The total distance traveled in one period is (e) 0.20 m
The total distance the particle travels in one period of simple harmonic motion (SHM) is twice the amplitude. Therefore, if a particle undergoes SHM with amplitude 0.10 m, the total distance it travels in one period is 2 x 0.10 m = 0.20 m.Another way to think about it is that the particle starts at its equilibrium position, moves to the maximum displacement of 0.10 m in one direction, moves back through the equilibrium position to a maximum displacement of 0.10 m in the opposite direction, and then returns to the equilibrium position. So it travels a distance of 0.10 m in each direction, for a total distance of 0.20 m.Learn more about the SHM:
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a submarine hovers at yards below sea level. if it ascends yards and then descends yards, what is the submarine’s new position, in yards, with respect to sea level?
Given, The submarine hovers at `y` yards below sea level and it ascends `a` yards and then descends `d` yards. We need to find the new position of the submarine from the sea level. Therefore, the submarine’s new position, in yards, with respect to sea level is `y - a - d` yards.
So, the submarine was at a depth of `y` yards and it ascends to `a` yards. Therefore, the submarine is now at a depth of `y - a` yards from the sea level.
Now, the submarine again descends `d` yards from the new position.
So, the new position of the submarine from sea level
`= (y - a) - d` yards`= y - a - d` yards,
which is the required answer to the given problem. Therefore, the submarine’s new position, in yards, with respect to sea level is `y - a - d` yards.
The given problem states that a submarine is hovering at `y` yards below sea level. If it ascends `a` yards and then descends `d` yards, we need to find the submarine’s new position with respect to the sea level. We know that the distance between a submarine and the sea level is measured in yards.
Let's find the answer step by step.
Based on the problem, the submarine was initially at a depth of `y` yards and it ascends to `a` yards.
Therefore, the submarine is now at a depth of `y - a` yards from the sea level.
That means the submarine is currently `y - a` yards deep.
Now, the submarine descends again by `d` yards.
Therefore, the new position of the submarine from sea level `= (y - a) - d` yards`= y - a - d` yards.
This is the required answer to the given problem.
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when a solid melts into a liquid, do the chemical bonds between molecules expand or break? I thought that the bonds would break when the solid is broken(like when glass breaks). When a solid melts into a liquid, wouldn't the bond between them just grow weaker and stretch out a bit more?
Pls detail.
Answer: The added heat or thermal energy leads to the molecular bonds breaking which leads to a change of state of solid to a liquid, then eventually gas. Solids melt when they absorb enough thermal energy.
Which situation has the greatest magnitude of net force along incline? (By magnitude we mean |F|.)
1. the net force is always the same for an incline, f = mgsinθ
2. when the cart is going uphill
3. the net force is always the same for an incline, f = mgcosθ
4. when the cart is going downhill
The situation with the greatest magnitude of net force along the incline is when the cart is going uphill (Option 2).
When an object is on an inclined plane, the net force acting along the incline can be determined by resolving the force of gravity into components parallel and perpendicular to the incline.
For Option 2 (cart going uphill), the force of gravity component acting parallel to the incline helps to counteract the force required to move the cart upwards.
In this case, the net force is the sum of the force of gravity component parallel to the incline and the applied force (if any) in the same direction.
The magnitude of the net force (|F|) in this case can be calculated using the formula:
|F| = |mgsinθ + F_applied|
where m is the mass of the cart, g is the acceleration due to gravity, θ is the angle of the incline, and F_applied is any additional applied force.
In this situation, the force of gravity component parallel to the incline is working against the motion of the cart, resulting in a greater net force compared to the other options.
The cart going uphill (Option 2) experiences the greatest magnitude of net force along the incline.
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A measuring station detects an earthquake has occurred. The P-waves arrive at 15 km/s and the S-waves arrive at 10 km/s, with a time delay between them of 10 seconds. How far is the epicenter of the earthquake from the measuring station?
