If her center of mass accelerates only horizontally, the magnitude of the minimum static friction needed so she won't slip is 52 lb.
To find the speed increase of the player, we can utilize the equation a = Δv/Δt, where Δv = 15 ft/s (last speed) and Δt = 2 s (time taken to arrive at that speed). In this way, a = 7.5 ft/s².
To work out the base static erosion expected to forestall slipping, we want to consider the powers following up on the player. The player's weight acts descending (W = mg = 5.59 slug x 32.2 ft/s² = 180 lb), and the power of static erosion acts on a level plane the other way to the course of movement. The greatest power of static grinding is equivalent to the coefficient of static contact (μs) increased by the ordinary power (N = W).
Since the player isn't slipping, the power of static erosion should be equivalent to or more noteworthy than the power expected to evenly speed up her. Accordingly, we can set the power of static contact equivalent to ma*ma (mass x speed increase), which gives:
μsN = ma*ma
μs(W) = ma*ma
μs(180 lb) = (5.59 slug)(7.5 ft/s²)
μs = 0.29
In this manner, the extent of the base static grating required so she won't slip is 0.29 times the heaviness of the player, or around 52 lb.
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the one property of a main-sequence star that determines all its other properties is its: question 20 options: luminosity. temperature. mass spectral type.
The one property of a main-sequence star that determines all its other properties is its mass. Option c is correct.
The main-sequence is a sequence of stars in which they spend most of their lifetimes. The position of a star on the main-sequence depends on its mass, which determines its luminosity, temperature, spectral type, and other properties. The more massive a star is, the hotter and brighter it is, and the shorter its lifetime.
Conversely, less massive stars are cooler and dimmer, and live much longer. The mass of a star also determines the reactions that occur in its core and the elements that are produced, shaping its evolution and eventual fate. Therefore, the mass of a main-sequence star is a fundamental property that determines most of its other characteristics. Hence option c is correct.
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the planet mars is, on average, about 228 million km from the sun. how long does it take light from the sun to reach mars? (recall that the speed of light is about 300,000 km/s.) group of answer choices about 8.4 minutes about 12.7 minutes about 1.52 light seconds about 1.52 hours
When the planet Mars is, on average, about 228 million km from the Sun, the correct option is about 1.52 hours.
The time it takes light from the Sun to reach Mars when the planet Mars is, on average, about 228 million km from the Sun is about 12.7 minutes.
The given question can be solved using the formula; Time = Distance / Speed of light
Given that Distance of Mars from the Sun = is 228 million km
The speed of light = 300,000 km/sNow, let's plug in the values in the formula.
Time = Distance / Speed of light = 228 × 106 km / 300,000 km/s = 760 secondsTherefore, the time taken for light to reach Mars from the Sun is 760 seconds.1 hour is equal to 3600 seconds.
Therefore, the time taken for light to reach Mars from the Sun is about 760 / 3600 hours = 0.21 hours or about 1.52 hours.
Hence, the correct option is about 1.52 hours.
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in the circuit on the right the capacitor is initially uncharged. describe what is observed when the switch is closed
When the switch is closed, the capacitor will start to charge. The bulb will light up and its brightness will gradually decrease as the capacitor charges. If the capacitor was twice as large, it would take longer for it to charge and the bulb would stay bright for a longer period of time.
If the bulb had half as much resistance, it would allow more current to flow through it and the bulb would be brighter. The capacitor would also charge faster. When the switch in the circuit is closed, the capacitor will start charging up from zero voltage. Once the capacitor is fully charged, the current in the circuit stops, and the voltage across the capacitor is equal to the voltage of the battery.
When the switch is closed, the capacitor starts charging up from zero voltage, and the voltage across the capacitor increases while the current flowing through the circuit decreases. The resistor in the circuit limits the flow of current, causing the charging process to take some time.
During this charging process, the voltage across the capacitor increases, while the current flowing through the resistor and the capacitor decreases. As time goes on, the voltage across the capacitor approaches the voltage of the battery in the circuit, which is 12 volts in this case. Once the capacitor is fully charged, the current flowing through the circuit stops since no more charge can be stored in the capacitor.
During the charging process, the voltage across the resistor decreases, as the voltage across the capacitor increases. Once the capacitor is fully charged, the voltage across the resistor becomes zero, and the voltage across the capacitor is equal to the voltage of the battery.
