a 75.8 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion. if the period of oscillation is 7.25 s, what is the spring constant of the bungee cord?

When a hybrid car brakes to a stop, much of its kinetic energy is transformed into different forms of energy. One of the primary forms of energy that is produced during this process is heat. As the brakes of the car are applied, the friction between the brake pads and the wheels creates heat. This heat is then dissipated into the air around the car, resulting in a loss of energy.

Hybrid cars are designed to capture some of this lost energy and convert it into useful forms of energy that can be used to power the car. In many cases, the kinetic energy that is lost during braking is converted into electric potential energy, which is then stored in the car's battery. This energy can then be used to power the car's electric motor, which in turn can help reduce the car's overall fuel consumption.

It is through the conversion of kinetic energy into electric potential energy or the conversion of energy into work, hybrid cars are a great example of how technology can be used to improve the efficiency of vehicles and reduce their environmental impact.

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The temperature of the [tex]CH_4[/tex] gas is 195 K (Option C).

We can use the ideal gas law to solve this problem: PV = nRT

where P is the pressure in atm, V is the volume in L, n is the number of moles, R is the gas constant (0.0821 L·atm/mol·K), and T is the temperature in K.

First, we need to calculate the number of moles of [tex]CH_4[/tex] using its molar mass:

Molar mass of [tex]CH_4[/tex] = 12.01 g/mol (C) + 4(1.01 g/mol) = 16.05 g/mol

Number of moles of [tex]CH_4[/tex] = 10.34 g / 16.05 g/mol = 0.644 mol

Now, we can rearrange the ideal gas law to solve for T:

T = PV / (nR)

Substituting the given values, we get:

T = (1.33 atm) x (50.0 L) / (0.644 mol x 0.0821 L·atm/mol·K) = 195 K

Therefore, the temperature of the [tex]CH_4[/tex] gas is 195 K (Option C).

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The mass of the other ball is 170 g, and its initial speed is 33.5 m/s. To solve this problem, we can use the conservation of momentum and the conservation of kinetic energy.

We know that the momentum before the collision is equal to the momentum after the collision, and that the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

Let m be the mass of the other ball, and v be its initial velocity. Then we have:

Momentum before = Momentum after
85 g * 8.30 m/s = m * (v/2) + 85 g * (-16.0 m/s)

Solving for m, we get m = 170 g.

Next, we can use the conservation of kinetic energy to find v:

Kinetic energy before = Kinetic energy after
(1/2) * 85 g * (8.30 m/s)^2 = (1/2) * m * (v/2)^2 + (1/2) * 85 g * (16.0 m/s)^2

Solving for v, we get v = 33.5 m/s.

Therefore, the mass of the other ball is 170 g, and its initial speed is 33.5 m/s.

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This means the ball drops about 0.87 meters (about 2 feet, 10 inches) by the time it reaches the catcher.

To answer this question, we need to use the equation for projectile motion, which is:
y = yo + voyt + 1/2at^2

where y is the vertical displacement, yo is the initial vertical position, voy is the initial vertical velocity, t is time, and a is the acceleration due to gravity.

In this case, the ball is thrown horizontally, so there is no initial vertical velocity (voy = 0), and the initial vertical position is also zero (yo = 0). We know the distance the ball travels horizontally (17 m) and the initial speed (41 m/s), so we can use the equation:
x = vot + 1/2at^2

where x is the horizontal displacement and vo is the initial horizontal velocity. Solving for t, we get:
t = x / vo = 17 / 41 = 0.4146 s

Now we can use the equation for vertical displacement to find how much the ball drops during that time. Since we know the acceleration due to gravity is -9.8 m/s^2 (downward), we get:
y = 1/2at^2 = 1/2(-9.8)(0.4146)^2 = -0.8735 m

This means the ball drops about 0.87 meters (about 2 feet, 10 inches) by the time it reaches the catcher. It's important to note that this calculation assumes the ball is thrown perfectly horizontally, with no vertical component to its motion. In reality, a pitched ball will have some degree of arc, which will affect its trajectory and how much it drops before reaching the catcher.

