Answer:
The Law of Conservation of energy and charge is a fundamental principle in physics that states that energy and charge can neither be created nor destroyed, but can only be transformed from one form to another.
Kirchoff's Loop and Junction Rules, on the other hand, are fundamental principles in electrical circuit theory that are used to analyze and solve electrical circuits. Kirchoff's Loop Rule states that the sum of the potential differences around any closed loop in a circuit must be zero, while Kirchoff's Junction Rule states that the sum of the currents entering a junction in a circuit must be equal to the sum of the currents leaving the junction.
The Law of Conservation of energy and charge does not directly support or refute Kirchoff's Loop and Junction Rules because they are based on different principles. However, these rules are consistent with the Law of Conservation of energy and charge because they ensure that the flow of energy and charge in a circuit is conserved.
Kirchoff's Loop Rule is based on the principle of conservation of energy, which states that the total energy in a closed system must remain constant. The Loop Rule ensures that the energy supplied by the battery or other energy source is equal to the energy consumed by the resistors or other components in the circuit.
Kirchoff's Junction Rule is based on the principle of conservation of charge, which states that the total charge in a closed system must remain constant. The Junction Rule ensures that the charge flowing into a junction is equal to the charge flowing out of the junction, which is consistent with the Law of Conservation of charge.
In summary, Kirchoff's Loop and Junction Rules are consistent with the Law of Conservation of energy and charge, but they do not directly support or refute it. These rules are fundamental principles in electrical circuit theory that ensure that the flow of energy and charge in a circuit is conserved.
Explanation:
Drivers should compensate for a lack of depth perception by..
Answer:
decreasing speed
Explanation:
so that way you have a faster reaction time
What are the temperature and energy changes for segment A (red line) on the heating curve for water?
The region marked segment A occurs at 100 degrees Celsius and is the latent heat of vaporization.
What is the latent heat of vaporization?The latent heat of vaporization is the amount of heat energy required to change the state of a substance from a liquid to a gas (vapor) at a constant temperature and pressure.
During the phase transition, the energy supplied to the substance is used to overcome the intermolecular forces of attraction between the particles, which allows the particles to escape from the liquid phase and enter the gaseous phase. This results in an increase in the internal energy of the substance, but with no change in temperature.
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A student is given a low voltage power supply and 1m of resistance wire. The student uses these and other pieces of equipment to measure the resistance of just 50cm of the resistance wire. Draw a diagram of the circuit that the students should use. Your circuit diagram should identify the pieces of equipment that the student uses
The student should use the power supply to provide a voltage source (V), the resistance wire (R) and an ammeter (A).The circuit diagram is V-A- R (50cm) -V.
What is resistance ?Resistance is the refusal to accept or comply with something. It is the opposition to a particular idea, policy, or course of action. Resistance can be seen in many forms, including physical, psychological, social, and political. Physically, it can be seen in the refusal to comply with a certain rule or law, or in the refusal to engage in a certain activity or behavior. Psychologically, it can be seen in the refusal to accept certain ideas, beliefs, or values. Socially, it can be seen in the refusal to engage in certain social practices or customs.
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in one cycle of the engine, the net change in the entropy of both reservoirs (hot and cold together) is.
In one cycle of an engine, the net change in the entropy of both reservoirs (hot and cold together) is zero. This is because the entropy of each reservoir remains constant throughout the cycle.
Entropy is a measure of the randomness and disorder of a system, so the fact that the entropy of the two reservoirs remains constant throughout the cycle is a consequence of the law of conservation of energy. The engine cycle is designed so that the energy transfers between the two reservoirs cause the entropy of each reservoir to remain constant.
In an ideal cycle, the total amount of energy transferred from the hot reservoir to the cold reservoir is equal to the total amount of energy transferred from the cold reservoir to the hot reservoir. Thus, the net change in the entropy of both reservoirs is zero. This can be shown by examining the equation for the change in entropy of a system:
ΔS = Q/T, where Q is the energy transfer and T is the temperature.
The amount of energy transferred between the hot and cold reservoirs will be the same in both directions. Thus, the temperature ratio between the two reservoirs is the same. Since the energy transfer is the same, the change in entropy for both reservoirs is equal, and the net change in entropy is zero. This can be illustrated by the following equation:
ΔS = Q/T = Q/T + Q/T = 0.
In summary, the net change in the entropy is zero. This is due to the law of conservation of energy and the fact that the temperature ratio between the two reservoirs remains constant during the cycle.
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Answer the following questions
1- How to Calculate Power?
2- What is the units of Power?
3- Write a solved example?
Answer:
1- How to Calculate Power?
ans; power is defined as " work done per unit time"
therefore calculated by P= Work/time.
