The transmission of the electric fields of light by a polarizing sheet C. Only the components parallel to the polarizing axis of the sheet are transmitted.
A polarizing sheet is a material that selectively transmits light waves based on their orientation. The sheet has a specific polarizing axis, which is a direction in the material that allows certain components of the electric fields of light to pass through. When light encounters a polarizing sheet, it consists of both components parallel and perpendicular to the polarizing axis.
The polarizing sheet only transmits the components of the electric fields of light that are parallel to its polarizing axis. This occurs because the sheet's molecules preferentially absorb the light components that are perpendicular to the polarizing axis, effectively blocking them from passing through. On the other hand, the components parallel to the polarizing axis are allowed to continue without being absorbed.
By transmitting only the parallel components, the polarizing sheet effectively polarizes the light, which means the light waves become more aligned in a single direction. This property of polarizing sheets is useful in various applications, such as reducing glare in sunglasses and improving image clarity in optical instruments. Therefore the correct option is C
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A 4.0 m radius circular platform rotates with a constant angular acceleration of 20.0 rad/s^2. What is the acceleration of a point on the edge of the disk at the instant that its angular speed is 1.0 rev/s?
The acceleration of a point on the edge of a circular platform with a 4.0 m radius is 127 m/s² when its angular velocity is 1.0 rev/s and has a constant angular acceleration of 20.0 rad/s².
How to find the acceleration of a point?We can use the kinematic equations of rotational motion to solve this problem.
The first kinematic equation for rotational motion is:
ωf = ωi + αt
where ωf is the final angular velocity, ωi is the initial angular velocity, α is the angular acceleration, and t is the time.
We can use this equation to find the time it takes for the platform to reach an angular velocity of 1.0 rev/s:
1.0 rev/s = ωi + (20.0 rad/s²)t
ωi = 0 (initially at rest)
t = 1.0 rev/s / 20.0 rad/s²
t = 0.05 s
Next, we can use the second kinematic equation for rotational motion:
θ = ωit + 1/2 αt²
where θ is the angular displacement.
We can use this equation to find the angular displacement of a point on the edge of the platform during the time it takes to reach an angular velocity of 1.0 rev/s:
θ = (0)(0.05 s) + 1/2 (20.0 rad/s²)(0.05 s)²
θ = 0.0125 rad
Finally, we can use the third kinematic equation for rotational motion:
ωf² = ωi² + 2αθ
We can use this equation to find the acceleration of a point on the edge of the platform at the instant its angular velocity is 1.0 rev/s:
ωf = 1.0 rev/s = 2π rad/s
ωi = 0
θ = 0.0125 rad
a = (2π rad/s)² / 2(0.0125 rad)
a = 127 m/s²
Therefore, the acceleration of a point on the edge of the platform at the instant its angular velocity is 1.0 rev/s is 127 m/s².
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What is the most common net charge of an atom?
a) Negative
b) Positive
c) Neutral
d) Dipole
The most common net charge of an atom is Neutral. So, option c) is correct.
An atom consists of protons, neutrons, and electrons. Protons carry a positive charge, electrons carry a negative charge, and neutrons have no charge. In a neutral atom, the number of protons equals the number of electrons, which results in the charges canceling each other out.
Therefore, the net charge of the atom is zero or neutral. This is the most common charge state for an atom in its natural state.
In certain circumstances, atoms can gain or lose electrons and become charged particles called ions. If an atom loses electrons, it will have more protons than electrons and carry a positive charge, becoming a positive ion or cation.
Conversely, if an atom gains electrons, it will have more electrons than protons and carry a negative charge, becoming a negative ion or anion. While these charged states do occur, they are less common than neutral atoms.
In summary, the most common net charge of an atom is neutral, as the number of protons and electrons in the atom balance each other out, resulting in a net charge of zero. Therefore, option c) is correct.
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What skeletal muscle lengths generated passive force? (Provide a range.)
Passive force is generated in skeletal muscles when they are stretched beyond their resting length.