The epicenter of the earthquake is approximately 300 kilometers away from the measuring station.
How to solve for the distance
To determine the distance to the epicenter of the earthquake, we can use the formula:
Distance = Velocity × Time
First, let's calculate the time it took for the P-waves to reach the measuring station:
Time (P-wave) = Distance / Velocity (P-wave) = ? / 15 km/s
Next, we'll calculate the time it took for the S-waves to reach the measuring station:
Time (S-wave) = Distance / Velocity (S-wave) = ? / 10 km/s
Given that there is a time delay of 10 seconds between the arrival of the P-waves and S-waves, we can set up the following equation:
Time (S-wave) - Time (P-wave) = 10 seconds
Now, let's substitute the formulas for time and solve for distance:
(Distance / 10 km/s) - (Distance / 15 km/s) = 10 seconds
To simplify the equation, we can find the common denominator, which is 30 km/s:
[(3 * Distance) - (2 * Distance)] / (30 km/s) = 10 seconds
Distance / (30 km/s) = 10 seconds
Multiplying both sides of the equation by 30 km/s:
Distance = 10 seconds * 30 km/s
Distance = 300 kilometers
Therefore, the epicenter of the earthquake is approximately 300 kilometers away from the measuring station.
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Which refers to the pressure found in a water distribution system during normal consumption demands?
The term that refers to the pressure found in a water distribution system during normal consumption demands is called the normal operating pressure. The normal operating pressure of a water distribution system refers to the range of pressures that occur in the system during average consumption demands.
It is measured in pounds per square inch (psi).The normal operating pressure for a water distribution system is usually between 30 and 80 psi. The exact pressure range will depend on the specific system and the location. The normal operating pressure is important to maintain in order to ensure that the system operates effectively and efficiently. If the pressure is too high, it can cause damage to the pipes and fixtures, and if it is too low, it can result in poor water flow and inadequate supply to consumers. Therefore, it is important to regularly monitor and maintain the normal operating pressure of a water distribution system to ensure the system functions properly.
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PLS ANWSER FAST WILL GIVE BRAINL!!!!
Explain to me in YOUR own words, how convection currents create a cycle? In other words, tell me how heat and cooling create a cycle.
Answer:
because it get the energy from the heat and the cold mixed together
Explanation:
A block of wood is floating in a pool of water. One third of the block is above the surface of the water. Discuss the buoyant force that is acting on the log. Is the buoyant force greater than, less than or equal to the weight of the block? Explain your answer. Is the volume of water displaces by the block greater than, less than or equal to the volume of the block? Explain your answer after watching the following video.
The required,
The buoyant force is equivalent to the weight of the block.The volume of water displaced by the block is equal to the volume of the submerged portion of the block, which is two-thirds of the total volume of the block.The buoyant force functioning on the block of wood is equal to the weight of the water displaced by the block. According to Archimedes' principle, when an object is submerged in a fluid, it experiences an upward buoyant force equal to the weight of the fluid it displaces.
In this case, one-third of the block is above the surface of the water, which signifies two-thirds of the block is submerged. The buoyant force is equal to the weight of the water displaced by the submerged portion of the block.
Since the block is floating, it is in equilibrium, which means the buoyant force is equal to the weight of the block. If the buoyant force were more significant than the weight of the block, the block would rise to the surface and float higher. If the buoyant force were less than the weight of the block, the block would sink.
Therefore, the buoyant force is equal to the weight of the block. Both forces have the same magnitude but act in opposite directions. The weight of the block acts downward, while the buoyant force acts upward.
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A guitar string 61 cm long vibrates with a standing wave that has three antinodes.
1-Which harmonic is this? Express your answer using one significant figure.
2-What is the wavelength of this wave? Express your answer using two significant figures.
1. The length of the guitar string can be related to the wavelength by the following equation: L = (nλ) / 2, where n is the harmonic number, and λ is the wavelength of the wave.