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Complete question:
In the circuit on the right, the capacitor is initially uncharged. a. Describe what is observed when the switch is closed. b. How would your observations be changed capacitor were twice as large? c. How would your observations be changed if the bulb had half as much resistance?
two masses, m1 and m2 are separated by a distance d. What changes in variables will result in no change in the gravitational force between two masses ?
a)m1 is doubled and d is doubled
b)m2 is tripled and d is quadrupled c)both m1 is tripled and d is tripled d)None of them I need the answer urgently
The changes in the variables that will result in no change in the gravitational force between m1 and m2 is if m1 is doubled and d is doubled. Option A.
Gravitational forceThe gravitational force between two masses depends on the masses and the distance between them, according to the equation F = G(m1m2)/d^2, where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses, and d is the distance between them.
To have no change in the gravitational force between two masses, we need to keep the value of F constant. This can be achieved by changing the variables in the following way:
If m1 is doubled and d is doubled, then the value of F will be unchanged, as (2m1)(m2)/(2d)^2 = m1m2/d^2.If m2 is tripled and d is quadrupled, then the value of F will change, as (m1)(3m2)/(4d)^2 = 3m1m2/16d^2.If both m1 and d are tripled, then the value of F will change, as (3m1)(m2)/(3d)^2 = m1m2/3d^2.None of them will result in no change in the gravitational force between two masses.Therefore, the correct answer is a), where m1 is doubled and d is doubled.
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dylan has a weight of 600 n when he is standing on the surface of the earth. what would his weight (the gravitational force due to the earth) be if he tripled his distance from the center of the earth by flying in a spacecraft?
Dylan's weight (the gravitational force due to the earth) would decrease to 1/9 of its original value if he tripled his distance from the center of the earth by flying in a spacecraft.
The force of gravity, which determines the weight of an object, is proportional to the mass of the two objects involved (in this case, Dylan and the Earth) and inversely proportional to the square of the distance between them.
The formula for the gravitational force is F = GmM/r², where G is the gravitational constant, m and M are the masses of the two objects, and r is the distance between them.
If Dylan tripled his distance from the center of the Earth, his distance from the center would become three times larger than before. Using the inverse square law, the gravitational force he experiences would be nine times smaller (3²) than it was before.
Therefore, his weight would decrease to 1/9 of its original value, which is 600 N/9 = 66.67 N.
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you are working in a lab on your magnetic levitation experiment. you are standing directly in front of two magnets. another scientist near you has a malfunction with their equipment and beta radiation is emitted directly at you, passing through the magnets first. you are wearing a lab coat and goggles. are you safe? a. no, it was beta radiation and very dangerous. b. no, beta radiation is not deflected by magnets so it would have continued straight and hit you . c. yes, beta radiation is strongly deflected by magnets so it would have missed you. d. yes, beta radiation is easily blocked by the lab coat.
When you are working on a magnetic levitation experiment in a lab and beta radiation is emitted directly at you, passing through the magnets first, you are not safe because it was beta radiation and very dangerous.
The correct answer is option a.
Beta radation is a form of ionizing radiation that is made up of high-energy beta particles. Beta particles are high-energy, high-speed electrons or positrons that are emitted by certain radioactive isotopes. The emission of beta radiation from radioactive materials is referred to as beta decay.
Beta particles, unlike alpha particles, have a longer range and can penetrate through paper, skin, and other low-density materials but can be blocked by high-density materials.
Therefore, option a is correct.
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what is the magnitude of the radial component of acceleration of the point just as the centrifuge begins its decelerati
The magnitude of the radial component of acceleration is 1.25 m/s^2.
To solve this problem, we can use the formula for radial acceleration:
a_r = r * (Δω / Δt)
where a_r is the radial acceleration, r is the distance from the axis of rotation, Δω is the change in angular velocity, and Δt is the time interval over which the change occurs.
Plugging in the given values, we get:
a_r = 0.75 m * ((10 rad/s - 15 rad/s) / 3 s)
a_r = 0.75 m * (-1.67 rad/s^2)
a_r = -1.25 m/s^2
Note that the negative sign indicates that the acceleration is directed inward, towards the axis of rotation. So the magnitude of the radial component of acceleration is:
|a_r| = 1.25 m/s^2
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--The complete question is, What is the magnitude of the radial component of acceleration of a point located at a distance of 0.75 meters from the axis of rotation of a centrifuge, just as it begins its deceleration from an angular velocity of 15 rad/s to 10 rad/s over a time period of 3 seconds?--
you have a mass-spring-damper system as described by where the unit of newtons (n) is equivalent to if you were to displace the mass by 0.2 m from its equilibrium position, how would you characterize the response of the mass once you let go? (hint: you need to determine the value of the damping ratio).