The range of skeletal muscle lengths that generate passive force is typically greater than the resting length, up to the maximum elongation the muscle can tolerate without tearing. This range is also known as the muscle's "physiological range of motion". The amount of passive force generated by the muscle increases as it is stretched further beyond its resting length, until it reaches a maximum at the point of muscle failure.
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A body of mass 2.2 kg makes an elastic collision with another body at rest and continues to move in the original direction but with 1/4 of its original speed. (a) What is the mass of the other body? (b) What is the speed of the two-body center of mass if the initial speed of the 2.2 kg body was 6.8 m/s?
(a). The mass of the other body is: 4.4 kg. (b). The speed of the two-body center of mass if the initial speed of the 2.2 kg body was 6.8 m/s is 2.27 m/s.
In an elastic collision, both momentum and kinetic energy are conserved. Using these principles, we can solve for unknowns in this problem:
(a) Let the mass of other body be m. From conservation of momentum, we have:
2.2 kg × 6.8 m/s = (2.2 kg + m) × 1.7 m/s
Solving for m, we get m = 4.4 kg.
(b) The velocity of center of mass is given by:
[tex]Vcm = (m1v1 + m2v2)/(m1 + m2)[/tex]
We know m2 = 4.4 kg and v2 = 0 m/s.
Substituting the given values, we get:
[tex]Vcm = (2.2 kg * 6.8 m/s)/(2.2 kg + 4.4 kg) = 2.27 m/s[/tex]
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On an average diet, the consumption of 10 liters of oxygenreleases how much energy? (4.8 kcal are released per liter ofoxygen consumed.)
a. 48 kJ
b. 200 kJ
c. 4.2 kJ
d. 4200 kJ
e. 120 kJ
On an average diet, the consumption of 10 liters of oxygen releases 120 kJ of energy. So, the correct answer is option e.
When 10 litres of oxygen are consumed, 120 kJ (kilojoules) of energy are released.
This is computed by dividing the volume of oxygen consumed (10 litres) by the amount of energy released (4.8 kcal or 4.2 kJ) per litre of oxygen ingested.
Therefore, 4.2 kJ x 10 liters, or 120 kJ, is the total energy released by the ingestion of 10 litres of oxygen. The body uses this energy for a number of metabolic processes, including digestion, respiration, and physical activity.
The body's primary energy source, ATP (adenosine triphosphate), is produced only using oxygen. In conclusion, using 10 litres of oxygen daily will result in the release of 120 kJ of energy.
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when the oscillation of the particles in a medium is parallel to the direction of the wave's motion, what type of wave is this?
When the oscillation of the particles in a medium is parallel to the direction of the wave's motion, this type of wave is called a longitudinal wave.
In a longitudinal wave, the particles of the medium vibrate back and forth along the direction of the wave's motion, rather than moving perpendicular to it. This creates areas of compression and rarefaction in the medium, which travel through the medium as the wave moves.
A common example of a longitudinal wave is a sound wave. When sound is produced, it causes the air particles to vibrate back and forth in the same direction that the sound is traveling. As these vibrations travel through the air, they create areas of compression and rarefaction, which our ears perceive as sound.
Another example of a longitudinal wave is a seismic wave, which is produced by earthquakes and travels through the Earth's crust. Seismic waves cause the particles in the Earth's crust to vibrate back and forth parallel to the direction of the wave's motion.
Therefore, a longitudinal wave is a wave in which the oscillation of particles in the medium is parallel to the direction of the wave's motion. Sound waves and seismic waves are examples of longitudinal waves.
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there must be equal amounts of mass on both side of the center of mass of an object. there must be equal amounts of mass on both side of the center of mass of an object. true false
The statement "There must be equal amounts of mass on both sides of the center of mass of an object" is true.
The center of mass is a point in an object where the mass is equally distributed on all sides.
It is the point at which an object would balance if it were placed on a pivot.
This is due to the equal distribution of mass around the center of mass, which ensures the object's stability.
Therefore, the statement "There must be equal amounts of mass on both sides of the center of mass of an object" is true.
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The fraction of energy carried by the reflected sound is small if the surface is
The fraction of energy carried by the reflected sound is small if the surface is smooth and hard.