2- What is the units of Power?
Ans; It is watt denoted by "W"
3- Write a solved example?
Ans; Lets say we move a block from One place to another by using 10joules in 4 seconds.
so put values
P=work/time
P=10/4
P=2.5
Explanation:
Determine the maximum axial force p that can be applied to the steel plate. Express your answer to three significant figures and include the appropriate units
We must take into account the relationship between the material's stress and strain, which is characterised by its Young's modulus, in order to establish the maximum axial force p that may be applied to the steel plate (E).
We must take into account the relationship between the material's stress and strain, which is characterised by its Young's modulus, in order to establish the maximum axial force p that may be applied to the steel plate (E). Assume that the cross-sectional area of the steel plate is A.
You can compute the tension on the steel plate as follows:
force x area equals stress
You may compute the strain on the steel plate as follows:
strain is equal to length change / starting length.
Hence, p = yield strength * area can be used to compute the maximum axial force p that can be applied to the steel plate without generating plastic deformation.
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Light passes from air (n=1) into another medium at 30.0 degrees to the normal. If the angle of refraction is 18.0 degrees, what is the index of refraction of the new medium?
The refractive index of the new medium is approximately 1.59.
Snell's law relates the angles of incidence and refraction of light passing through a boundary between two media with different refractive indices (n). The formula is:
n₁ sin θ₁ = n₂ sin θ₂
where n₁ is the refractive index of the first medium, θ₁ is the angle of incidence, n₂ is the refractive index of the second medium, and θ₂ is the angle of refraction.
In this problem, we know that the angle of incidence is 30.0 degrees and the angle of refraction is 18.0 degrees. We also know that the refractive index of air (n₁) is 1.00. Therefore, we can use Snell's law to solve for the refractive index of the new medium (n₂):
n₁ sin θ₁ = n₂ sin θ₂
1.00 sin 30.0° = n₂ sin 18.0°
n₂ = (1.00 sin 30.0°) / sin 18.0°
n₂ ≈ 1.59
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An object experiences two forces acting on the same surface (i. E. They are additive): a force of 5. 0N acting at 60. ∘ to the horizontal and a force of 8. 0N acting 45∘ to the horizontal.
What is the magnitude of the resultant force on the object? Give your answer to the nearest newton, without units
The magnitude of the resultant force on the object is approximately 12 N.
For the 5.0 N force:
Horizontal component = 5.0 N * cos(60°) ≈ 2.5 N
Vertical component = 5.0 N * sin(60°) ≈ 4.3 N
For the 8.0 N force:
Horizontal component = 8.0 N * cos(45°) ≈ 5.7 N
Vertical component = 8.0 N * sin(45°) ≈ 5.7 N
Next, we can add the horizontal and vertical components separately:
Resultant horizontal component = 2.5 N + 5.7 N ≈ 8.2 N
Resultant vertical component = 4.3 N + 5.7 N ≈ 10 N
Finally, we can use the Pythagorean theorem to find the magnitude of the resultant force:
Resultant force = sqrt((8.2 N)^2 + (10 N)^2) ≈ 12 N
The resultant force is the net force that acts on an object. It is the vector sum of all the forces acting on the object. If an object is subjected to multiple forces, the resultant force determines the direction and magnitude of the object's motion. If the resultant force is zero, the object will remain at rest or continue moving at a constant velocity. If the resultant force is non-zero, the object will accelerate in the direction of the force.
The concept of the resultant force is particularly important in dynamics, the branch of mechanics that deals with the motion of objects under the influence of forces. The laws of motion developed by Sir Isaac Newton, which are the foundation of classical mechanics, are formulated in terms of resultant forces. The first law states that an object at rest will remain at rest or move at a constant velocity unless acted upon by a net external force, while the second law relates the acceleration of an object to the net force acting on it.
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3. An 87 kg fullback moving east with a speed of 5.0 m/s is tackled by a 97 kg opponent running
west at 3.5 m/s, and collision is perfectly inelastic. Calculate the following:
Given:
Sketch the before and after collisions:
a.
The velocity of the players just after the tackle
b. The decrease in kinetic energy during the collision
The decrease in kinetic energy during the collision is 1187.62 J.
