ))
For a lagging power factor, cos(φ) = 0.8, so
Iphase = 10,000 / (3 x 380 x 0.8) = 10.46 A
The direct axis component (Id) and the quadrature axis component (Iq) make up the armature current. Using the given values of Xd, Xq, and Ra, we can calculate these components as follows:
Id = (VL - IaRa) / Xd
Iq = (VL - IaRa) / Xq
where Ia is the magnitude of the armature current, which is equal to the magnitude of the line current divided by √3. Thus,
Ia = Iphase / √3 = 10.46 / √3 = 6.03 A
Substituting the given values:
Id = (380 - 6.03 x 1) / 12 = 30.28 A
Iq = (380 - 6.03 x 1) / 8 = 46.92 A
(ii) Excitation voltage of the generator:
The excitation voltage (E) of the generator is given by:
E = Vphase + IqXq
Substituting the given values:
E = 380 + 46.92 x 8 = 765.36 V
Therefore, the direct axis and quadrature axis components of armature current are 30.28 A and 46.92 A respectively, and the excitation voltage of the generator is 765.36 V.
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Why would you want to build a house with materials having a high specific heat
Materials with high specific heat include materials such as concrete, masonry, adobe, and some types of stone. These materials can be used in walls, floors, or roofs to provide thermal mass and improve indoor comfort, energy efficiency, and sustainability.
A piece of metal at the bottom of a beaker is 0.50m below the eye. If water is poured into the cylinder to a depth of 0.24m . how much closer does the metal appear? (Refractive index of water = 4/3)
If a piece of metal at the bottom of a beaker is 0.50m below the eye, and water is poured into the cylinder to a depth of 0.24m, Then the metal appears closer by 0.38m.
To calculate how much closer the metal appears, we can use the lens maker's formula, which relates the object distance, image distance, and focal length of a lens or in this case, the refractive index of water.
The formula is:
1/f = (n - 1) x (1/R1 - 1/R2)
where f is the focal length, n is the refractive index of the medium (water in this case), R1 is the radius of curvature of the first surface of the medium (in this case, the bottom surface of the beaker), and R2 is the radius of curvature of the second surface of the medium (in this case, the air-water interface at the top of the water level).
Assuming the radius of curvature of both surfaces is large and the surfaces are nearly flat, we can approximate R1 and R2 as infinity, so the formula simplifies to:
1/f = (n - 1) x (1/R2)
Since the piece of metal is at the bottom of the beaker, it acts as the object, and the eye acts as the observer. Initially, the object distance is the distance from the bottom of the beaker to the eye, which is 0.50m.
When water is poured into the beaker to a depth of 0.24m, the new object distance is the distance from the bottom of the beaker to the water level, which is 0.50m - 0.24m = 0.26m.
To calculate the image distance, we can use the thin lens equation:
1/f = 1/o + 1/i
where o is the object distance, and i is the image distance.
Solving for i, we get:
i = (1/f - 1/o)^-1
Since we know that the object distance o is 0.26m, we just need to calculate the focal length f using the lens maker's formula.
1/f = (n - 1) x (1/R2) = (4/3 - 1) x (1/0.24m) = 1/0.72m
f = 0.72m
Plugging in the values, we get:
i = (1/0.72m - 1/0.26m)^-1 = 0.12m
Therefore, the metal appears closer by 0.50m - 0.12m = 0.38m.
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A swimmer swims the 50m length of a swimming pool in 15.91 seconds. He makes the return swim in 18 seconds. What is the swimmer's average speed for the full swim?
If a swimmer swims the 50m length of a swimming pool in 15.91 second and makes the return swim in 18 seconds then the swimmer's average speed for the full swim is 2.948 metres per second.
To find the swimmer's average speed for the full swim, we need to calculate the total distance swum and the total time taken.
The swimmer swam the 50m length of the pool twice, once in 15.91 seconds and once in 18 seconds. So the total distance swum is 50m + 50m = 100m.
The total time taken is the sum of the time taken for each length: 15.91 seconds + 18 seconds = 33.91 seconds.
To find the swimmer's average speed, we divide the total distance by the total time:
Average speed = total distance ÷ total time
= 100m ÷ 33.91 seconds
= 2.948 metres per second
Therefore, the swimmer's average speed for the full swim is 2.948 metres per second. This is a relatively fast speed, but it's important to note that the swimmer's speed may vary depending on their stroke technique, physical fitness, and other factors. Nonetheless, it's an impressive achievement for the swimmer to complete the full 100m in under 34 seconds.