This is because when sound waves hit a smooth and hard surface, they bounce back in a very predictable manner, following the law of reflection.
This means that the angle of incidence is equal to the angle of reflection, and the sound waves are directed away from the surface in a way that does not scatter or disperse the energy.
As a result, the amount of energy that is reflected is relatively small compared to the energy that is absorbed or transmitted through the surface.
In contrast, rough and soft surfaces tend to scatter and absorb sound waves, which can result in a higher fraction of energy being reflected back.
This is why soundproofing materials are often designed to be soft and porous, in order to absorb and dampen sound waves rather than reflect them back into space.
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Question 1-7: Does the temperature change produced by one pulse depend on how warm the water is? Why or why not?
The temperature rise produced by one pulse may not be as noticeable or may not even occur if the energy input is not sufficient to cause a measurable temperature change.
The temperature change produced by one pulse does not depend on how warm the water is.
When a pulse of energy is applied to a sample of water, it is absorbed by the water molecules and causes them to vibrate more quickly. This increased vibrational energy results in an increase in temperature of the water, known as the temperature rise. The temperature rise is dependent on the amount of energy that is applied, as well as the mass and specific heat capacity of the water.
The initial temperature of the water will not have an effect on the temperature rise produced by one pulse, as long as the temperature is above the freezing point and below the boiling point of water. This is because the temperature rise is determined solely by the energy applied to the water, and not by the initial temperature.
However, if the initial temperature of the water is near its boiling point or freezing point, the temperature rise produced by one pulse may not be as noticeable or may not even occur if the energy input is not sufficient to cause a measurable temperature change.
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A mass M is placed on a spring with a constant K and is pulled back a distance x to allow the spring to oscillate horizontally on a friction less surface with a period T. What factor can must be changed to increase the period of oscillation?
To increase the period of oscillation (T) in a horizontal mass-spring system, you can either increase the mass (M) or decrease the spring constant (K).
The time taken for an oscillating particle to complete one cycle of oscillation is known as the Period of oscillating particle.
To increase the period of oscillation (T) of a mass (M) placed on a spring with a spring constant (K) and pulled back a distance (x) on a frictionless surface, you can change one of the following factors:
1. Increase the mass (M): Increasing the mass will cause the period of oscillation to increase because the spring will take more time to complete one oscillation cycle.
2. Decrease the spring constant (K): Reducing the spring constant will make the spring less stiff, allowing the mass to oscillate more slowly, thus increasing the period of oscillation.
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A gyroscope has a moment of inertia of 0.14kg×m2 and an initial angular speed of 15 rad/s.Friction in the bearings causes its speed to reduce to zero in 30s. What is the value of the average frictional torque?
a.
3.3 ´ 10-2N×m
b.
8.1 ´ 10-2N×m
c.
14 ´ 10-2N×m
d.
7.0 ´ 10-2N×m
A gyroscope has a moment of inertia of 0.14kg×m2 and an initial angular speed of 15 rad/s. Friction in the bearings causes its speed to reduce to zero in 30s.The closest answer choice to this value is 3.3 * 10^-2 N*cm, so the answer is (a).
The average frictional torque formula is
α = τ / I
Where α is the angular acceleration, τ is the torque, and I is the moment of inertia.
We can rearrange this formula to solve for τ:
τ = α * I
The gyroscope starts with an initial angular speed of 15 rad/s, and comes to a stop after 30 seconds. We can use this information to find the angular acceleration:
α = (0 - 15 rad/s) / 30 s
α = -0.5 rad/s^2
Now we can use the moment of inertia given (0.14 kg*m^2) and the calculated angular acceleration to find the frictional torque:
τ = (-0.5 rad/s^2) * (0.14 kg*m^2)
τ = -0.07 N*m
The frictional torque is negative because it is acting in the opposite direction of the gyroscope's initial angular velocity. To find the average frictional torque, we can assume that the frictional torque is constant over the 30 seconds that it takes for the gyroscope to come to a stop:
τ_avg = τ / t
τ_avg = (-0.07 N*m) / (30 s)
τ_avg = -2.33 * 10^-3 N*m
We need to convert this to N*cm to match the units in the answer choices:
τ_avg = -2.33 * 10^-3 N*m * (100 cm / 1 m)
τ_avg = -0.233 N*cm
The closest answer choice to this value is 3.3 * 10^-2 N*cm, so the answer is (a).