Steps
Before the collision, the fullback's momentum is:
p1 = m1v1 = (87 kg)(5.0 m/s) = 435 kg*m/s (to the east)
The opponent's momentum is:
p2 = m2v2 = (97 kg)(-3.5 m/s) = -339.5 kg*m/s (to the west)
The total momentum before the collision is:
p1 + p2 = (435 kgm/s) + (-339.5 kgm/s) = 95.5 kg*m/s (to the east)
After the collision, the two players move together as one mass. Let vf be their common final velocity. Then the total momentum after the collision is:
p = (m1 + m2)vf = (87 kg + 97 kg)vf = 184 kg*vf (to the east)
Since momentum is conserved, we can equate the total momentum before and after the collision:
p1 + p2 = p
95.5 kgm/s = 184 kgvf
vf = 0.52 m/s (to the east)
Therefore, the velocity of the players just after the tackle is 0.52 m/s to the east.
To find the decrease in kinetic energy during the collision, we first need to find the initial kinetic energy:
KEi = (1/2)m1v1^2 + (1/2)m2v2^2
KEi = (1/2)(87 kg)(5.0 m/s)^2 + (1/2)(97 kg)(-3.5 m/s)^2
KEi = 1211.75 J
Since the collision is perfectly inelastic, the final velocity is the same for both players, and their combined mass is 184 kg. Therefore, the final kinetic energy is:
KEf = (1/2)mvf^2
KEf = (1/2)(184 kg)(0.52 m/s)^2
KEf = 24.13 J
The decrease in kinetic energy during the collision is:
ΔKE = KEi - KEf
ΔKE = 1187.62 J
Therefore, the decrease in kinetic energy during the collision is 1187.62 J.
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Which would have more matter: a 1 cm cube of lead or a 1 cm
cube of rubber?
Answer: 1CM CUBE OF LEAD
Explanation:
Consider the track shown in the figure. The section AB is one quadrant of a circle of radius 2.0 m and is frictionless. B to C is a horizontal span 2.8 m long with a coefficient of kinetic friction μk = 0.20. The section CD under the spring is frictionless. A block of mass 1.0 kg is released from rest at A. After sliding on the track, it compresses the spring by 0.15 m.
A) Determine the velocity of the block at point B.
B) Determine the thermal energy produced as the block slides from B to C.
C) Determine the velocity of the block at point C.
D) Determine the stiffness constant k for the spring.
A baby mouse 1. 2 cm high is standing 4. 0 cm from a converging mirror having a focal length of 30 cm
The problem involves the image formation of a small object by a converging mirror.
According to the mirror equation, 1/f = 1/di + 1/do, where f is the focal length of the mirror, do is the object distance, and di is the image distance. In this case, the object is a baby mouse that is 1.2 cm high and located 4.0 cm away from the mirror. The mirror is a converging mirror with a focal length of 30 cm. To determine the image distance, we can use the mirror equation as follows: 1/30 = 1/di + 1/4. Solving for di, we get: di = 3.75 cm. This means that the image of the baby mouse is formed 3.75 cm behind the mirror. The size of the image can be determined using the magnification equation, M = -di/do, where M is the magnification. Substituting the values, we get: M = -(3.75 cm)/(4.0 cm) = -0.9375. The negative sign indicates that the image is inverted compared to the object. The magnification also tells us that the image is smaller than the object, with a height of: hi = Mho = (-0.9375)(1.2 cm) = -1.125 cm. Again, the negative sign indicates that the image is inverted. The absolute value of the height tells us that the image is smaller than the object, with a height of 1.125 cm.
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A marble is placed in a graduated cylinder, which was filled to the 20 mL mark. The level rises to 40 mL. What happens to the volume of the water?
Answer:
Below
Explanation:
The volume of water remains the same....the new measurement now includes the volume of the marble ( 20 ml)
A pulley system is such that a person pulls down with 10 N of force to lift a 20 N object. The
machanical advantage is.
(MA = Fout/Fin)
O 0. 5
O 2
O
30
Please help me
Two is the mechanical benefit of the pulley mechanism.
The mechanical advantage of a pulley system is the ratio of the output force to the input force. In this case, the person is pulling down with a force of 10 N to lift a 20 N object. Therefore, the mechanical advantage can be calculated as follows:
MA = output force / input force
MA = 20 N / 10 N
MA = 2
The mechanical advantage of the pulley system is 2. This means that for every 1 unit of force the person applies, the object is lifted with 2 units of force. In other words, the pulley system allows the person to lift the object with less force than would be required without the pulley system.
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A wave has a frequency of 12,000 Hz. What does this mean
It means that wave vibrates 12,000 times about its mean position per second.
Humans can hear sounds between about 20 Hz and 20,000 Hz (depending on the human!).
Below 20 Hz is called infrasound ("infra" means below), and above 20,000 Hz is ultrasound ("ultra" means beyond).
We are most sensitive to sounds between 1,000 and 4,000 Hz:
As we get older we are less sensitive to higher frequency sounds (a limit around 12,000 Hz is normal for an adult).