By shooting past it's equilibrium position and then oscillate before settling to the equilibrium position is characterize the response of the mass once you let go. So, option A is corect choice.
The given mass-spring-damper system can be represented by the differential equation:
[tex]my'' + \mu_fy' + k\timesy = F_{ext}[/tex]
where y is the displacement of the mass from its equilibrium position, and [tex]F_{ext}[/tex] is any external force acting on the system.
To determine the response of the mass once you let go, we need to solve the above differential equation for the initial condition y(0) = 0.2 and y'(0) = 0 (assuming that the mass is released from rest).
To solve the differential equation, we first need to determine the damping ratio, which is given by:
damping ratio (ζ) = [tex]\mu_f / (2 \times \sqrt{(k \times m)})[/tex]
Substituting the given values, we get:
damping ratio (ζ) = [tex]7 / (2 \times \sqrt{160 \times 80})[/tex] = 0.1106
Since the damping ratio is less than 1, the system is underdamped.
Therefore, the response of the mass once you let go will oscillate with a decreasing amplitude until it reaches its equilibrium position.
The frequency of oscillation (ω) can be determined using the following formula:
ω = [tex]\sqrt{(k / m - \zeta ^2 times (\mu_f^2 / 4 \times m^2))}[/tex]
Substituting the given values, we get:
ω = [tex]\sqrt{160 / 80 - 0.1106^2 \times (7^2 / (4 \times 80^2)}[/tex]= 4.352 rad/s
The time period of oscillation (T) can be determined using the formula:
T = 2π / ω
Substituting the value of ω, we get:
T = 2π / 4.352 = 1.444 s
Therefore, once you let go, the mass will oscillate around its equilibrium position with a decreasing amplitude and a time period of 1.444 s until it eventually comes to rest.
Hence, the correct answer is option A: "It would shoot past its equilibrium position and then oscillate before settling to the equilibrium position."
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Question:-
you have a mass spring damper system as descried by
[tex]m\frac{d^2y}{dt^2}+\mu_f\frac{dy}{dt} +ky =F_{ext}[/tex]
where
m= 80 kg
[tex]\mu_f[/tex] = 7 N*s/m
k = 160 N/m
The unit of Newtons (N) is equivalent to kg m/s²
If you were to displace the mass by 0.2 m from its equilibrium position, how would you characterize the response of the mass once you let go? (Hint: you need to determine the value of the damping ratio).
a. It would shoot past it's equilibrium position and then oscillate before settling to the equilibrium position
b. It will approach the equilibrium position very slowly but not oscillate.
c. The answer depends on the value of the time constant
d. The system will be critically damped.
what must the charge (sign and magnitude) of a particle of mass 1.46 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 670 n/c ?
To remain stationary in a downward-directed electric field of magnitude 670 N/C, the electric force acting on the charged particle must be equal in magnitude and opposite in direction to the gravitational force acting on the particle.
The gravitational force acting on the particle can be calculated using the formula Fg = mg, where m is the mass of the particle and g is the acceleration due to gravity (9.81 m/s^2).
Fg = mg = (1.46 g)(9.81 m/s^2) = 14.33 × 10^-3 N
The electric force acting on the charged particle can be calculated using the formula Fe = qE, where q is the charge of the particle and E is the electric field strength. Fe = qE = (q)(670 N/C)
To remain stationary, the electric force and gravitational force must be equal and opposite, so we can set them equal to each other and solve for the charge q: Fe = Fg
[tex](q)(670 N/C) = 14.33 × 10^-3 N[/tex]
[tex]q = (14.33 × 10^-3 N) / (670 N/C)[/tex]
[tex]q = 2.14 × 10^-5 C[/tex]
Since the particle is stationary in a downward-directed electric field, the charge must be negative, so the charge of the particle is[tex]-2.14 × 10^-5 C.[/tex]
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how does water pressure 2m below the surface of a small pond compared with water pressure 2m below the surface
Water pressure 2m below the surface of a small pond is the same as water pressure 2m below the surface of any other body of water, which is determined by the weight of the water above it.