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Look at the net of a right circular cylinder.
1.32 cm
1.32 cm
What is the surface area of the cylinder?
6.25 cm
According to the question the radius of the cylinder is 1.32 cm and the height of the cylinder is also 1.32 cm.
What is radius?Radius is a term used to describe the distance from the center to the circumference of a circle. It can also be used to measure the distance from the center of a sphere to its surface. The radius is half of the diameter, which is the distance from one side of the circle or sphere to the other.
The surface area of a right circular cylinder can be calculated using the formula A = 2πrh + 2πr2, where r is the radius of the cylinder and h is the height of the cylinder. In this case, the radius of the cylinder is 1.32 cm and the height of the cylinder is also 1.32 cm. Plugging these values into the formula, we get: A = 2π(1.32) x (1.32) + 2π(1.32)2 = 6.25 cm2.
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Which type of torque will increase the angular velocity of a system?
Answer:
Propulsive
Explanation:
Friction grabs the rim at the point of contact and pulls it backwards. This causes a torque that causes angular acceleration; thus increasing angular velocity over time.
How is sound produced? Please respond in 1-2 complete sentences using your best grammar.
Answer: Sounds are made when objects vibrate. The vibration makes the air around the object vibrate and the air vibrations enter your ear, You hear them as sounds
Explanation:
What does the frequency spectrum of noise energy look like?
The audio frequency spectrum represents the range of frequencies that the human ear can interpret. Sound frequency is measured in Hertz (Hz) unit. This audible frequency range, in the average person at birth, is from 20Hz to 20000Hz, or 20 kHz. The audio frequency spectrum is also known as sound frequency spectrum.
After the Sun leaves the constellation Sagittarius, how long until it returns to this constellation?
After the Sun leaves the constellation Sagittarius, it takes approximately one year, or 365.25 days, for it to return to this constellation.
This is because the Sun follows a path through the 12 zodiac constellations over the course of one year, known as the ecliptic. As the Earth orbits the Sun, the Sun appears to move through each of these constellations in turn.
Sagittarius is one of the constellations along this path, and the Sun typically passes through it between November 22 and December 21 each year. Once the Sun moves past Sagittarius, it continues on its journey through the remaining zodiac constellations until it completes its orbit and returns to Sagittarius approximately one year later.
It is worth noting that due to variations in the Earth's orbit, the exact dates on which the Sun enters and leaves each zodiac constellation can vary slightly from year to year. However, the overall period of one year remains consistent, meaning that the Sun will always return to Sagittarius approximately one year after it last passed through the constellation.
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a block of mass m slides with speed v0 at the bottom of a ramp of negligible friction that has a height h , as shown. how do the total mechanical energy of the block alone and the total mechanical energy of the block-earth system change when the block slides up the ramp to point p ?
When the block slides up the ramp to point p, the total mechanical energy of the block remains constant and the total mechanical energy of the block-earth system decreases.
What is mechanical energy?
Mechanical energy of a body is the sum of the Kinetic energy and potential energy.
As the block slides up the ramp to point P, its mechanical energy decreases due to the work done by gravity against the block's motion. This work is equal to the change in the block's potential energy, which is mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height of the ramp.
At the same time, the block gains kinetic energy due to its motion up the ramp. This gain in kinetic energy is equal to the work done by the component of the block's weight parallel to the ramp, which is mgh sin(theta), where theta is the angle of the ramp with respect to the horizontal.
Therefore, the total mechanical energy of the block alone remains constant since the loss in potential energy is equal to the gain in kinetic energy.
However, the total mechanical energy of the block-earth system decreases as the block gains potential energy and the Earth gains an equal amount of potential energy in the opposite direction. This is because the block and Earth together form a closed system, and the total mechanical energy of a closed system remains constant only if there are no external forces acting on it. In this case, gravity is an external force that does work on the block and transfers energy to the Earth, causing a decrease in the system's total mechanical energy.