According to the problem, the length of the guitar string is 61 cm, and the wave has three antinodes.
We can therefore substitute these values into the equation and solve for n:61 cm = (3λ) / 2λ = (2 × 61 cm) / 3λ = 40.7 cm (rounded to one significant figure)
Therefore, the wavelength is 40.7 cm (rounded to two significant figures).
2. The wavelength: We can now use the above value of λ and the formula
v = fλ to calculate the frequency of the wave.
However, the velocity of a wave in a string is given by the formula
v = √(T/μ), where T is the tension in the string and μ is its linear mass density (mass per unit length).
These values are not given in the problem, so we cannot solve for frequency.
Instead, we can use another equation that relates the wavelength to the length of a string:λ = 2L / n,
where L is the length of the string and n is the harmonic number. Substituting the given values: L = 61 cm, n = 3λ = 40.7 cm (from part a).
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A 1.80 kg monkey wrench is pivoted 0.250 m from its center of mass and allowed to swing as a physical pendulum. The period for small-angle oscillations is 0.940 s. Correct Correct answer is shown. Your answer 0.099 kg⋅m 2
was either rounded differently or used a different number of significant figures than required for this part. Part B If the wrench is initially displaced 0.400rad from its equilibrium position, what is the angular speed of the wrench as it passes through the equilibrium position? Express your answer in radians per second.
The angular speed of the wrench as it passes through the equilibrium position is approximately 3.17 radians per second.
To calculate the angular speed of the wrench as it passes through the equilibrium position, we can use the formula for the period of a physical pendulum, which is T = 2π√(I/mgd), where T is the period, I is the moment of inertia, m is the mass, g is the acceleration due to gravity, and d is the distance from the pivot to the center of mass.
Given:
Mass of the wrench (m): 1.80 kg
Period of small-angle oscillations (T): 0.940 s
Displacement from equilibrium (θ): 0.400 rad
First, we need to find the moment of inertia (I) of the wrench. The correct answer provided is 0.099 kg·m^2.
Now, we can use the formula T = 2π√(I/mgd) to solve for the angular speed (ω).
Rearranging the formula:
T = 2π√(I/mgd)
√(I/mgd) = T / (2π)
I/mgd = (T / (2π))^2
ω = √(gd/I)
Substituting the given values:
g = 9.8 m/s^2 (acceleration due to gravity)
d = 0.250 m (distance from pivot to center of mass)
I = 0.099 kg·m^2 (moment of inertia)
ω = √(9.8 * 0.250 / 0.099) ≈ 3.17 rad/s
Therefore, the angular speed of the wrench as it passes through the equilibrium position is approximately 3.17 radians per second.
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a nucleus emits a gamma ray of energy 0.511 mev from a state that has a lifetime of 1.0 ns. (a) What is the uncertainty in the energy of the gamma ray? (b) The best gamma-ray detectors can measure gamma-ray energies to a precision of no better than a few eV. Will this uncertainty be directly measurable?
The uncertainty of energy of the gamma ray is 3.34 x 10⁻²⁵MeV.
Energy of the gamma ray emitted by the nucleus, E = 0.511 MeV
Lifetime of the nucleus, Δt = 1 ns = 10⁻⁹s
a) The expression for the uncertainty of energy of the gamma ray is given by,
ΔE = h/(2Δt)
ΔE = 6.67 x 10⁻³⁴/(2 x 10⁻⁹)
ΔE = 3.34 x 10⁻²⁵MeV
b) Detection of gamma rays is carried out photon by photon. By studying the impact they have on materials, gamma rays can be found.
Gamma rays can either drive an electron to a higher energy level (photoelectric ionization) or crash with it and scatter off of it like a pool ball.
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Birds sitting on a single power line don't get shocked. But if they were to place one foot on each of two lines, ___________ would flow between them and they would receive a terrible shock
When birds sit on a single power line, they are not grounded, which means they do not provide a path for current to flow from the high voltage power line through their body to the ground. But if they were to place one foot on each of two lines, they would complete a circuit, and current would flow through their body, and they would receive a terrible shock.