Water pressure is determined by the depth of water, the density of the water, and gravity. The pressure at a particular depth is determined by the weight of the water column above that point. Therefore, the water pressure 2m below the surface of a small pond is greater than the water pressure 2m below the surface of a larger body of water, such as a lake or ocean. This is because the weight of the water column above a given point is greater in a larger body of water.
Additionally, water pressure increases with depth regardless of the size of the body of water. This is due to the increasing weight of the water column above and the corresponding increase in the force of gravity acting on the water molecules.
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) A photo emissive surface has a threshold frequency of 4.02 x 10¹4Hz. Calculate the (i) threshold wavelength. (ii) work funtion. (iii) Kinetic energy of the emitted photoelectrons. (c = 3.0 x 108ms ¹; h=6.63 x 10-³4 Js).
what is the maximum kinetic energy in ev of electrons ejected from a certain metal by 442 nm em radiation, given the work function is 2.34 ev?
The maximum kinetic energy of the ejected electrons is approximately 1.16 eV. To find the maximum kinetic energy (KE) of ejected electrons, we can use the photoelectric effect equation:
KE = hf - W
where h is Planck's constant [tex](4.14 × 10^(-15) eV·s)[/tex], f is the frequency of the electromagnetic radiation, W is the work function (2.34 eV), and KE is the maximum kinetic energy.
First, we need to find the frequency (f) using the wavelength (λ) given:
c = λf
where c is the speed of light [tex](3 × 10^8 m/s)[/tex].
1. Convert the wavelength to meters: [tex]442 nm × 10^(-9) m/nm = 4.42 × 10^(-7) m[/tex]
2. Rearrange the equation to solve for [tex]f: f = c / λ[/tex]
3. Calculate [tex]f: f = (3 × 10^8 m/s) / (4.42 × 10^(-7) m) ≈ 6.79 × 10^14 Hz[/tex]
Now, we can find the maximum kinetic energy (KE) using the photoelectric effect equation:
4. Calculate KE: [tex]KE = (4.14 × 10^(-15) eV·s × 6.79 × 10^14 Hz) - 2.34 eV ≈ 1.16 eV[/tex]
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a ball is projected horizontally above level ground from the top of a vertical cliff. the ball strikes the level ground 1 km from the base of the cliff 4 seconds after it is fired. the heigh of the cliff is:
i lost my mind help
[tex]h=v_i_yt+\frac{1}{2}gt^2\\ \\h=\frac{1}{2}(9.81m/s^2)(4s)^2\\[/tex]
h = 78.48 m
Answer:
78.48 m.Explanation:
Since the ball is projected horizontally, its initial vertical velocity is zero. Therefore, we can use the following equation to find the height of the cliff:where h is the height of the cliff, g is the acceleration due to gravity, and t is the time it takes for the ball to hit the ground.
We know that the ball hits the ground 1 km from the base of the cliff, so the horizontal distance traveled by the ball is 1 km. We can use the following equation to find the time t it takes for the ball to travel this distance:where d is the distance traveled, v is the initial horizontal velocity, and t is the time it takes to travel the distance.
Since the ball is projected horizontally, its initial horizontal velocity is constant, and we can assume it is the same as the speed at which it hits the ground. Therefore, we can write:
t = d/vSubstituting in the given values, we get:
t= 1000 m / 250 m/st = 4 sTherefore, it takes the ball 4 seconds to hit the ground. Now we can use the first equation to find the height of the cliff:h = (1/2)gt^2h = (1/2)(9.81 m/s^2)(4 s)^2h = 78.48 mTherefore, the height of the cliff is 78.48 m.a 11.5 kg rifle fires a 0.050 kg bullet with a muzzle velocity of 600 m/s. what is the magnitude of the momentum of the rifle due to recoil after the gun has been fired?
The magnitude of the momentum of the rifle due to recoil after firing is 0.13 kg*m/s. The negative sign indicates that the rifle recoils in the opposite direction of the bullet's motion, by the law of conservation of momentum.