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Pressure is applied from a gas cylinder to one side of a U-tube manometer with constant internal diameter “D", as shown below. The manometer is filled with fresh water (density = 999 kg/m3) and upon application of the pressure, the water column on one side rises by 4.00 cm relative to the other. Draw and label a free-body diagram representing the forces in the manometer. Calculate the pressure above atmospheric pressure applied by the gas.
Answer:
P = 39200.76
Explanation:
formula is:
P = pgh
P = pressure
p = density of fluid
g = acceleration due to gravity ( 9.81 m/s2 for earth)
h = height above the point we are looking for
P = pgh
P = 999x9.81x4
P = 39200.76
a bus with a maximum speed of 20m/s takes 21sec to travel 270m from stop to stop. Its acceleration is twice as great as its deceleration. Find: A, the acceleration B, the distance traveled at maximum speed
Answer:
We can use the kinematic equations to solve this problem.
Let a be the acceleration and d be the distance traveled at maximum speed.
First, we can use the equation:
d = vt + (1/2)at^2
where v is the maximum speed, t is the time traveled at maximum speed, and a is the acceleration.
We know that the bus takes 21 seconds to travel 270 meters, so the time traveled at maximum speed is:
21 - 2a = t
We also know that the acceleration is twice as great as the deceleration, so we can write:
a = 2d
Then, we can substitute these expressions into the first equation:
d = (20)(21 - 2a) + (1/2)(2d)(21 - 2a)^2
Simplifying and solving for d, we get:
d = 210 - 5.25a + 0.25a^2
To find the acceleration, we can use the fact that the maximum speed is reached at the midpoint of the trip, so the distance traveled at maximum speed is half the total distance:
d = 1/2(270)
Solving for d, we get:
d = 135
Substituting this value into the equation for d, we get:
135 = 210 - 5.25a + 0.25a^2
Simplifying and solving for a, we get:
a = 8 m/s^2
Finally, we can use the equation:
d = vt
to find the distance traveled at maximum speed:
d = (20)(21 - 2a)
Substituting the value of a, we get:
d ≈ 188.4 m
Therefore, the acceleration of the bus is 8 m/s^2 and the distance traveled at maximum speed is approximately 188.4 meters.
a cello plays a c# note with a frequency of 346.2hz. if the speed of sound in air is 225m/s, what is the wavelength of this note?
The wavelength of the note produced by the cello is 0.65 m.
The frequency of the note played by cello, f = 346.2 Hz
Speed of the sound in air, v = 225 m/s
Speed of a wave is calculated by taking the product of its frequency and wavelength.
The expression for the speed of the sound wave is given by,
v = fλ
where λ is the wavelength of the note produced.
Therefore, the wavelength of the note produced by the cello,
λ = v/f
Applying the values of v and f,
λ = 225/346.2
λ = 0.65 m
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A car manufacturer wants to change its car's design to increase the car's acceleration. Which changes should the
engineers consider making to the design?
O increase the force that the engine provides
O decrease the force that the engine provides
O increase the mass of the car
O decrease the mass of the car
O increase the top velocity the car can travel
decrease the top velocity the car can travel
To increase a car's acceleration, the engineers should consider decreasing the car's mass and/or increasing the force that the engine provides. Therefore, options A and D are the correct answers.
Acceleration is directly proportional to the net force acting on an object and inversely proportional to its mass. This means that a smaller mass or a larger force will result in a greater acceleration. Therefore, reducing the mass of the car will require less force to achieve the same acceleration, or the same force will result in a greater acceleration.
To decrease the mass of the car, the engineers could consider using lighter materials for the car's body and frame or removing unnecessary components. On the other hand, to increase the force that the engine provides, the engineers could consider upgrading the engine or increasing the fuel intake.
Option B, which suggests decreasing the force that the engine provides, would have the opposite effect and result in a slower acceleration. Option C, which suggests increasing the mass of the car, would also have the opposite effect and require more force to achieve the same acceleration.
Option E, which suggests increasing the top velocity the car can travel, is not directly related to acceleration. Top velocity refers to the maximum speed that a car can reach, whereas acceleration refers to the rate of change of velocity. While increasing the car's top velocity may indirectly affect acceleration, it is not a direct solution to increasing acceleration.
Option A and D.
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A child throws a small rubber ball at a heavier, larger basketball that is sitting still. The small ball bounces backward off the basketball. Describe the motion of the basketball after the small ball bounces back. Does it move at all? Does it move faster or slower than the small ball? what direction does it move? How does newtons third law affect your answer?