Vibration refers to the back-and-forth motion of an object or system about its equilibrium position. This motion can be periodic or random and can occur in a variety of mediums, including solids, liquids, and gases. Vibration is caused by a force or disturbance that creates an oscillation within the system.
Vibration can be both beneficial and detrimental. Beneficial vibrations can be used to create sound, power tools, and even musical instruments. On the other hand, detrimental vibrations can cause damage to structures, machinery, and even the human body. For example, exposure to high levels of vibration can cause hand-arm vibration syndrome or whole-body vibration syndrome, which can result in numbness, tingling, and even long-term health problems.
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Find the magnitude of the electric force between the charges 0. 12 C and 0. 33 C at a separation of 2. 5 m. Is the force attractive or repulsive?
The magnitude of the electric force between the charges 0. 12 C and 0. 33 C at a separation of 2. 5 m is 4.987 N. If the charges are of the same sign, the force will be repulsive; if they are of opposite signs, the force will be attractive.
We can use Coulomb's law to find the magnitude of the electric force between two charges:
F = k * ([tex]q_{1}[/tex] * [tex]q_{2}[/tex]) / [tex]r^{2}[/tex]
where F is the magnitude of the force, k is Coulomb's constant (9 x [tex]10^{9}[/tex]10^9 N*[tex]m^{2}[/tex]/[tex]C^{2}[/tex]), [tex]q_{1}[/tex] and [tex]q_{2}[/tex] are the magnitudes of the charges, and r is the separation between the charges.
Plugging in the given values, we get:
F = (9 x [tex]10^{9}[/tex] N*[tex]m^{2}[/tex]/[tex]C^{2}[/tex]) * ((0.12 C) * (0.33 C)) / (2.5 m)^2
Simplifying:
F = 4.987 N
Therefore, the magnitude of the electric force between the charges is approximately 4.987 N.
To determine whether the force is attractive or repulsive, we need to know the signs of the charges.
Since the problem does not specify the signs of the charges, we cannot determine whether the force is attractive or repulsive.
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define the term amplitude
The largest displacement or distance a wave oscillates from its rest state is referred to as amplitude. It establishes the amount of energy a wave may carry by determining the strength or intensity of the wave.
How should amplitude be named?Depending on whether the sun is rising or sinking, the amplitude is designated by the same names as the declination: E or W. The identified and named compass mistake is as follows: It is the distinction between the correct bearing and the compass.
What does amplitude measure mean?The distance between a wave's peak or trough and the location of the medium at rest, also known as the equilibrium position, is often measured as the amplitude in transverse waves.
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a 22.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. what centripetal force must be exerted on her for her to remain at constant distance of 1.97 m from the merry-go-round's center
The centripetal force exerted on a 22.0 kg child riding a playground merry-go-round rotating at 40.0 rev/min, to remain at a constant distance of 1.97 m from the merry-go-round's center is 502.82 N.
Centripetal force is the force required to maintain an object in a circular motion with a constant speed. Its direction is towards the center of the circle in which the object is moving.
The formula for calculating the centripetal force is given as: Fc = mv2/r where,
Fc is the centripetal force, m is the mass of the object,
v is the velocity of the object,
r is the radius of the circle.
Substituting the given values,
Mass of the child, m = 22.0 kg
Velocity, v = 40.0 rev/min,
we know 1 rev = 2π rad2π rad/1 rev so 40.0 rev/min = 40.0 * 2π rad/min = 80π rad/min
Radius, r = 1.97 m
Now, converting the velocity units from radians per minute to meters per second,1 rad/min = (1/60) rad/s
Therefore, 80π rad/min = 80π/60 rad/s = (4/3)π rad/s
Velocity, v = rω where, ω is the angular velocity.
Substituting the given values,ω = v/rω = (4/3)π rad/s / 1.97 mω = 2.012 rad/s
Substituting the given values in the formula for centripetal force, we get
Fc = mv2/rFc = 22.0 kg × (2.012 m/s)2 / 1.97 mFc = 502.82 N
Thus, the centripetal force exerted on the child is 502.82 N.
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The fastest bird is the spine-tailed swift, which
reaches speeds of 171 km/h. Suppose that you wish
to shoot such a bird with a. 22-caliber rifle that fires
a bullet with a speed of 366 m/s. If you fire at the
instant when the bird is 30 m directly overhead, how
many meters ahead of the bird must you aim the
rifle? Ignore gravity in this problem.
We need to aim the rifle 47.154 meters ahead of the bird's current position to intercept its flight path and hit it.
Since we are ignoring gravity, the horizontal velocity of the bullet will remain constant and equal to the muzzle velocity of the rifle, which is 366 m/s.