According to the law of conservation of momentum, the total momentum before and after an event must be the same. In this case, the momentum of the rifle and the bullet before firing is zero because they are at rest. After firing, the momentum of the bullet is given by the product of its mass and velocity, which is:
p_bullet = m_bullet * v_bullet
= 0.050 kg * 600 m/s
= 30 kg*m/s
The momentum of the rifle due to recoil can be found by multiplying the negative of the bullet's momentum by the mass ratio of the rifle to the bullet since the total momentum must be zero:
p_rifle = -m_bullet/m_rifle * p_bullet
= -(0.050 kg)/(11.5 kg) * 30 kgm/s
= -0.13 kgm/s
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pv what size tank would be needed to contain this same amount of helium at atmospheric pressure (1 atm )?
Values for P1, V1, and P2 to find the size of the tank needed to contain the helium at atmospheric pressure.
To determine the size of the tank needed to contain the helium at atmospheric pressure (1 atm), we need to use the ideal gas law formula:
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Moles of gas
R = Ideal gas constant (0.0821 L atm/mol K)
T = Temperature (in Kelvin)
From the student question, we know that the initial pressure (P1) and volume (V1) are given. We also know that the final pressure (P2) is 1 atm. We need to find the final volume (V2).
Step 1: Calculate the moles of helium gas (n) using the initial conditions.
P1V1 = nRT1
n = (P1V1) / (RT1)
Step 2: Calculate the final volume (V2) using the moles of helium gas (n) and atmospheric pressure (1 atm).
P2V2 = nRT2
V2 = (nRT2) / P2
Since we want the tank size for the same amount of helium at atmospheric pressure, we can assume the temperature remains constant (T1 = T2). Therefore, you can simply use the initial conditions to find the final volume:
V2 = (P1V1) / P2
Plug in the given values for P1, V1, and P2 to find the size of the tank needed to contain the helium at atmospheric pressure.
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the rotational velocity of a merry-go-round increases at a constant rate from 2.5 rad/s to 18.6 rad/s in a time of 12.5 s. what is the rotational acceleration of the merry-go-round?
The merry-go-rotational round's acceleration may be computed using the following formula: (final rotational velocity - starting rotational velocity) / time Equals rotational acceleration.
Using the provided values, we get: Rotational acceleration = (12.5 s x (18.6 rad/s - 2.5 rad/s 1.368 rad/s2 rotational acceleration As a result, the merry-go-rotational round's acceleration is 1.368 rad/s2. This suggests that the merry-go-rotational round's velocity is growing at a rate of 1.368 rad/s2. In other words, the rotational velocity of the merry-go-round increases by 1.368 radians per second for every second that passes. Understanding rotational acceleration is crucial in engineering and physics because it is used to describe the motion of spinning items like gears and wheels, which can affect their performance and stability.
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the magnetometer will measure the total magnetic field. which of the components should be zero when the dipole is oriented along the y-axis, based on what you learned in the previous lab about the direction of magnetic field lines due to a dipole magnet?
The component of the magnetic field along the y-axis should be zero when the dipole is oriented along the y-axis.
This is because the magnetic field lines of a dipole magnet are perpendicular to the axis of the magnet. When the dipole is oriented along the y-axis, the axis of the dipole is also along the y-axis. Therefore, the magnetic field lines of the dipole are oriented in the x-z plane and are perpendicular to the y-axis. As a result, the component of the magnetic field along the y-axis is zero.
In contrast, when the dipole is oriented along the x-axis, the axis of the dipole is parallel to the x-axis, and the magnetic field lines of the dipole are perpendicular to the x-axis. Therefore, the component of the magnetic field along the x-axis is zero. Similarly, when the dipole is oriented along the z-axis, the axis of the dipole is parallel to the z-axis, and the magnetic field lines of the dipole are perpendicular to the z-axis. Therefore, the component of the magnetic field along the z-axis is zero.
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A satellite circles planet Roton every 6.3 h
in an orbit having a radius of 3.5 × 107 m.
If the radius of Roton is 1.75 × 107 m, what
is the magnitude of the free-fall acceleration
on the surface of Roton?
Answer:
The period (T) of a satellite in a circular orbit can be determined using the formula:
T = 2πr/v
where r is the radius of the orbit and v is the speed of the satellite.
We can rearrange this formula to solve for v:
v = 2πr/T
Substituting the given values, we get:
v = 2π(3.5 × 10^7 m)/(6.3 h × 3600 s/h) ≈ 5,099 m/s
The centripetal force (F) that keeps the satellite in orbit is given by:
F = mv^2/r
where m is the mass of the satellite.