Electricity flows through a circuit when there is a path for current to flow from a power source to the ground. The power lines carry high voltage electricity, which can be dangerous to living organisms, including birds. However, birds sitting on a single power line don't get shocked because they are not providing a path for current to flow from the power line through their body to the ground.The reason birds are not grounded when they sit on a single power line is that they have only one point of contact with the power line.
Therefore, the current cannot flow through their body and reach the ground. In other words, they are not part of the circuit.In conclusion, birds sitting on a single power line do not get shocked because they are not grounded and do not provide a path for current to flow through their body to the ground. However, if they were to place one foot on each of two lines, they would complete a circuit, and current would flow through their body, resulting in a terrible shock.
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Three resistors R1= 3 Ω, R2= 5 Ω, and R3 = 2 Ω are connected with 150 V power supply. What is the voltage across R2 ?
a.75 V
b.15 V
c.10 V
d.32 V
If the resistors are in series, it's 75 volts.
If they're in parallel, it's 150 volts.
You never told us series or parallel.
In 10 billion years, the peak of the spectrum emitted from the cosmic microwave background radiation (CMB) will ____. A) remain the same. B) shift to shorter wavelengths. C) shift to longer wavelengths. D) continue to redshift until it reaches infinitely long wavelengths
In 10 billion years, the peak of the spectrum emitted from the cosmic microwave background radiation (CMB) will shift to longer wavelengths.
What is cosmic microwave background radiation (CMB)?
The cosmic microwave background radiation (CMB) refers to a pervasive form of electromagnetic radiation that fills the entire universe. It is considered to be the afterglow of the Big Bang, the event that marked the beginning of our universe approximately 13.8 billion years ago.
This phenomenon is known as cosmological redshift. As the universe continues to expand, the wavelengths of the CMB radiation will stretch, causing the peak of the spectrum to shift toward longer wavelengths. This is consistent with the observed expansion of the universe and the redshift of light from distant galaxies. Therefore, the correct answer to In 10 billion years, the peak of the spectrum emitted from the cosmic microwave background radiation (CMB) is option C i.e. shift to longer wavelengths.
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Light of wavelength 503 nm in vacuum passes through a piece of fused quartz of index of refraction n = 1.458.
(a) Find the speed of light in fused quartz.
(b) What is the wavelength of this light in fused quartz?
(c) What is the frequency of the light in fused quartz?
(a) The wavelength of light in vacuum is 503 nm(b) The wavelength of light in fused quartz is 345.24 nm(c) The frequency of light in fused quartz is 8.702 × 10^14 Hz.
The speed of light in vacuum is a fundamental constant equal to 299,792,458 meters per second. The wavelength of light is the distance between two consecutive peaks or troughs in the wave pattern. The frequency of light is the number of cycles of the wave that pass a point in a second. The refractive index of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in that medium. The refractive index of fused quartz is 1.458.The wavelength of light in fused quartz is given by the formulaλquartz = λvacuum/ nquartz Substituting the values,λquartz = 503 nm / 1.458= 345.24 nm The frequency of light remains the same in vacuum and in the medium. Therefore, the frequency of light in fused quartz is the same as in vacuum, which is given by the formula, frequency = speed of light / wavelength Substituting the values, frequency = 299,792,458 / 345.24 × 10^-9= 8.702 × 10^14 Hz.
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True or False: Opaque materials allow no visible light through them.
PLEASE ANSWER QUICKLY.
Answer:
This is true.
Which of the following correctly describes the relationship between current and voltage as the voltage of a battery increases. Remember that Ohm's Law
states: I = x
As voltage increases, current decreases because current and voltage are inversely proportional.
o As voltage increases, current decreases because current and voltage are directly proportional
As voltage increases, current increases because current and voltage are directly proportional
Answer:_COC1\/2+_H\/2O>_HC1+CO\/2
Explanation:
Need help asap
Question:
(need answers now I have time)
A freely-falling object is accelerating.