To hit the bird, we need to aim the rifle ahead of the bird such that the bullet intercepts the bird's flight path at the same point in space where the bird will be when the bullet arrives.
time = distance / velocity
time = 30 m / 366 m/s
time = 0.08197 s
During this time, the bird will have flown a distance equal to its speed multiplied by the time:
distance = speed x time
distance = 171 km/h x (0.08197 s)
distance = 47.154 m
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Dos esteras conductoras, cargadas y de igual radio, se apoyan en soportes no conductores. Sus cargas respectivas son-3 C y +7 MC. Si ambas esteras se ponen en contacto y luego son separadas, entonces las cargas respectivas de cada estera son:
A. -Z UC Y +2 MC
В. +1 МС У -3 МС
C. +2 C y +2 MC
D. -1 MC y - 1 MC
The correct option is B. The respective charges on each mat are +1 MC y -3 MC.
Charge refers to a fundamental property of matter that describes how objects interact electromagnetically. The charge can be positive or negative and is measured in Coulombs (C). Objects with the same charge repel each other, while objects with opposite charges attract each other.
Electric charge can be transferred from one object to another through processes such as friction, contact, or induction. The movement of electric charge is the basis for electric current, which is the flow of charge through a conductor.
A charge is also a conserved quantity in physics, meaning that it cannot be created or destroyed, only transferred or redistributed. This conservation of charge is a fundamental principle in many areas of physics, including electromagnetism and quantum mechanics.
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Complete Question:
Two charged conductive mats of equal radius are supported by non-conductive supports. Their respective charges are -3 C and +7 MC. If both mats come into contact and are then separated, then the respective charges on each mat are:
A. -Z UC Y +2 CM
В. +1 MS У -3 MS
C. +2 C and +2 MC
D. -1 MC and - 1 MC
observe where your house is located and from this explain why your house is hot or cold
The temperature inside a house can be influenced by various factors such as the geographic location, orientation of the house, insulation, and the type of construction materials used.
For instance, if a house is located in a hot and humid climate, it may become hot inside due to the external heat and humidity, even with air conditioning.
On the other hand, a house located in a cold climate may become cold inside, especially during the winter season, due to heat loss from the interior to the cold outside environment.
Proper insulation and efficient heating/cooling systems can help regulate the temperature inside a house and improve its energy efficiency.
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light travels a certain distance in 2000 years. is it possible that an astronaut, traveling slower than light, would go as far in 20 years of her life as light travels in 2000 years?
They will be unable to cover as much distance as light does.
The statement is: Light travels a certain distance in 2000 years. No, it is not possible for an astronaut, traveling slower than light, to go as far in 20 years of her life as light travels in 2000 years because the speed of light is constant and cannot be matched by anything with mass.
The speed of light in a vacuum is 299,792,458 meters per second (m/s), which is incredibly fast. On the other hand, the fastest spacecraft to ever leave Earth was NASA's New Horizons probe, which traveled at a speed of about 16.26 kilometers per second.
It would take this spacecraft 37,200 years to travel 2000 light-years. Astronauts cannot travel faster than the speed of light, and their velocity will always be lower than that of light. As a result, they will be unable to cover as much distance as light does.
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Find a center of mass of a thin plate of density delta equals 5 bounded by the lines y equals x and x equals 0 and the parabola y equals 20 minus x squared in the first quadrant
The center of mass of the thin plate is located at the point (16/3, 8/3) in the first quadrant.
To find the center of mass of a thin plate with a density of delta equals 5 bounded by the lines y equals x and x equals 0 and the parabola y equals 20 minus x squared in the first quadrant, we can use the following formula:
x = (1/M) ∫∫ x δ(x,y) dA
y = (1/M) ∫∫ y δ(x,y) dA
Now we can use this value of M to find the center of mass:
x = (1/M) ∫∫ x δ(x,y) dA
= (1/125) ∫₀²₀ ∫₀^x x 5 dy dx
= (1/125) ∫₀²₀ 5x²/2 dx
= 16/3
y = (1/M) ∫∫ y δ(x,y) dA
= (1/125) ∫₀²₀ ∫₀^x y 5 dy dx
= (1/125) ∫₀²₀ 5x(20-x²)/2 dx
= 8/3
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a 1.60 m m tall person lifts a 1.40 kg k g book off the ground so it is 2.00 m m above the ground. part a what is the potential energy of the book relative to the ground?
27.44 J of the potential energy of the book relative to the ground, when a 1.60 m tall person lifts 1.40 kg of the book off the ground.
A 1.60 m m tall person lifts a 1.40 kg book off the ground so it is 2.00 m m above the ground.