The gravitational force (Fg) between the satellite and Roton is given by:
Fg = GmM/R^2
where G is the gravitational constant, M is the mass of Roton, and R is the radius of Roton.
Since the satellite is in a circular orbit, the centripetal force is equal to the gravitational force:
mv^2/r = GmM/R^2
Simplifying and solving for M, we get:
M = v^2r/GR^2
Substituting the given values, we get:
M = (5,099 m/s)^2 × 3.5 × 10^7 m/(6.674 × 10^-11 m^3/kg s^2) × (1.75 × 10^7 m)^2 ≈ 8.35 × 10^23 kg
The free-fall acceleration (g) on the surface of Roton is given by:
g = GM/R^2
Substituting the calculated value of M and the given value of R, we get:
g = (6.674 × 10^-11 m^3/kg s^2) × (8.35 × 10^23 kg)/(1.75 × 10^7 m)^2 ≈ 8.73 m/s^2
Therefore, the magnitude of the free-fall acceleration on the surface of Roton is approximately 8.73 m/s^2
the current in a coil changes from 3.70 a to 2.20 a in the same direction in 0.520 s. if the average emf induced in the coil is 16.0 mv, what is the inductance of the coil?
The inductance of the coil is 0.028 H.
To find the inductance of the coil, first, we need to determine the change in current (∆I) and the rate of change of current (dI/dt).
1. Calculate the change in current: ∆I = I_final - I_initial = 2.20 A - 3.70 A = -1.50 A.
2. Calculate the rate of change of current: dI/dt = ∆I / ∆t = -1.50 A / 0.520 s = -2.885 A/s.
3. Use Faraday's Law: |emf| = L * |dI/dt|, where emf is the average induced electromotive force, and L is the inductance.
4. Solve for L: L = |emf| / |dI/dt| = 16.0 mV / 2.885 A/s = 0.028 H.
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Which of the following best represents
why does an inflated beach ball pushed beneath the water surface swiftly shoot above the water surface when released
An inflated beach ball pushed beneath the water surface will swiftly shoot above the water surface when released due to the principle of buoyancy.
Buoyancy is a physical phenomenon that describes the upward force this is exerted on an item submerged in a fluid, inclusive of water or air. This pressure is a result of the pressure differences between the top and backside of an item in a fluid, and it is referred to as buoyant pressure.
Here, the beach ball is pushed beneath the water surface, it displaces a certain amount of water which will weight more than the beach ball itself. This cause acceleration upwards towards the surface when ball released.
If the object is less dense than the liquid, it will float. if the object is more dense than the liquid, it will sink.
Beach ball has a lower density than the water it displaces. So the buoyant force is so higher than a solid object of the same size and weight. The combination of buoyancy and the low density of air inside the beach ball causes it to shoot up quickly to the water surface when released.
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Temperature and Location Data for Leadville and Kansas City
Leadville, Colorado Kansas City, Missouri
Latitude 39.2°N 39.1°N
Elevation 3,094 m 277 m
Average Yearly High Temperature 5°C 22°C
Leadville, Colorado is at almost the same latitude as Kansas City, Missouri. However, the two cities have different elevations and average yearly high temperatures.
Based on the information in the table, average yearly high temperature _______ as elevation _______.
A.
stays the same; decreases
B.
decreases; increases
C.
increases; increases
D.
decreases; decreases
Average yearly high temperature decreases as elevation increases.
What is the likely reason for the difference in average yearly high temperatures between Leadville and Kansas City?The difference in elevation between the two cities results in a difference in atmospheric pressure and thus a difference in temperature.
What effect would a decrease in latitude have on the average yearly high temperatures of Leadville and Kansas City?A decrease in latitude would likely increase the average yearly high temperatures of both cities due to the increased amount of solar radiation they receive.
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a ball whose mass is 0.3 kg hits the floor with a speed of 4 m/s and rebounds upward with a speed of 1 m/s. if the ball was in contact with the floor for 2 ms (210-3 s), what was the average magnitude of the force exerted on the ball by the floor?
The force the floor typically applies to the ball is on average magnitude is 750 N.