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When the block slides up the ramp to point p, the total mechanical energy of the block remains constant and the total mechanical energy of the block-earth system decreases.
What is mechanical energy?The mechanical energy of a body is the sum of the Kinetic energy and potential energy.
The mechanical energy of the block diminishes as it slides up the ramp to point P due to the effort done by gravity against the block's motion. This effort is equivalent to the change in potential energy of the block, which is mgh, where m is the mass of the block, g is gravity's acceleration, and h is the height of the ramp.
Simultaneously, the block accumulates kinetic energy as it moves up the ramp. This increase in kinetic energy is equal to the work done by the block's weight component parallel to the ramp, which is much sin(theta), where theta is the ramp's angle with respect to the horizontal.
Therefore, the total mechanical energy of the block alone remains constant since the loss in potential energy is equal to the gain in kinetic energy.
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Two pendulums, P and Q, are set up alongside each other. The period of P is 1.90 s and the period of Q is 1.95 s. How many oscillations are made by pendulum Q between two consecutive instants when P and Q move in phase with each other?A) 19B) 38C) 39D) 78
This means that Q completes 41 oscillations while P completes 40 oscillations the answer is (A) 1 oscillation.
The pendulums will move in phase with each other when the time taken for each oscillation is an integer multiple of the common time period. Let T be the common time period between P and Q.
Then, the time taken for P to complete one oscillation is 1.90 s. Therefore, we have:
[tex]1.90 s = n * T[/tex], where n is an integer.
Similarly, the time taken for Q to complete one oscillation is 1.95 s, so:
[tex]1.95 s = m * T[/tex], where m is an integer.
To find the number of oscillations made by Q between two consecutive instants when P and Q move in phase with each other, we need to find the smallest integers n and m such that their ratio m/n is close to 1. Let's divide the second equation by the first:
[tex]1.95 s / 1.90 s = m * T / n * T[/tex]
Simplifying, we get:[tex]1.026 = m/n[/tex]
Now we need to find the smallest integers m and n such that m/n is close to 1.026. One way to do this is to use continued fractions. We have:
[tex]1.026 = 1 + 1/(1 + 1/40.15)[/tex]
Therefore, the best approximation for m/n is:
[tex]m/n = 41/40[/tex]
This means that Q completes 41 oscillations while P completes 40 oscillations. The number of oscillations made by Q between two consecutive instants when P and Q move in phase with each other is therefore:
[tex]41 - 40 = 1[/tex]
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T/F A braking torque is always negative and leads to a decrease in angular velocity
Yes, a braking torque is always negative and leads to a decrease in angular velocity. The given statement is true.
Torque is defined as the product of force and the lever arm, which is the perpendicular distance between the line of action of the force and the axis of rotation. Mathematically, torque can be expressed as:
τ = r × F
where τ is the torque, r is the lever arm, and F is the force.
When a braking torque is applied to an object, it is always opposite in direction to the direction of motion or rotation. The braking torque is applied to slow down or stop the rotation of the object. The direction of the torque is determined by the right-hand rule, which states that if you curl your fingers in the direction of the rotation of the object, then your thumb points in the direction of the torque. If the direction of the braking torque is opposite to the direction of the angular velocity, then the torque is negative.
The negative sign of the torque indicates that it is opposing the direction of the motion or rotation of the object. In other words, the braking torque acts to decrease the angular velocity of the object. The magnitude of the torque depends on the magnitude of the force and the distance between the force and the axis of rotation. A larger force or a greater lever arm will result in a larger torque and a greater slowing down of the object's rotation.
Therefore, a braking torque is always negative and leads to a decrease in angular velocity. It is the opposite of a driving torque, which is applied to increase the angular velocity of an object.
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(E) The net charge on the two spheres is +Q so when they touch and separate, the charge on each sphere (divided equally) is ½ Q. F â Q1Q2 so before contact F â (2Q)(Q) = 2Q^2 and after contact F â (½ Q)(½ Q) = ¼ Q^2 or 1/8 of the original force^
Two identical conducting spheres are charged to +2Q and -Q. respectively, and are separated by a distance d (much greater than the radii of the spheres) as shown above. The magnitude of the force of attraction on the left sphere is F1. After the two spheres are made to touch and then are reseparated by distance d, the magnitude of the force on the left sphere is F2. Which of the following relationships is correct?