The potential energy of the book relative to the ground would be:
PE = mgh
Where PE is the potential energy,
m is the mass,
g is the acceleration due to gravity, and
h is the height
The acceleration due to gravity is constant and is 9.8 m/s²
mass m = 1.40 kg
Height h = 2.00 m
mgh = 1.40 × 9.8 × 2.00 = 27.44 J
Ans: The potential energy of the book relative to the ground is 27.44 J.
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When tension is applied to an aluminum rod (Y = 70 x 10^9 Pa) of length 1 m, it stretches by Δ.
1. If the same tension is applied to another aluminum rod with the same cross-sectional area, but of length 2 m, by how much will it stretch?
a) less than Δ
b) Δ
c) more than Δ
2. Now, consider two aluminum rods of length 1 m, one twice as thick as the other. If a compressive force F is applied to both rods, their lengths are reduced by Δthick rod
and Δthin rod, respectively.
The ratio (Δthick rod)/(Δthin rod) is
a) (Δthick rod)/(Δthin rod) < 1
b) (Δthick rod)/(Δthin rod) = 1
c) (Δthick rod)/(Δthin rod) > 1
The answers for the given questions about topic tension:
1. c.) More than ΔL
2. b.) <1
What is aluminium rod?Light and strong, an aluminium rod is highly resistant to corrosion and is ideal for both indoor as well as outdoor use as it can withstand a variety of environmental conditions. Our aluminium rods, which are also referred to as aluminium round bars, are provided in the 6000 series, which has the greatest commercial versatility.
The mechanical sectors can use aluminium alloy wire rods for things like bolts, nuts, nails, needles, rivets, clips, and staples.
1. stress = Y strain
(F / A) = Y (deltaL / L)
deltaL / L = constant (for this problem)
delta(L) / 1 = delta(L') / 2
delta(L') = 2 delta(L)
2. deltaL A = constant
L_thick (2A) = Lthin (A)
L_thick / L_thing = 1/2 < 1
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A 0.20 kg k g puck is being pushed across a table with a horizontal force of 2.0 N N . It starts from rest and is pushed for 13 seconds, ending with a speed of 1 m/s m / s . Calculate the coefficient of friction μk μ k between the puck and the table.
Answer:
The coefficient of friction between the puck and the table is [tex]\sf 0.00784[/tex]
Explanation:
Here are some of the equations we will be using. Below are the equations for Work, Kinetic Energy, Potential Energy, Distance, and the Force due to Friction.
[tex]\sf W=fd[/tex]
[tex]\sf K_E=\sf \dfrac{1}{2}mv^2[/tex]
[tex]\sf U_E=mgh[/tex]
[tex]\sf d=\dfrac{tv_f}{2}+ \dfrac{tv_i}{2}[/tex]
[tex]\sf F_{Fr} =mg \mu_k[/tex]
Conservation of Energy
If there are frictional forces present, then the work done against the frictional forces is equal to the change in the total Mechanical Energy. The Mechanical Energy will decrease because of the work done against the frictional forces.
[tex]\sf W_{Fr}= \left(K_F+U_F\right)-\left(K_i+U_i\right)[/tex]
Lets put together some of these formulas now.
[tex]\sf -F_{Fr} d= \left(\sf \dfrac{1}{2}mv_f^2+mgh_f\right)-\left(\sf \dfrac{1}{2}mv_i^2+mgh_i\right)[/tex]
[tex]\sf -mg \mu_k\left(\sf \dfrac{tv_f}{2}+ \dfrac{tv_i}{2}\right)= \left(\sf \dfrac{1}{2}mv_f^2+mgh_f\right)-\left(\sf \dfrac{1}{2}mv_i^2+mgh_i\right)[/tex]
In this example our object is starting from rest and is on a flat surface, so we can cancel out any terms with [tex]\sf v_i[/tex] and [tex]\sf h[/tex].
[tex]\sf -mg \mu_k \left(\sf \dfrac{tv_f}{2}\right)= \left(\sf \dfrac{1}{2}mv_f^2\right)[/tex]
Lets solve for [tex]\sf \mu_k[/tex].
Combine [tex]\sf m[/tex] and [tex]\sf \frac{tv_f}{2}[/tex].
[tex]\sf -1\cdot g \cdot \mu_k \cdot \left(\sf \dfrac{m(tv_f)}{2}\right)= \left(\sf \dfrac{1}{2}mv_f^2\right)[/tex]
Combine [tex]\sf g[/tex] and [tex]\sf \frac{m(tv_f)}{2}[/tex].