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At a given rotational speed, linear (or tangential) speed increases as the distance from the axis increases.

This relationship is described by the equation:

v = rω

where v is the tangential velocity, r is the distance from the axis (i.e., the radius), and ω is the angular velocity.

This means that for a given rotational speed (i.e., angular velocity), objects farther from the axis will be moving faster than objects closer to the axis. This relationship is demonstrated in everyday examples such as the rotation of a bicycle wheel, where the speed of the outer edge is much greater than the speed of the hub.

It's important to note that this relationship assumes a constant angular velocity. If the angular velocity changes, then the linear speed will also change, regardless of the distance from the axis.

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The instantaneous velocity is zero at t = 0 seconds.(a)

To plot x as a function of t, we can substitute values of t from 0 to 3.0 seconds into the equation x = 34 + 10 − 2t2. This gives us the following values of x: at t=0, x=34+10=44; at t=1.0, x=34+10−2(1.0)2=42; at t=2.0, x=34+10−2(2.0)2=30; at t=3.0, x=34+10−2(3.0)2=10. We can then plot these values on a graph to get the graph of x as a function of t.

(b) The average velocity of the object between 0 and 3.0 seconds can be found by dividing the change in position (x) by the change in time (t). The change in position is x(3.0) − x(0) = 10 − 44 = −34 meters, and the change in time is 3.0 − 0 = 3.0 seconds. Therefore, the average velocity is −34/3 = −11.3 m/s.

(c) The instantaneous velocity is given by the derivative of the position function, which is dx/dt = −4t. The instantaneous velocity is zero when −4t = 0, or when t = 0. Therefore, the instantaneous velocity is zero at t = 0 seconds.

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Both Type Ia and Type II supernovae are important astronomical events that provide us with valuable information about the life cycle of stars and the formation of the universe.

Type Ia and Type II supernovae are two different types of supernovae that are caused by different types of stars. Type Ia supernovae are caused by white dwarf stars, which are the remnants of stars that have exhausted all of their nuclear fuel and have collapsed to a very small size. These stars are typically in a binary system with another star, and they can accrete matter from their companion star. When a white dwarf reaches a certain mass, it can undergo a runaway nuclear reaction that causes it to explode as a supernova.
On the other hand, Type II supernovae are caused by much more massive stars, which have exhausted their nuclear fuel and can no longer support their own weight. These stars undergo a series of complex nuclear reactions that result in the production of heavier elements, and eventually, they collapse under their own gravity and explode as supernovae.
Overall, both Type Ia and Type II supernovae are important astronomical events that provide us with valuable information about the life cycle of stars and the formation of the universe. By studying these explosions and their remnants, astronomers can learn more about the composition and evolution of the universe, as well as the origins of the chemical elements that make up our world.

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Objects that are not actively moving but have the capacity to do so possess a.) kinetic energy.

So, the correct answer is a. kinetic energy.

Kinetic energy is the energy that an object possesses due to its motion. Even if an object is not currently in motion, if it has the capacity to move, it possesses potential kinetic energy. This energy can be released when the object is set into motion. Entropy potential energy and living energy are not relevant in this context. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body when decelerating from its current speed to a state of rest.

So, Objects that are not actively moving but have the capacity to do so possess a.) kinetic energy.

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The relevance of the first law of thermodynamics to biology. Energy can be freely transformed among different forms as long as the total energy is conserved. Correct option is B.

The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed, only transformed from one form to another. This principle is relevant to biology because living organisms are constantly converting energy from one form to another.

For example, photosynthetic organisms convert light energy into chemical energy stored in glucose molecules, which can then be used by other organisms through cellular respiration to produce ATP, the main source of energy for cellular processes.

The first law of thermodynamics also applies to other energy transformations that occur in biological systems, such as the breakdown of food molecules during digestion and the use of stored energy to power muscle contractions. Overall, this law emphasizes the importance of energy conservation and efficient energy use in biological systems.

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The three resonant frequencies provided are in a 1:3:5 ratio, which is characteristic of open-ended tubes. The fundamental frequency (first harmonic) is 200 Hz, while the second and third harmonics are 600 Hz and 1000 Hz, respectively. These harmonics indicate the presence of an open-ended tube.