(A) 2Fâ = Fâ (B) Fâ = Fâ (C) Fâ = 2Fâ (D) Fâ = 4Fâ (E) Fâ = 8Fâ
The relationship is F2 = 1/8 F1. The correct option is E.
When the two spheres are separated by distance d, the electrostatic force of attraction between them is given by Coulomb's law:
F1 = (k * Q^2) / d^2
where k is the Coulomb constant.
When the spheres are touched and then separated again by distance d, they will share their charges, so each sphere will have a charge of +Q/2 and -Q/2. The electrostatic force between them will be:
F2 = (k * (+Q/2)^2) / d^2 = (k * Q^2) / (4 * d^2)
Therefore, the ratio of F2 to F1 is:
F2/F1 = [(k * Q^2) / (4 * d^2)] / [(k * Q^2) / d^2] = 1/4
So, F2 is 1/4 of F1.
However, the question asks for the relationship between F2 and F1 in terms of the force on the left sphere (not the total force of attraction between the spheres). When the spheres are separated, the force on the left sphere is F1/2, since the two spheres are identical and the force is evenly distributed between them. Similarly, when the spheres are re-separated, the force on the left sphere is F2/2. Therefore, the relationship between F2 and F1 in terms of the force on the left sphere is:
F2/2 = (1/8) (F1/2)
or
F2 = (1/8) F1
The other options are not true because:
(A) 2F1 = F1 - This is not true because the force of attraction between the spheres decreases when they are re-separated, so 2F1 would be greater than F1, not equal to it.
(B) F1 = F1 - This is always true, but it does not provide any information about the relationship between F2 and F1.
(C) F1 = 2F1 - This is not true because the force of attraction between the spheres decreases when they are re-separated, so F1 would be greater than 2F1, not equal to it.
(D) F1 = 4F1 - This is not true because the force of attraction between the spheres decreases when they are re-separated, so F1 would be greater than 4F1, not equal to it.
Therefore, the correct option is (E) F2 = 1/8 F1.
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For a given element, the black lines on an absorption spectrum appear at
For a given element, the black lines on an absorption spectrum appear at specific wavelengths.
Where the energy of the incoming light matches the energy required to excite electrons in the element's atoms from their ground state to higher energy levels. These wavelengths correspond to the specific electronic transitions that are possible within the element's atomic structure. Each element has a unique set of absorption lines that can be used to identify it, making absorption spectroscopy a powerful tool for chemical analysis and identification. For a given element, the black lines on an absorption spectrum appear at specific wavelengths corresponding to the energy levels of the element's electrons. These lines are called absorption lines and they occur when electrons absorb energy and transition from lower to higher energy levels.
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in the derivation of the wave equation, a simplifying approximation is made. what is it? select answer from the options below the maximum displacement is small. the wave speed is small. newton's second law is valid. the wave shape is sinusoidal. the wave frequency is small.
The simplifying approximation made in the derivation of the wave equation is that the maximum displacement is small.
What is Wave?
A wave is a disturbance that propagates through space or a medium, transporting energy from one point to another without the transfer of mass. Waves can take many forms, including electromagnetic radiation, sound waves, water waves, seismic waves, and more.
In the derivation of the wave equation, the wave is assumed to be a small disturbance from its equilibrium state. This means that the displacement of the wave from its equilibrium position is small relative to the wavelength of the wave. This assumption allows us to linearize the equation of motion and apply Newton's second law to the system.
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carolina is hanging christmas lights from a tree in the center of her yard. she wants the lights to be straight similar to the diagram below. she knows the lights are feet long. she stands tall and holds the lights away from the point where they will secure into the ground. using the diagram, how far up the tree should she place the hook to hold the lights?
By using the Pythagoras theorem to determine the height where Carolina should place the hook to hold the lights, we find that she should place them at a height of 12.81 ft.