The initial momentum of the ball:
[tex]p_1 = m*v_1 = 0.3 kg * 4 m/s = 1.2 kg m/s[/tex]
The final momentum of the ball:
[tex]p_2 = m*v_2 = 0.3 kg * (-1 m/s) = -0.3 kg m/s[/tex]
The difference in momentum of the ball is therefore:
Δp = [tex]p_2 - p_1 = -1.5 kg m/s[/tex]
Δp = FΔt
Δt = 2×10^-3 s.
F = Δp/Δt = (-1.5 kg m/s)/(2×10^-3 s) = -750 N
|F| = 750 N
Magnitude is a term used to describe the size, amount, or extent of something. In science and mathematics, magnitude often refers to the numerical value or quantity of a measurement or vector. For example, the magnitude of a force is the amount of force being applied, while the magnitude of an earthquake is a measure of the energy released.
In astronomy, magnitude is used to describe the brightness of celestial objects such as stars and galaxies. The magnitude scale was developed by ancient Greek astronomers, with lower magnitudes indicating brighter objects. Today, astronomers use the apparent magnitude scale, which takes into account both the intrinsic brightness of an object and its distance from Earth. In general usage, magnitude can refer to the importance, impact, or significance of something.
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which car has traveled farther after 10 s ? which car has traveled farther after 10 ? car a car b both cars travel the same distance. request answer part e after 10 s which car has a larger kinetic energy? view available hint(s)for part e after 10 which car has a larger kinetic energy? car a car b both cars have the same kinetic energy. part f after 10 s which car has a larger momentum? view available hint(s)for part f after 10 which car has a larger momentum? car a car b both cars have the same momentum. provide feedback correct. no additional followup.
1. Car A will travel farther than Car B after 10 s.
2. Car A will have a larger kinetic energy due to its greater mass.
3. Car A will have a larger momentum due to its greater mass.
Assuming both cars have the same constant acceleration, the car with the greater weight (Car A) will travel farther after 10 s according to the equation d = 0.5at^2, where d is the distance, a is the acceleration, and t is the time. Therefore, Car A will travel farther than Car B after 10 s.
The kinetic energy of a moving object is given by the equation KE = 0.5mv^2, where KE is the kinetic energy, m is the mass, and v is the velocity. Both cars have the same acceleration, so after 10 s, the car with the higher velocity will have a larger kinetic energy. Assuming both cars accelerate uniformly, Car A will have a larger kinetic energy due to its greater mass.
The momentum of an object is given by the equation p = mv, where p is the momentum, m is the mass, and v is the velocity. After 10 s, the car with the higher velocity will have a larger momentum. Assuming both cars accelerate uniformly, Car A will have a larger momentum due to its greater mass.
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--The complete question is, Car A is 1000g in weight and B is 800g. Both car begin from rest and start at same position.
1. which car has traveled farther after 10 s?
2. after 10 s which car has a larger kinetic energy?
3. after 10 s which car has a larger momentum?--
What does a correlation of -0.9 mean?
1.there is no relationship between the 2 variables
2.as one variable increase the other variable decreases
3.as one variable decreases the other variable decreases
4.as one variable increases the other variable increases
A correlation of -0.9 mean : 2.) as one variable increases, the other variable decreases.
What does a correlation of -0.9 mean?A correlation of -0.9 indicates a strong negative correlation between two variables, which means that as one variable increases, then other variable decreases. Closer is the correlation coefficient to -1, stronger is the negative correlation between the variables. Hence, option 2)as one variable increases, the other variable decreases is the correct answer.
Since correlation is symmetrical and represents the strength of the association between two variables, it follows that the correlation between variables A and B and B and A are identical.
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1. A machine with an input power of 2kW is made in part of 40kg iron and it experiences a rise in temperature after 2. minutes of use. If the machine is 80% power efficient, calculate the rise in temperature (Specific heat capacity of iron = 500J/kgK) A. 2,4°c B. 5.4°c C. 4.8°c D. 4.2°C
The rise in temperature of the iron part of the machine is approximately 9.6°C.
What is heat transfer?Heat transfer is the movement of thermal energy from one object or system to another due to a temperature difference. Thermal energy, which is the energy associated with the motion of particles within a substance, always moves from higher temperature regions to lower temperature regions until thermal equilibrium is achieved.
Equation:To calculate the rise in temperature of the iron part of the machine, we can use the following formula:
Q = m × c × ΔT
where Q is the heat absorbed by the iron part of the machine, m is the mass of the iron part, c is the specific heat capacity of iron, and ΔT is the rise in temperature.