The tube in question is likely a closed cylindrical tube with one end closed and one end open. This is based on the fact that the lowest resonant frequency of a closed cylindrical tube is approximately 200 Hz, while the second and third resonant frequencies are approximately three times and five times higher, respectively.  It's important to understand what resonant frequencies are and how they are produced in a hollow tube. Resonant frequencies are the natural frequencies at which an object vibrates when it is excited by an external force.

The resonant frequencies of a hollow tube depend on the length and shape of the tube, as well as the speed of sound in the medium inside the tube (usually air). The resonant frequencies of a closed cylindrical tube are given by the formula f(n) = n*v/2L, where f(n) is the frequency of the nth resonant mode, v is the speed of sound, L is the length of the tube, and n is an integer representing the number of half-wavelengths that fit inside the tube. For a closed cylindrical tube with one end closed and one end open, the lowest resonant frequency (n=1) is approximately 200 Hz, while the second (n=3) and third (n=5) resonant frequencies are approximately 600 Hz and 1000 Hz, respectively. Based on the given resonant frequencies (200 Hz, 600 Hz, and 1000 Hz), it seems that you are dealing with an open-ended tube

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Answer:

One must consider the phase difference between each side of the slit:

One can write Δ = W sin θ       where Δ is the phase difference between light leaving opposite sides of the slit, W = slit width, and θ is the angle of difraction

As the slit gets wider (W increases) and for a  given phase difference the angle of difraction will decrease

The bands on the screen will be closer for a wider slit

The objective of such an approach would be to minimize the potential impact and ensure that any damage is contained to the extent possible.would be to minimize the potential impact and ensure that any damage is contained to the extent possible.

A potential impactor with a maximum torino ranking means that there is a significant risk of it hitting the Earth and causing substantial damage. In such a scenario, it would be necessary to take swift and decisive action to prevent or mitigate the impact. While there are various options available, such as striking the object with a nuclear missile or attempting to alter its orbital path, it would be advisable to consider a comprehensive approach that involves all of the above. This means organizing a regional evacuation to a far removed place on the globe while also attempting to alter the trajectory of the potential impactor. The objective of such an approach would be to minimize the potential impact and ensure that any damage is contained to the extent possible. It is essential to recognize that the potential impact of an object with a maximum torino ranking could have far-reaching consequences for the planet, and we must take all necessary measures to prevent such an eventuality.

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False. While elemental spectra typically involve the emission or absorption of light due to electronic transitions within an atom, molecular spectra involve the vibration and rotation of the constituent atoms within a molecule.

Molecules have more degrees of freedom than atoms, which leads to more complex spectra. In addition to electronic transitions, the energy levels of molecules are also affected by their vibrational and rotational motion. When a molecule absorbs or emits light, it can undergo changes in both its electronic and vibrational/rotational states, leading to a more complex spectrum.

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The acceleration of the sprinter during the first 75.0 m was 1.44 m/[tex]s^2.[/tex]

We can use the kinematic equations of motion to solve this problem. Let's assume that the sprinter has an initial velocity of zero at the starting point, and a final velocity of v at the end of the 75.0 m distance. We can also assume that the time taken to cover the first 75.0 m is [tex]t_1,[/tex] and the time taken to cover the last 25.0 m is [tex]t_2[/tex].

For the first 75.0 m, we can use the following kinematic equation:

[tex]d = (1/2)at1^2[/tex]

where d is the distance covered, a is the acceleration, and t1 is the time taken to cover the distance.

For the last 25.0 m, we can use the following kinematic equation:

[tex]d = vt_2[/tex]

where d is the distance covered, v is the final velocity, and [tex]t_2[/tex] is the time taken to cover the distance.

We can also use the following kinematic equation for the entire 100.0 m distance:

[tex]d = (1/2)at^2[/tex]

where d is the distance covered, a is the acceleration, and t is the total time taken to cover the distance.