Let us assume the points on the given triangle as following:
A = the point where the lights will be secured into the ground
B = the position where Carolina is standing, holding the lights
C = the point on the tree where the hook will be placed to hold the lights
D = the bottom of the tree trunk
Now, we want to determine the height of point C on the tree because we are aware that the length of the lights is feet long.
Using the Pythagoras theorem,
AB² + BC² = AC²
Since Carolina is 10 feet away from the tree, we know that AB = 10 feet, and BC = 8 feet because the lights are 8 feet long and we want them to hang straight down. We can change these numbers in the equation to:
=10² + 8² = AC²
=100 + 64 = AC²
= AC² = √164
= AC ≈ 12.81 feet
Therefore, Carolina should place the hook about 12.81 feet up the tree to hold the lights straight.
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Astronomers use the gravitational wobble of a star to indicate the presence of planets outside our solar system. Astronomers specifically look for planets in the habitable zone of 1.6x10^11 m away from the star. Calculate the mass of a planet that exerts a 1.5 x 10^13 force on a 5.4 x 10^30 kg star.
The mass of the planet that exerts a 1.5 x 10^13 N force on a 5.4 x 10^30 kg star is approximately 1.34 x 10^25 kg.
To calculate the mass of a planet, we can use Newton's law of universal gravitation. The formula is:
F = G * (m1 * m2) / r^2
where F is the gravitational force, G is the gravitational constant (6.674 × 10^-11 N*(m/kg)^2), m1 and m2 are the masses of the two objects, and r is the distance between them.
In this case, F = 1.5 x 10^13 N, m1 (mass of the star) = 5.4 x 10^30 kg, and r = 1.6 x 10^11 m. We want to find m2, the mass of the planet. Rearrange the formula to solve for m2:
m2 = (F * r^2) / (G * m1)
Now plug in the given values:
m2 = (1.5 x 10^13 N * (1.6 x 10^11 m)^2) / (6.674 × 10^-11 N*(m/kg)^2 * 5.4 x 10^30 kg)
After performing the calculations, you get:
m2 ≈ 1.34 x 10^25 kg
So, the mass of the planet that exerts a 1.5 x 10^13 N force on a 5.4 x 10^30 kg star is approximately 1.34 x 10^25 kg.
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sound waves in the thin martian atmosphere travel at 245 m/s . what is the period of a 125 hz sound wave in the martian atmosphere? what is the frequency of a sound wave in the martian atmosphere that has wavelength 3.00 m ?
Answer:
Sound waves in the thin Martian atmosphere travel at 245 m/sm/s.
Explanation:
The number of overtones, and their relative intensities, is associated with what property of the tone generated by a musical instrument?
a. attack pattern
b. interference pattern
c. quality
d. range
The number of overtones and their relative intensities are associated with the quality of the tone generated by a musical instrument. The correct option is c.
Quality, also known as timbre, is the characteristic sound of an instrument that distinguishes it from others. Overtones are additional frequencies that are produced along with the fundamental frequency of a note when it is played.
The number and intensity of these overtones determine the unique timbre of the instrument.
The attack pattern refers to the initial transient sound produced when a note is played, while interference pattern refers to the interference of sound waves from multiple sources.
Range, on the other hand, refers to the span of notes an instrument can produce.
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Can you please help with these 2 questions??
A) The solution has a molarity of 0.1176 M. B) The pipet contains 0.00353 moles of copper(II) nitrate and C) The new solution has a molarity of 0.0147 M.
Calculation-A) Cu(NO3)2 has a molar mass of 187.55 g/mol.
Mass / molar mass = number of moles
5.52 g / 187.55 g/mol equals the number of moles.
Molecular weight: 0.0294 mol
Molarity is equal to the moles of solute per litre of solution.
250.0 mL = 0.2500 L
Molarity is equal to 0.0294 mol/0.2500 L.
Molarity equals 0.1766 M
The solution's molarity is 0.1176 M as a result.
B) The solution's molarity, which we determined in section (a):
Liquid volume divided by the molarity gives the moles of Cu(NO3)2
Molecules of Cu(NO3)2 are equal to 0.1176 M and 0.0300 L, respectively.