A. True
B. False
Answer:
the answer is true.
Explanation:
hope it will help you
An electron and a proton are fixed at a separation distance of 911 nm. Find the magnitude and direction of the electric field at their midpoint Magnitude: Number 1.388 x 104 N/ C Direction: O Toward the electrorn O Toward the proton Perpendicular to the line of the particles O Cannot be determined
The direction of the electric field at the midpoint is toward the proton.
Given that the separation distance between the electron and the proton is 911 nm (9.11 x 10^-7 m) and the charges of an electron and a proton are equal in magnitude but opposite in sign, we can consider the electric field created by both charges separately.Using Coulomb's law, the magnitude of the electric field created by each charge at the midpoint is calculated as E = k * (|q| / r^2), where k is the electrostatic constant (8.99 x 10^9 N m^2/C^2), |q| is the magnitude of the charge, and r is the separation distance.For each charge, |q| = 1.6 x 10^-19 C, and the separation distance is half of the initial distance, i.e., 0.5 * 9.11 x 10^-7 m = 4.555 x 10^-7 m.Calculating the electric field magnitude for each charge and adding them together, we have E = k * (|q| / r^2) + k * (|q| / r^2) = 2 * k * (|q| / r^2) ≈ 1.388 x 10^4 N/C. Thus, the magnitude of the electric field at the midpoint is approximately 1.388 x 10^4 N/C. Now, to determine the direction of the electric field at the midpoint, we consider the forces experienced by a positive test charge placed at that point. Since opposite charges attract each other, the electric field points toward the positive charge. In this case, the proton is positively charged, so the electric field is directed toward the proton. Therefore, the direction of the electric field at the midpoint is toward the proton.
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to take up and store energy without reflecting or transmitting that energy
Answer:
Absorbed
Explanation:
Hope this helped!!!
During a football workout two linemen are pushing on the coach and the sled. The combined mass of the sled and the coach is 300 kg the coefficient of friction of between the sled and the grass is. 800. The sled accelerates at a rate of. 580 m/s/s. Determine the force applied to the sled by the lineman
A football workout involves two linemen who are pushing on the coach and the sled. The sled and the coach's combined mass is 300 kg, while the coefficient of friction between the sled and the grass is .800. The sled accelerates at a rate of .580 m/s/s. We need to determine the force applied to the sled by the linemen.
The total force acting on the sled is:Force = Mass × AccelerationF = 300 kg × .580 m/s/s = 174 NSince the sled and the grass's coefficient of friction are known, we can determine the force applied to the sled by the linemen using the following equation:
Frictional Force = Coefficient of Friction × Normal Force
The normal force is equal to the weight of the sled and the coach. Thus,Normal Force = Mass × GravityN = 300 kg × 9.81 m/s/s = 2943 NFrictional Force = .800 × 2943 N = 2354.4 NThe force applied to the sled by the linemen is the difference between the total force acting on the sled and the frictional force.
Force Applied by Linemen = Total Force − Frictional ForceF = 174 N − 2354.4 N = −2180.4 NThe linemen are exerting a force of −2180.4 N on the sled in the opposite direction to the sled's movement. This is because the frictional force is greater than the total force acting on the sled.
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You push on a 30 kg box with a force of 120 N. What is the acceleration of the box2
Answer:
6
Explanation:
120/30=6
The mean orbital radius of the earth around the sun 1.5 × 108 km. Calculate the
mass of the sun if G = 6.67 × 10-11 Nm2/kg -2?
Answer:
M = 1.994 × 10^(30) kg
Explanation:
We are given;
Orbital radius; r = 1.5 × 10^(8) km = 1.5 × 10^(11) m
Gravitational constant; G = 6.67 × 10^(-11) N.m²/kg²
If the orbit is circular, the it means the gravitational force is equal to the centripetal force.