Using the given values of distance and time, we can write the following three equations:

75.0 m = [tex](1/2)at1^2[/tex] (equation 1)

25.0 m =[tex]vt_2[/tex] (equation 2)

100.0 m = [tex](1/2)at^2[/tex] (equation 3)

Since the sprinter covers the last 25.0 m at constant velocity, we know that his final velocity, v, is the same as his average velocity over the last 25.0 m. Therefore, we can write:

v = 25.0 m / t2

Substituting this expression for v into equation 3, we get:

100.0 m = [tex](1/2)at1^2[/tex] + 25.0 m / [tex]t_2[/tex]

Simplifying this equation, we get:

200.0 m = at1^2 + 50.0 m / [tex]t_2[/tex]

Now we can use equation 1 to eliminate t1:

[tex]t_1 =\sqrt(2d/a)[/tex]

Substituting this expression for t1 into equation 2, we get:

25.0 m = [tex]v(\sqrt(2d/a))[/tex]

Simplifying this equation, we get:

[tex]v^2[/tex]= 50.0ad

Substituting this expression for [tex]v^2[/tex] into the previous equation, we get:

200.0 m = (a/2)(2d/a) + (d/a) [tex]v^2[/tex]

Simplifying this equation, we get:

200.0 m = d(1/2 + 1/2)

or

d = 200.0 m

Substituting this value of d into equation 3, we get:

200.0 m = [tex](1/2)at^2[/tex]

Simplifying this equation, we get:

a =[tex](2d/t^2)[/tex]

Substituting the given values of distance and time, we get:

a = (2 x 75.0 m / (9.87 [tex]s)^2)[/tex]

a = 1.44 m[tex]/s^2[/tex]

Therefore, the acceleration of the sprinter during the first 75.0 m was 1.44 m/[tex]s^2.[/tex]

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The two major lines of evidence that support the Big Bang theory and have not been successfully explained by alternative models are:

The existence and specific characteristics of the observed cosmic microwave background (CMB): The CMB is a form of radiation that fills the entire universe and is believed to be the leftover heat from the early universe when it was hot and dense. The Big Bang theory predicts the existence of the CMB and its specific characteristics, such as its uniformity and specific temperature fluctuations, which have been observed and confirmed through various experiments.

The observed overall chemical composition of the universe: The Big Bang theory predicts that the early universe was primarily composed of hydrogen and helium, with trace amounts of other elements. Observations of the chemical composition of the universe have confirmed this prediction and provide strong support for the Big Bang theory.

The other options provided in the question are not accurate as they do not represent the two major lines of evidence that support the Big Bang theory. The expansion of the universe is actually a consequence of the Big Bang theory, rather than evidence for it. The observed ratio of spiral to elliptical galaxies is not a major line of evidence for the Big Bang theory. The episode of inflation and separation of gravity from other forces are both aspects of the Big Bang theory itself and not independent lines of evidence supporting it. Lastly, the statement "the early universe was hot and dense" is a prediction of the Big Bang theory and not a line of evidence supporting it.

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The magnitude of the sum of two vectors A and B is 4.13 m and the angle of the resultant vector is 10.86°.

From the given,

A = 3.14 m

B = 2.71 m

The resultant vector C= A + B

Vector A is resolved into its vertical and horizontal components,

Aₓ = 3.14 cos(30) = 2.71 m

Ay = 3.14 sin (30) = 1.57 m

Vector B is resolved into its vertical and horizontal components,

Bx = 2.71 cos(60) = 1.355 m

By = ₋2.71 sin (60) = -2.35 m

C = A + B

  = (2.71+1.355) x + (1.57 -2.35) y

 = 4.064 i - 0.78 j

the magnitude of C = √(4.06)² + ( 0.78)² = 4.13 m

The angle, tan α = 0.78 / 4.06

 α = 10.8°

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Answer:

Wavelength is inversely proportional to the frequency.

Explanation:

V=fλ, where V is the speed of wave , f is the frequency of wave and lamda is the Wavelength of wave.

So rearranging the above formula : λ=v/f so λ∝ 1/f

The component of the net force, fnet,x, in the x-direction can be expressed as the sum of all forces acting in that direction: fnet,x = ΣFx.