First, we need to calculate the heat generated by the machine in 2 minutes, using its input power and efficiency:
P = 2 kW
Efficiency = 80% = 0.8
t = 2 minutes = 120 seconds
Energy generated by the machine = P × t × efficiency
Energy generated by the machine = 2 kW × 120 s × 0.8
Energy generated by the machine = 192 kJ
Next, we can calculate the mass of the iron part of the machine using its density:
Density of iron = 7.87 g/cm³ = 7870 kg/m³
Mass of iron part = 40 kg
Now, we can substitute the values into the formula and solve for ΔT:
Q = m × c × ΔT
ΔT = Q / (m × c)
ΔT = 192000 J / (40 kg × 500 J/kgK)
ΔT = 9.6°C
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The given options to the question must also include 9.6°C.
a 2 uf and a 1 uf capacitor are connected in parallel and a potential differene is applied across the combination. the 2 uf capacitor has:
When a 2 uF and a 1 uF capacitor are connected in parallel and a potential difference is applied across the combination, the 2 uF capacitor has the highest charge. The total capacitance of the capacitors connected in parallel is the sum of their individual capacitances.
When a potential difference is applied across two capacitors connected in parallel, the potential difference across each capacitor is the same. In this case, the 2 uF and 1 uF capacitors are in parallel, so the potential difference across each capacitor is the same as the potential difference applied to the combination. Since the capacitance of the 2 uF capacitor is larger than the 1 uF capacitor, it will store more charge for the same potential difference applied across the combination. This means that the 2 uF capacitor will have a larger charge stored on its plates compared to the 1 uF capacitor The charge stored on a capacitor is directly proportional to its capacitance, so the 2 uF capacitor will have twice the charge stored on its plates as the 1 uF capacitor.
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The x-velocity of a horizontal projectile is always ___. The initial y-velocity of a horizontal projectile is always ___
Answer: constant and horizontal; changing (usually increasing) and downward
Explanation:
in a projectile, there is not horizontal acceleration (unless friction), which means the velocity is constant horizontally. vertically, there is gravity so that the vertical velocity would be increasing, zero at the top, and then decreasing, and the y vector pointing downwards
if two joggers have similar levels of commitment and purpose but one is clearly better at jogging than the other, what is the MOST likely difference between the two joggers? A. fitness goals B. body composition C. motivation D. body positivity
Answer:
Fitness Level
Explanation:
The most likely difference between the two joggers who have similar levels of commitment and purpose but one is clearly better at jogging than the other is their fitness level. Fitness level refers to the level of physical fitness and conditioning of an individual, which is usually determined by their exercise habits, lifestyle, and genetics. A person with a higher level of fitness will generally be able to perform physical activities, such as jogging, more easily and with greater proficiency than a person with a lower level of fitness. Therefore, the jogger who is better at jogging is likely to have a higher level of fitness than the other jogger.
While factors such as body composition, motivation, and body positivity can also impact jogging performance, they are less likely to be the most significant difference between the two joggers in this scenario.
Answer:
The most likely difference between the two joggers is
B. body composition.
Body composition refers to the ratio of fat mass to lean mass in the body, and it plays an important role in athletic performance. A jogger with a lower percentage of body fat and a higher percentage of lean muscle mass is likely to have better endurance and speed than a jogger with a higher percentage of body fat and lower muscle mass, even if both have similar levels of commitment and motivation. Fitness goals, motivation, and body positivity can also play a role in jogging performance, but body composition is the most likely factor in this scenario.
to stretch a spring a distance of 0.3 m from the equilibrium position, 120 j of work is done. what is the value of the spring constant k?
It takes 120 j of effort to extend a spring 0.3 m from its equilibrium state. The value of the spring constant k is 2666.67 N/m.
The work done to stretch a spring is given by the formula:
W = 0.5 × k × x^2
where W is the work done, k is the spring constant, and x is the distance the spring is stretched from its equilibrium position.
In this problem, we know that the work done is 120 J and the distance the spring is stretched is 0.3 m. Substituting these values into the formula, we get:
120 = 0.5 × k × 0.3^2
Simplifying the equation, we get:
k = 120 / (0.5 × 0.3^2)
k = 2666.67 N/m
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