Thus; F_g = F_c
GMm/r² = mv²/r
Simplifying gives;
GM/r = v²
M = v²r/G
Now, v is the speed of the earth around the sun and from online sources it has a value of around 29.78 km/s = 29780 m/s
Thus;
M = (29780^(2) × 1.5 × 10^(11))/6.67 × 10^(-11)
M = 1.994 × 10^(30) kg
A box weighing 18 newtons requires a force of 6 newtons to drag it. the coefficient of kinetic friction is
When a box weighs 18 newtons, a force of 6 newtons is required to drag it, the coefficient of kinetic friction is 0.333. Friction is the force that opposes motion when an object is pushed along a surface or in contact with another object.
It always acts in the opposite direction to the direction of movement. There are two types of friction, kinetic friction and static friction. The friction acting on an object that is already moving is kinetic friction. Friction acting on an object that is at rest is called static friction. The coefficient of kinetic friction is the ratio of the friction force between two objects and the force pressing them together. It's a dimensionless scalar quantity. To be precise, the formula for the coefficient of kinetic friction is given as: Coefficient of Kinetic Friction = Frictional Force / Normal Force. Where, Normal Force = The perpendicular force exerted by a surface on an object in contact with it. The force required to drag the box is 6N, so the kinetic frictional force on the box is 6N. The formula for coefficient of kinetic friction is :Coefficient of Kinetic Friction = Frictional Force / Normal Force. If the force required to drag the box is 6N, then the normal force acting on the box is 18N. So, the coefficient of kinetic friction will be: Coefficient of Kinetic Friction = 6N / 18N = 0.333
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compute the kinetic energy of a proton (mass 1.67×10−27kg ) using both the nonrelativistic and relativistic expressions for speed of 9.00×107 m/s . enter your answers in joules separated by a comma.
The kinetic energy of the proton is 7.515 × 10⁻¹¹ J (nonrelativistic) and 2.144 × 10⁻¹¹ J (relativistic).
To compute the kinetic energy of a proton using both the nonrelativistic and relativistic expressions, we can use the following formulas:
1. Nonrelativistic expression:
The kinetic energy (K) of a particle is given by the formula:
K = (1/2) * m * v²
where m is the mass of the proton and v is its velocity.
Substituting the values into the formula:
K = (1/2) * (1.67 × 10⁻²⁷ kg) * (9.00 × 10⁷ m/s)²
Calculating the kinetic energy using the above formula, we get:
K = 7.515 × 10⁻¹¹ J
2. Relativistic expression:
The relativistic expression for kinetic energy takes into account the effects of special relativity and is given by the formula:
K = [(γ - 1) * m * c²]
where γ is the Lorentz factor, m is the mass of the proton, and c is the speed of light.
The Lorentz factor (γ) is given by:
γ = 1 / √(1 - (v²/c²))
Substituting the values into the formulas:
γ = 1 / √(1 - [(9.00 × 10⁷ m/s)² / (3.00 × 10⁸ m/s)²])
γ = 2.029
K = [(2.029 - 1) * (1.67 × 10⁻²⁷ kg) * (3.00 × 10⁸ m/s)²]
Calculating the kinetic energy using the above formula, we get:
K = 2.144 × 10⁻¹¹ J
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1. What is the law of reflection? Does it apply to rough surfaces as well as smooth? 2) Distinguish between regular and irregular reflection. Give an example of each. 3) Why is light refracted when it passes from one medium into another? 4) Is an image formed by reflection real or virtual? Where is it located?
1. The law of reflection states that the angle of incidence is equal to the angle of reflection when light rays incident to the surface. This law applies to both smooth and rough surfaces. However, on rough surfaces, the reflection is scattered in different directions, leading to diffuse reflection.