To find the component of net force, fnet,x, in the x-direction, we need to consider all the forces acting in that direction. Let's assume there are n forces acting in the x-direction. Then we can write:

ΣFx = F1x + F2x + F3x + ... + Fnx

where F1x, F2x, F3x, ..., Fnx are the x-components of the individual forces.

We can then substitute the expressions for each of the individual x-components of the forces. For example, if we have a force F1 acting at an angle θ1 to the x-axis, we can use trigonometry to find its x-component:

F1x = F1 cos(θ1)

Similarly, for all the other forces, we can find their x-components and add them up to get the total sum of forces in the x-direction. This gives us the expression for the component of the net force, fnet,x, in the x-direction:

fnet,x = ΣFx

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The density of the metal can be calculated using the formula:

density = mass / volume

where mass is the mass of the metal and volume is the volume of water displaced by the metal.

Given that the mass of the metal is 0.6 kg and it displaces 1 liter (1000 cubic centimeters) of water, we can substitute these values into the formula:

density = 0.6 kg / 1000 cubic centimeters

Simplifying, we get:

density = 0.0006 kg/cubic centimeter

Therefore, the density of the metal is 0.0006 kg/cubic centimeter.

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The electron configuration of a ground-state Be atom is 1s² 2s². Each electron in the atom is described by a set of quantum numbers, including:

Principal quantum number (n): The first electron has n = 1 and the second electron has n = 2.

Azimuthal quantum number (l): For n = 1, the only possible value of l is 0 (s orbital), and for n = 2, the possible values of l are 0 (s orbital) and 1 (p orbital). Therefore, the first electron has l = 0 and the second electron has l = 0 or l = 1.

Magnetic quantum number (m): For l = 0, the only possible value of m is 0, and for l = 1, the possible values of m are -1, 0, and 1. Therefore, the first electron has m = 0, and the second electron has m = -1, 0, or 1.

Spin quantum number (s): Each electron has s = +1/2 or -1/2.

Thus, the set of quantum numbers for each electron in a ground-state Be atom is:

Electron 1: n = 1, l = 0, m = 0, s = +1/2 or -1/2

Electron 2: n = 2, l = 0 or 1, m = -1, 0, or 1, s = +1/2 or -1/2

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The image will be formed 60 cm behind the lens, on the same side as the object, but it will be a virtual image, which means that it cannot be projected onto a screen.

To determine the location of the image formed by a lens, we can use the thin lens equation:

1/f = 1/di + 1/do

where f is the focal length of the lens, di is the distance from the lens to the image, and do is the distance from the lens to the object. We can rearrange this equation to solve for di:

1/di = 1/f - 1/do

Plugging in the values given in the problem, we have:

1/-20 = 1/di + 1/30

Simplifying and solving for di, we get:

di = -60 cm

The negative sign for the image distance indicates that the image is formed on the opposite side of the lens from the object, which means it is a virtual image.

Therefore, the image will be formed 60 cm behind the lens, on the same side as the object, but it will be a virtual image, which means that it cannot be projected onto a screen.

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Answer: 5. On the Moon.

Explanation: Weight is a measure of the force of gravity acting on an object.  The weight of an object depends on the mass of the object and the strength of the gravitational force at a particular location.

On Earth, the weight of an object is determined by the mass of the object and the strength of Earth's gravitational force. At the equator, the weight of an object is slightly less compared to the poles due to the centrifugal force caused by the Earth's rotation. This force counteracts a small portion of the gravitational force, resulting in a slightly lower weight.

At the North and South poles, the weight of an object is slightly higher compared to the equator due to the shape of the Earth. The Earth is not a perfect sphere but slightly flattened at the poles, which causes objects at the poles to be closer to the center of the Earth and experience a slightly stronger gravitational force.

However, on the Moon, the weight of an object is significantly less compared to Earth.  The Moon has a much smaller mass and weaker gravitational force than Earth, resulting in objects weighing less on the lunar surface.

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Answer:

D

Explanation:

A column of alcohol will be longer by a factor which is the ratios of the densities 13600/789.  