When a small rubber ball bounces off a larger, heavier basketball, the basketball will move slightly in the opposite direction, but it will move much slower than the rubber ball due to its larger mass.
According to Newton's third law of motion, every action has an equal and opposite reaction. In the case of the small rubber ball bouncing off the basketball, the rubber ball exerts a force on the basketball, and the basketball exerts an equal and opposite force back on the rubber ball.
As a result, the small rubber ball bounces back in the opposite direction, while the basketball experiences a force in the opposite direction.
The motion of the basketball after the small ball bounces back depends on the mass and velocity of the two objects. Since the basketball is much larger and heavier than the rubber ball, it will not move much, if at all.
In fact, if the rubber ball is light enough and bounces back with enough force, it may cause the basketball to move slightly in the opposite direction. However, the basketball will move much slower than the rubber ball due to its larger mass and slower acceleration.
In terms of direction, the basketball will move in the opposite direction of the rubber ball, as dictated by the conservation of momentum. Since the total momentum of the system before and after the collision must be conserved, the basketball will move in the opposite direction of the rubber ball to balance out the momentum.
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Two waves traveling in the same medium and
having the same wavelength interfere to create a
standing wave. What is the distance between two
consecutive nodes on this standing wave?
The distance between two consecutive nodes on this standing wave if the two waves traveling in the same medium is half of the wavelength.
The standing wave is defined as a stationary wave in which a combination of two waves moves in opposite directions with the same amplitude and same frequency. This phenomenon results in Interference. When two waves are superimposed the output waves are added up to each other and cancel out each other.
From the given, the two waves traveling in the same medium and having the same wavelength (λ) interfere to create the standing wave. The distance between two consecutive nodes is half of the wavelength and hence the distance is λ/2.
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Could you please help me out and explain?
A liner object is placed parallel to a mirror. The object rotates 40° clockwise, in a perpendicular line on the mirror. How does the objects image change?
A linear object is placed parallel to a mirror. The object rotates 40° clockwise, in a perpendicular line on the mirror. The object's image would appear rotated counterclockwise by 40° compared to its initial position.
When a linear object is placed parallel to a mirror and rotates 40° clockwise in a perpendicular line on the mirror, the image of the object undergoes a change. The image formed in a mirror follows the law of reflection, which states that the angle of incidence is equal to the angle of reflection.
Initially, when the object is parallel to the mirror, its image is also parallel and located at the same distance behind the mirror as the object is in front of it. However, as the object rotates clockwise, the incident angle between the object and the mirror changes, resulting in a different reflected angle.
As the object rotates, the image rotates by the same angle but in the opposite direction. In this case, the image would rotate 40° counterclockwise. Therefore, the image of the linear object would appear rotated counterclockwise by 40° compared to its initial position.
It is important to note that the size of the image and the distance between the object and its image remain unchanged, as long as the object maintains the same distance from the mirror throughout the rotation. Only the orientation or rotation of the image changes.
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. Date Find the speed of a sound in the air If it has a frequency of 1120hz and A mavelength of 30cm.
Answer:
Therefore, the speed of sound in air, with a frequency of 1120 Hz and a wavelength of 30 cm, is approximately 336 meters per second.
Explanation:
To find the speed of sound in air, we can use the formula:
speed = frequency × wavelength
Given:
Frequency (f) = 1120 Hz
Wavelength (λ) = 30 cm
First, we need to convert the wavelength to meters since the speed of sound is commonly expressed in meters per second.
1 meter = 100 centimeters
Wavelength in meters (λ) = 30 cm ÷ 100 = 0.3 meters
Now we can substitute the values into the formula:
speed = frequency × wavelength
= 1120 Hz × 0.3 meters
Calculating the value:
speed = 336 meters per second
Why do diving seabirds need to account for the bending of light?
2-3 sentence response.
Diving seabirds need to account for the bending of light because water has a higher refractive index than air.
When light passes through the air-water interface, it slows down and changes direction, causing the image of an object to appear distorted. This phenomenon is known as refraction. For diving seabirds, the refraction of light can make it challenging to accurately locate their prey when diving, which is essential for their survival.
To account for the bending of light, diving seabirds have evolved specialized eyes that allow them to adjust their focus and perceive images differently than humans. These adaptations enable the birds to see through the distorted image caused by refraction and accurately locate their prey while diving. One adaptation that diving seabirds have is a flattened cornea, which allows them to maintain a sharp focus while diving.
Overall, the ability of diving seabirds to account for the bending of light is essential for their survival, as it enables them to accurately locate and catch their prey while diving in the often murky and challenging conditions of the ocean.
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