2. Regular reflection occurs when light waves are reflected from a smooth and flat surface, resulting in a clear and sharp image. An example of regular reflection is when you see your reflection in a mirror. Irregular reflection, also known as diffuse reflection, occurs when light waves are reflected from a rough or uneven surface, causing the light to scatter in various directions. An example of irregular reflection is when you see the reflection of light on a piece of paper.
3. Light is refracted when it passes from one medium into another due to the change in its speed. The speed of light changes as it enters a medium with a different optical density, which causes the light waves to bend or change direction. This bending of light is understood as refraction.
4. An image created by reflection can be either real or virtual. A real image is formed when light rays actually converge at a specific point, allowing the image to be projected onto a screen. Real images can be captured and seen by placing a screen at the location where the light converges. On the other hand, a virtual image is formed when light rays appear to diverge from a specific point, giving the illusion of an image, but it cannot be projected onto a screen. Virtual images are formed when light rays appear to come from a certain location, such as in a mirror.
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calculate the mass of each of a solid with a volume of 1.68ft^3 and a density of 9.2g/ml
The mass of the solid with a volume of 1.68 ft and density of 9.2g/ml is 14.85 kilograms.
To calculate the mass of the solid, we need to use the formula: Mass = Density × Volume.
First, we need to convert the volume from cubic feet to milliliters, as the density is given in grams per milliliter.
1 cubic foot is equal to 28,316.8466 milliliters (ml). So, the volume of the solid is 1.68 ft^3 × 28,316.8466 ml/ft^3 = 47,594.768 ml.
Now, we can calculate the mass by multiplying the density (9.2 g/ml) by the volume (47,594.768 ml).
Mass = 9.2 g/ml × 47,594.768 ml = 437,186.38 grams.
Finally, we convert grams to kilograms by dividing by 1,000, resulting in a mass of approximately 437.19 kilograms or rounded to 14.85 kilograms.
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A 3. 00-kilogram mass is thrown vertically upward
with an initial speed of 9. 80 meters per second.
What is the maximum height this object will
reach? [Neglect friction. ]
(1) 1. 00 m (3) 9. 80 m
(2) 4. 90 m (4) 19. 6 m
A 3.00-kilogram mass is thrown vertically upward with an initial speed of 9.80 meters per second. [Neglect friction.]When an object is thrown vertically upward, the initial velocity is positive, and the acceleration due to gravity is negative, directed downward.
We can use the following formula to calculate the maximum height, also known as the maximum displacement, reached by the object:
[tex]v_f^2 = v_i^2 + 2ad[/tex]
where v_f is the final velocity, [tex]v_i[/tex] is the initial velocity, a is the acceleration, and d is the displacement. At the maximum height, the final velocity is zero, so we can simplify the equation to:
[tex]d = (v_f^2 - v_i^2) / (2a)[/tex]
Substituting the given values:
[tex]d = (0 - 9.80^2) / (2 x -9.81)d = 4.90 m[/tex]
Therefore, the maximum height reached by the object is 4.90 m.
Hence, the correct option is (2) 4.90 m.Note: In the above calculations, a negative value is used for the acceleration due to gravity, because it is acting downward, while the upward direction is taken as positive.
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a very long, thin wire has a uniform linear charge density of 91 µc/m. what is the electric field (in n/c) at a distance 8.0 cm from the wire? (enter the magnitude.)
Therefore, the electric field (magnitude only) at a distance of 8.0 cm from the wire is approximately 3.24 x 10^4 N/C.
The electric field of a long, thin wire can be determined by Coulomb's law. Coulomb's Law states that the electric force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
For a long, thin wire, the electric field is given by;
E = λ/2πε₀r
Where;
λ = linear charge density = 91 µC/m,
ε₀ = permittivity of free space = 8.85 x 10^-12 C^2/Nm^2
r = distance from the wire = 8.0 cm = 0.08 m.
Substitute the given values into the formula to find the electric field;
E = (91 x 10^-6)/(2 x π x 8.85 x 10^-12 x 0.08)
E≈ 32433.8 N/C
E≈ 3.24 x 10^4 N/C.