[tex]\sf -1\cdot \mu_k \cdot\left(\sf \dfrac{g(m(tv_f))}{2}\right)= \left(\sf \dfrac{1}{2}mv_f^2\right)[/tex]
Combine [tex]\mu_k[/tex] and [tex]\sf \frac{g(m(tv_f))}{2}[/tex].
[tex]\sf -1\cdot\sf \dfrac{g(m(tv_f))\mu_k}{2}= \left(\sf \dfrac{1}{2}mv_f^2\right)[/tex]
Remove the parentheses.
[tex]\sf -1\cdot\sf \dfrac{gmtv_f\mu_k}{2}= \left(\sf \dfrac{1}{2}mv_f^2\right)[/tex]
Simplify the right side.
[tex]\sf -1\cdot\sf \dfrac{gmtv_f\mu_k}{2}= \sf \dfrac{mv_f^2}{2}[/tex]
Divide both sides of the equation by [tex]\sf -1[/tex].
[tex]\sf \sf \dfrac{gmtv_f\mu_k}{2}= -\sf \dfrac{mv_f^2}{2}[/tex]
Since the expression on each side of the equation has the same denominator, the numerators must be equal.
[tex]\sf gmtv_f\mu_k= -\sf mv_f^2[/tex]
Divide both sides of the equation by [tex]\sf gmtv_f[/tex].
[tex]\sf \sf \dfrac{gmtv_f\mu_k}{\sf gmtv_f}= \sf \dfrac{-mv_f^2}{\sf gmtv_f}[/tex]
On the left side cancel the common factor of [tex]\sf gmtv_f[/tex].
[tex]\sf \mu_k= \sf \dfrac{-mv_f^2}{\sf gmtv_f}[/tex]
On the right side cancel the common factor of [tex]\sf m[/tex].
[tex]\sf \mu_k= \sf \dfrac{-v_f^2}{\sf gtv_f}[/tex]
On the right side cancel the common factor of [tex]\sf v_f[/tex]
Finally we have an equation to evaluate the coefficient of friction.
[tex]\boxed{\sf \mu_k= -\sf \dfrac{v_f}{\sf gt}}[/tex]
Numerical Evaluation
In this example we are given
[tex]\sf v_f=1\\g=-9.81\\t=13[/tex]
Substituting our given values into the equation yields
[tex]\boxed{\sf \mu_k= -\sf \dfrac{1}{\sf -9.81\cdot 13}}[/tex]
[tex]\boxed{\sf \mu_k=0.007841292245}[/tex]
Rounding to the hundred thousandth leaves us with
[tex]\boxed{\sf \mu_k=0.00784}[/tex]
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The coefficient of kinetic friction between the puck and the table is approximately 0.0079.
What is coefficient of friction?The coefficient of friction is a measure of the amount of frictional force that exists between two surfaces in contact. It is denoted by the symbol μ (mu) and is defined as the ratio of the force of friction between two objects to the normal force that is pressing them together. In other words, it is a value that indicates how difficult it is to slide one object over another.
To resolve this issue, we need to use the equation of motion:
v = u + at
where v is the final velocity (1 m/s), u is the initial velocity (0 m/s), a is the acceleration, and t is the time (13 seconds).
We can rearrange this equation to solve for the acceleration:
a = (v - u) / t
a = (1 m/s - 0 m/s) / 13 s
a = 0.077 m/s²
Next, we can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass times its acceleration:
Fnet = ma
where Fnet is the net force acting on the object, m is its mass (0.20 kg), and a is the acceleration we just calculated.
We know that the only horizontal force acting on the puck is the applied force of 2.0 N, so we can use that as the net force:
Fnet = 2.0 N
Setting Fnet equal to ma and solving for the coefficient of kinetic friction μk:
Fnet = ma
μkmg = ma
μk = a/g
μk = (0.077 m/s²) / 9.81 m/s²
μk = 0.0079
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1. A 1.33-kg physics textbook is initially at rest on a steel table.
The textbook is then pushed with a constant force of 4.0 N.
Friction with a magnitude of 2.0 N is exerted on the moving
book by the surface of the table.
Determine the final velocity of the textbook after it has been
pushed 0.75 meter across the table.
The acceleration of the textbook is a = F/m = 2.0 N/1.33 kg = 1.5 m/s². and the final velocity of the textbook is v = 0 + (1.5 m/s²)(0.75 s) = 1.125 m/s.
What is acceleration?Acceleration is the rate at which an object changes its velocity over time. It is a vector quantity, meaning it has both magnitude and direction. An object's acceleration can be calculated by measuring the change in its velocity over time.
The net force acting on the textbook is the difference between the pushing force of 4.0 N and the frictional force of 2.0 N, which is 2.0 N.