As a result, the pipet contains 0.00353 moles of copper(II) nitrate.
C) The solution has been reduced by a factor of 8 by Mrs. Mandochino (240.0 mL / 30.0 mL). The new molarity is thus 1/8 of the initial molarity:
Molarity = 0.0147 M / Molarity = 0.1176 M / 8
As a result, the new solution has a molarity of 0.0147 M.
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Compare the force-time curve for the inelastic collision to that for the nearly elastic collision.
In an inelastic collision, the objects involved stick together after impact, causing a greater deformation and a longer duration of contact. This results in a relatively wider force-time curve with a lower peak force.
When we talk about collisions, we often use force-time curves to understand the behavior of the objects involved. In an inelastic collision, the objects involved stick together after colliding, meaning that some energy is lost as heat or sound. On the other hand, in a nearly elastic collision, the objects bounce off each other and conserve most of their kinetic energy.
Comparing the force-time curves for these two types of collisions, we would expect to see some differences. In an inelastic collision, the force-time curve will generally show a larger force over a longer period of time compared to a nearly elastic collision. This is because in an inelastic collision, the objects are deforming as they stick together, creating a larger force as they do so.
In a nearly elastic collision, the force-time curve will show a shorter duration of high force, as the objects bounce off each other and transfer most of their kinetic energy without deforming. The curve will also show a more gradual decrease in force as the objects separate, since there is less energy being dissipated as heat or sound.
Overall, the force-time curves for inelastic and nearly elastic collisions will look different due to the different ways that energy is transferred and dissipated during the collision.
In a nearly elastic collision, the objects rebound after impact with minimal deformation, leading to a shorter contact duration. This produces a steeper, narrower force-time curve with a higher peak force.
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A 1.8-kg block is projected up a rough 10° inclined plane. As the block slides up the incline, its acceleration is 3.8 m/s2 down the incline. What is the magnitude of the force of friction acting on the block?
1) 5.0 N
2) 3.8 N
3) 4.2 N
4) 4.6 N
5) 6.5 N
The magnitude of the force of friction acting on the block is approximately 4.2 N. The correct answer is 3) 4.2 N.
To find the magnitude of the force of friction acting on the block, we can use the equation:
f_friction = m * (g * sin(θ) - a)
Where:
- f_friction is the force of friction
- m is the mass of the block (1.8 kg)
- g is the acceleration due to gravity (9.81 m/s²)
- θ is the angle of the inclined plane (10°)
- a is the acceleration of the block down the incline (3.8 m/s²)
f_friction = 1.8 * (9.81 * sin(10°) - 3.8)
Calculating this, we get:
f_friction ≈ 4.2 N
So, the magnitude of the force of friction acting on the block is approximately 4.2 N. The correct answer is 3) 4.2 N.
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do all galaxies have spiral arms? if so, why? in not, what is different about spiral galaxies that gives them these spiral arms?
No, not all galaxies have spiral arms. The newly-formed stars and other luminous matter in the galaxy follow slightly elliptical orbits and they become concentrated along the spiral arms, making these arms more visible.
There are various types of galaxies, and they are generally classified into three main categories: spiral, elliptical, and irregular. Spiral galaxies have distinct spiral arms, while elliptical galaxies do not have any specific structure and are mostly spherical or elliptical in shape. Irregular galaxies also lack a well-defined structure and may have some chaotic appearance.
Spiral galaxies, like our own Milky Way, have spiral arms due to the density wave theory. This theory suggests that the spiral arms are formed as a result of gravitational interactions and the distribution of matter in the galaxy. The density wave moves through the galactic disk, compressing gas and dust, which eventually triggers the formation of new stars. As the newly-formed stars and other luminous matter in the galaxy follow slightly elliptical orbits, they become concentrated along the spiral arms, making these arms more visible.
The differences in galaxy types can be attributed to factors such as their formation history, the composition of their matter, and the various interactions they have undergone. Spiral galaxies typically have a flat rotating disk, a central bulge, and a surrounding halo of stars, whereas elliptical galaxies consist of a more homogeneous distribution of stars without a clear disk or spiral structure.
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