The length of the mercury column replaced by the alcohol is 75.6 - 73.2 cm.

Hence h = (75.6-73.2) x 13600/789 = 41.4 cm

True

Compared to Earth, the Moon lacks a hydrosphere (a system of water on its surface, including oceans, lakes, and rivers), an atmosphere and a magnetosphere (a region of space influenced by a planet's magnetic field).

The Moon's surface is dry and lacks significant amounts of water, it has an extremely thin or negligible atmosphere, and it has no substantial magnetic field to create a magnetosphere.

Here's a more detailed explanation:

Hydrosphere: The hydrosphere refers to the presence of water in liquid form on a planetary body. Earth has a significant hydrosphere, with about 71% of its surface covered by water in the form of oceans, seas, lakes, and rivers.

In contrast, the Moon lacks a substantial hydrosphere. While there is evidence of water ice in permanently shadowed regions near the Moon's poles, it is in the form of solid ice rather than liquid water.

Atmosphere: Earth has a dense atmosphere composed primarily of nitrogen (about 78%) and oxygen (about 21%), along with other trace gases. The atmosphere plays a crucial role in regulating temperature, supporting life, and protecting the planet from harmful radiation.

In contrast, the Moon has an extremely thin and tenuous atmosphere, often referred to as an exosphere. It consists of extremely low-density particles, such as atoms and ions, and is practically nonexistent compared to Earth's atmosphere.

Magnetosphere: Earth has a magnetic field generated by its liquid iron outer core. This magnetic field extends into space and creates a region around the planet known as the magnetosphere.

The magnetosphere protects Earth from the solar wind, a stream of charged particles emitted by the Sun. The Moon, however, lacks a significant magnetic field.

It does not have a liquid iron core like Earth, and thus, it does not generate a magnetosphere. As a result, the Moon is directly exposed to the solar wind and its associated radiation.

The absence of a hydrosphere, atmosphere, and magnetosphere on the Moon significantly influences its surface conditions and overall environment.

These factors contribute to the Moon's starkly different appearance and characteristics compared to Earth.

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The fairly flat, circular part of a galaxy is referred to as the "galactic disk" or the "stellar disk."

The galactic disk is one of the main components of a spiral galaxy and contains the majority of a galaxy's stars, as well as various interstellar materials like gas and dust.

The disk has a flattened shape, with stars and other objects orbiting the galaxy's central bulge. It is within the galactic disk that most of the star formation and ongoing stellar activity occur.

The galactic disk is often characterized by spiral arms, where regions of higher star density and star formation are observed.

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The phenomenon which made the first polythene rod heavier is because the electric charges on it made it attract more air molecules thus making it heavier.

A charged polythene rod can be used to transmit charge to an insulated conductor without contacting the two objects. The conductor is approached by bringing the negatively charged polythene rod close to it.

Electrons generate enough energy to 'rub off' onto the polythene rod and depart the atom. Electrons are rubbed off the acetate and onto the duster when the rod is replaced with a new substance, such as acetate. The rods and duster are both comprised of insulating materials.

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Full Question:

A student pit the charged polythene rod on to a balance.  The rod was separated from the metal pan of the balance by a thin block of ons laying material.  The student then held a second charged polythene rod above,  but not touching the rod.  The reading on the balance increased.  Explain

when the faster runner finishes the race, the slower runner still has 8.00 km - 6.30 km = 1.70 km left to run before reaching the finish line.

To find out how far from the finish line the slower runner is when the faster runner finishes the race, we need to first calculate how long it takes for the faster runner to finish the race.
Using the formula:
time = distance ÷ speed
We can calculate the time it takes for the faster runner to finish the race:
time = 8.00 km ÷ 14.1 km/h
time = 0.567 hours
Now we know that the faster runner finishes the race in 0.567 hours.
To find out how far the slower runner is from the finish line when the faster runner finishes the race, we need to calculate how far the slower runner has run in the same amount of time (0.567 hours).
distance = speed x time
For the slower runner:
distance = 11.1 km/h x 0.567 hours
distance = 6.30 km
Therefore, when the faster runner finishes the race, the slower runner still has 8.00 km - 6.30 km = 1.70 km left to run before reaching the finish line.
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If arrivals occur according to the Poisson distribution every 20 minutes, the statement that is NOT true is option, d: λ = 72 arrivals per day.