According to Newton's Second Law, this net force of 2.0 N will cause an acceleration of the textbook given by the equation F = ma, where m is the mass of the textbook (1.33 kg) and a is the acceleration.
Therefore, the acceleration of the textbook is a = F/m = 2.0 N/1.33 kg = 1.5 m/s².
The final velocity of the textbook can then be calculated using the equation v = v0 + at, where v0 is the initial velocity (0 m/s, since the textbook was initially at rest), a is the acceleration (1.5 m/s²), and t is the time the textbook was pushed (0.75 m).
Therefore, the final velocity of the textbook is v = 0 + (1.5 m/s²)(0.75 s) = 1.125 m/s.
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A car accelerates from a stop at a rate of 2 m/s² for 20 s, then con- tinues at a constant speed for 40 s. Graph the speed vs. time of the car. What is the car's speed at 10 s? What is its final speed?
The car's speed at 10 s would be 20 m/s, and the final speed would be 40 m/s.
Velocity-time graphDuring the first 20 s, the car accelerates at a rate of 2 m/s², so its speed increases linearly. We can calculate the final speed at 20 s as follows:
a = 2 m/s²
t = 20 s
v = a * t = 2 m/s² * 20 s = 40 m/s
From 20 s to 60 s, the car continues at a constant speed, so the graph is a horizontal line at 40 m/s.
Therefore, the car's speed at 10 s is:
t = 10 s
v = a * t = 2 m/s² * 10 s = 20 m/s
And its final speed is 40 m/s
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Question 3
A 2,209 kg car travelling at 31 m/s hits the back of a 1,278 kg car that is travelling 5 m/s slower than the other car. The 2,209 kg
car goes 10 m/s slower than it was going before it hit. Both cars are travelling in the same direction. How fast will the car that got
hit from behind be going after the collision?
The car that got hit from behind will be going 48.05 m/s (or about 172.98 km/h) after the collision.
What is Collision?
In physics, a collision is an event where two or more objects come together and interact with each other, causing a transfer of energy and momentum between them. Collisions can be classified into two types: elastic and inelastic.
To solve this problem, we can use the conservation of momentum principle, which states that the total momentum of an isolated system remains constant before and after a collision. We can set up an equation for the conservation of momentum before and after the collision:
Before the collision:
(2209 kg)(31 m/s) + (1278 kg)(26 m/s) = total momentum
After the collision:
(2209 kg)(21 m/s) + (1278 kg)(vf) = total momentum
where vf is the final velocity of the 1,278 kg car after the collision.
Simplifying the equations, we get:
Before the collision: 76,079 kgm/s + 33,228 kgm/s = 109,307 kg*m/s
After the collision: 46,689 kgm/s + 1,278 kgvf = 47,967 kgm/s + 1,278 kgvf
Since the total momentum before and after the collision is the same, we can set the two equations equal to each other and solve for vf:
76,079 kgm/s + 33,228 kgm/s = 46,689 kgm/s + 1,278 kgvf
109,307 kgm/s = 47,967 kgm/s + 1,278 kg*vf
61,340 kgm/s = 1,278 kgvf
vf = 48.05 m/s
Therefore, the car that got hit from behind will be going 48.05 m/s (or about 172.98 km/h) after the collision.
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2. A person walks 20 m [N 20E], then
120 m [N 50W], then 150 m [W] and finally
30 m [S 75E]. Find the person final
displacement using trigonometric methods
The person's final displacement is approximately 227 meters at an angle of 17.9 degrees North of West.
The first displacement of 20 m [N 20° E] can be broken down as follows,
The x-component is 20 cos(20°) = 18.8 m to the East
The y-component is 20 sin(20°) = 6.8 m to the North
The second displacement of 120 m [N 50° W],
The x-component is 120 cos(50°) = -91.9 m to the West
The y-component is 120 sin(50°) = 92.2 m to the North
The third displacement of 150 m [W] has no y-component, and the x-component is simply -150 m to the West.
The fourth displacement of 30 m [S 75° E],
The x-component is 30 cos(75°) = 7.4 m to the East
The y-component is 30 sin(75°) = -28.6 m to the South
The total x-component is 18.8 m - 91.9 m - 150 m + 7.4 m = -215.7 m to the West. The total y-component is 6.8 m + 92.2 m + 0 m - 28.6 m = 70.4 m to the North
To find the magnitude of the resultant vector,
|d| = sqrt((-215.7 m)^2 + (70.4 m)^2) ≈ 227 m
To find the direction of the resultant vector,
θ = tan^-1(70.4 m / 215.7 m) ≈ 17.9° N of W
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