Electric field refers to the force per unit charge that one object exerts on another object due to the electric charge present in the objects. It is a vector quantity and is measured in newtons per coulomb (N/C).
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A 150 kg. yak has an average power output of 120 W. The yak can climb a mountain 1.2 km high in (a) 25 min (b) 4.1 h (c) 13.3 h (d) 14.7 h.
I have worked this problem over and over and keep coming up with 14.7 h; however, the textbook tells me the answer is 4.1?
A 150 kg yak has an average power output of 120 W, then the yak can climb a mountain 1.2 km high in 14.7 h. So, option d is correct.
Power (P) is defined as the rate at which work is done, given by the formula: P = W/t, where W is the work done and t is the time taken. In this case, the power output of the yak is given as 120 W.
The work done (W) is calculated by multiplying the force applied by the distance traveled. Since the distance traveled is the height of the mountain (1.2 km), we need to find the force exerted by the yak to climb the mountain.
Force (F) is given by the formula: F = mg, where m is the mass of the yak (150 kg) and g is the acceleration due to gravity (9.8 m/s²).
Substituting the values, we find F = (150 kg)(9.8 m/s²) = 1470 N.
Now, we can calculate the work done:
W = F × d = (1470 N)(1.2 km) = 1764 kJ.
To find the time (t), we rearrange the power formula:
t = W/P = (1764 kJ)/(120 W) = 14.7 hours.
Therefore, the correct answer is (d) 14.7 hours.
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A 70 cm diameter wheel accelerates uniformly about its center from 130 rpm to 280 rpm in 4 s. Determine (a) its angular acceleration, and (b) the radial and tangential components of the linear acceleration of a point on the edge of the wheel 2 s after it has started accelerating. Show all work and formulas for best rating.
A 70 cm diameter wheel accelerates uniformly about its center from 130 rpm to 280 rpm in 4 s, the angular acceleration of the wheel is 3.93 rad/s², and the radial component of linear acceleration is approximately 1.375 m/s², and the tangential component is approximately 165.86 m/s².
(a) The angular acceleration of the wheel can be determined using the formula:
Angular acceleration (α) = (Final angular velocity - Initial angular velocity) / Time
Given:
Initial angular velocity (ω₁) = 130 rpm
Final angular velocity (ω₂) = 280 rpm
Time (t) = 4 s
First, we need to convert the angular velocities from rpm to radians per second (rad/s):
ω₁ = 130 rpm * (2π rad/1 min) * (1 min/60 s) = 13.61 rad/s
ω₂ = 280 rpm * (2π rad/1 min) * (1 min/60 s) = 29.33 rad/s
Substituting the values into the formula for angular acceleration:
α = (29.33 rad/s - 13.61 rad/s) / 4 s = 3.93 rad/s²
Therefore, the angular acceleration of the wheel is 3.93 rad/s².
(b) To determine the radial and tangential components of the linear acceleration of a point on the edge of the wheel after 2 s, we can use the following formulas:
Radial acceleration (ar) = r * α
Tangential acceleration (at) = r * ω²
Given:
Radius of the wheel (r) = 70 cm / 2 = 35 cm = 0.35 m
Angular acceleration (α) = 3.93 rad/s²
Angular velocity (ω) at t = 2 s can be found using the formula:
ω = ω₁ + α * t
Substituting the values:
ω = 13.61 rad/s + 3.93 rad/s² * 2 s = 21.47 rad/s
Now we can calculate the radial and tangential components of linear acceleration:
ar = r * α = 0.35 m * 3.93 rad/s² ≈ 1.375 m/s²
at = r * ω² = 0.35 m * (21.47 rad/s)² ≈ 165.86 m/s²
Therefore, 2 seconds after starting acceleration, the radial component of the linear acceleration is approximately 1.375 m/s², and the tangential component is approximately 165.86 m/s².
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