Given that arrivals occur according to the Poisson distribution every 20 minutes, we need to convert the rate to the appropriate time unit for each option:

a. λ = 20 arrivals per hour: This is true. The rate of 20 arrivals per hour is consistent with arrivals occurring every 20 minutes.

b. λ = 3 arrivals per hour: This is true. The rate of 3 arrivals per hour is consistent with arrivals occurring every 20 minutes.

c. λ = 1/20 arrivals per minute: This is true. The rate of 1/20 arrivals per minute is consistent with arrivals occurring every 20 minutes.

d. λ = 72 arrivals per day: This is NOT true. Since arrivals occur every 20 minutes, we need to convert the rate to arrivals per day. There are 24 hours in a day, and since arrivals occur every 20 minutes, there are 60 minutes / 20 minutes = 3 sets of 20 minutes in an hour. Therefore, there are 24 hours * 3 = 72 sets of 20 minutes in a day. The rate should be λ = 72 arrivals per day, not λ = 72 arrivals per day.

Therefore, option d is the statement that is NOT true.

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The answer to this question is a. 141 km/h. The airplane's speed and the crosswind speed are not added together to get the resultant speed because they are not in the same direction. Instead, we use the Pythagorean theorem to calculate the resultant speed.

To find the resultant speed, we need to use the Pythagorean theorem because the airplane's speed and the crosswind speed are perpendicular to each other. The Pythagorean theorem states that the square of the hypotenuse (resultant speed) is equal to the sum of the squares of the other two sides (airplane speed and crosswind speed). Using this formula, we can calculate the resultant speed as follows:
Resultant speed = √(80^2 + 60^2)
Resultant speed = √(6400 + 3600)
Resultant speed = √10000
Resultant speed = 100 km/h
Therefore, the answer is not d. 80 km/h, but rather a. 141 km/h.

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a) The resistance will be:

R = V / I = 12 V / 0.80 A = 15 Ω

b) E = P * t = 9.6 W * 60 s = 576 Joules

2) The 75 W lightbulb used at 120 V will have a relative brightness of 18.75 W 120 V bulbs.

a) To calculate the resistance, we can use Ohm's Law, which states that resistance (R) is equal to the voltage (V) divided by the current (I). Therefore, the resistance can be calculated as follows:

R = V / I = 12 V / 0.80 A = 15 Ω

b) The power (P) consumed by the battery can be calculated using the formula: P = V * I, where V is the voltage and I is the current. The energy (E) consumed in a given time period can be calculated by multiplying power (P) by time (t). In this case, since we want to find the energy consumed in a minute, the time is 60 seconds.

P = V * I = 12 V * 0.80 A = 9.6 W

E = P * t = 9.6 W * 60 s = 576 Joules

The power of the lightbulb remains constant regardless of the voltage applied, so the power rating of the lightbulb is still 75 W. However, the brightness of a lightbulb is typically measured in terms of its luminous flux, which is not directly proportional to power. Luminous flux is measured in lumens (lm).

To determine how bright the 75 W lightbulb will be at 120 V relative to 75 W 120 V bulbs, we need to compare the luminous flux. Assuming the lightbulb's resistance remains constant, we can use the formula for power (P) in terms of resistance (R) and voltage (V): P = V^2 / R.

For the 75 W lightbulb at 240 V:

P1 = 75 W

V1 = 240 V

For the lightbulb at 120 V:

P2 = ?

V2 = 120 V

Using the formula, we can solve for P2:

P1 / P2 = (V1 / V2)^2

75 W / P2 = (240 V / 120 V)^2

75 W / P2 = 2^2

75 W / P2 = 4

P2 = 75 W / 4 = 18.75 W

Therefore, the 75 W lightbulb used at 120 V will have a relative brightness of 18.75 W 120 V bulbs.

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