How many players are allowed to play for one team during a game of
5 points
Netball?

Answer:

1144.8

Explanation:

1 meter per second is equal to 3.6 kilometers per hour

To know more about the effects of doubling the radius of a wheel and solve the given problem:

When the radius of a wheel is doubled, the answers to the given questions are affected in the following ways:

The angular speed at 2.00 s is smaller: False. Doubling the radius does not directly affect the angular speed at a specific time. It depends on the angular acceleration and initial conditions.

The angle rotated through from t = 0 to t = 2.00 s is the same: True. The angle rotated depends on the angular speed and time, which are not directly affected by the radius.

The angular speed at t = 2.00 s is greater: False. The angular speed is not affected by doubling the radius, assuming the angular acceleration remains constant.

The angular speed at t = 2.00 s is the same: True. Doubling the radius does not directly impact the angular speed at a specific time, assuming other factors remain constant.

The angle rotated through from t = 0 to t = 2.00 s is greater: False. The angle rotated depends on the angular speed and time, which are not directly influenced by the radius.

The angle rotated through from t = 0 to t = 2.00 s is smaller: False. The angle rotated depends on the angular speed and time, which are not directly affected by the radius.

Now, let's solve the provided problem. We are given that the wheel has a constant angular acceleration of 3.40 rad/s² and an angular speed of 1.70 rad/s at t = 0.

(a) To find the angle rotated between t = 0 and t = 2.00 s, we can use the equation: θ = ω₀t + (1/2)αt², where ω₀ is the initial angular speed, α is the angular acceleration, and t is the time. Substituting the given values, we get θ = (1.70 rad/s)(2.00 s) + (1/2)(3.40 rad/s²)(2.00 s)². Evaluating this expression gives us the angle rotated in radians.

(b) To find the angular speed at t = 2.00 s, we can use the equation: ω = ω₀ + αt. Substituting the given values, we get ω = 1.70 rad/s + (3.40 rad/s²)(2.00 s).

(c) To find the angular displacement in revolutions when the angular speed doubles, we need to find the angular speed when t = 2.00 s and then double that value.

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Answer:

Both energies are equal when the rock has fallen 20 m or equivalently when it is at a height of 20 m.

Explanation:

Potential and Kinetic Energy

The gravitational potential energy is the energy an object has due to its height above the ground. The formula is

\(U=mgh\)

Where:

m = mass of the object

g = acceleration of gravity (9.8~m/s^2)

h = height

Note we can also use the object's weight W=mg into the formula:

\(U=Wh\)

The kinetic energy is the energy an object has due to its speed:

\(\displaystyle K=\frac{1}{2}mv^2\)

Where v is the object's speed.

Initially, the object has no kinetic energy because it's assumed at rest.

The W=30 N rock falls from a height of h=40 m, thus:

\(U=30*40=1,200 J\)

Since the sum of the kinetic and potential energies is constant:

U' + K' = 1,200 J

Here, U' and K' are the energies at any point of the motion. Since both must be the same:

U' = K' = 600 J

U'=Wh'=600

Solving for h':

\(\displaystyle h'=\frac{600}{W}=\frac{600}{30}=20~m\)

Both energies are equal when the rock has fallen 20 m or equivalently when it is at a height of 20 m.

Since the magnetic force and velocity are orthogonal, the velocity only changes in direction but not in magnitude . Circular motion is the end consequence. The force is in the opposite direction from the right-hand rule and has a negative charge.

Uniform In circular motion, an object moves in two dimensions with a constant speed in a fixed circular direction; however, because the object's direction changes at every point, the velocity also changes, and the direction at each point is always the tangent.

In physics, magnetic force is referred to as an object's maximal size and direction. Both vector and scalar values use magnetic force as a common factor. We understand that by definition

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Answer: The velocity simply changes in direction, not magnitude, as the magnetic force and velocity are orthogonal.

Explanation: The velocity simply changes in direction, not magnitude, as the magnetic force and velocity are orthogonal. The resultant motion is circular. The force has a negative charge and is directed in the opposite direction from the right-hand rule.

Uniform In a circular motion, an item moves in two dimensions with a constant speed in a fixed circular direction; yet, because the object's direction varies at each point, the velocity also alters, and the direction at each point is consistently the tangent.

As an object's maximum size and direction, the magnetic force is referred to in physics. Magnetic force is a factor that is present in both scalar and vector values. We comprehend that by nature.

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Answer: A.

It has to be A just from common knowldege.

The period of a simple pendulum of length 1m on a massive planet is 1 sec. The acceleration due to gravity on that planet is 39.48 m/s^2.

A simple pendulum's period is given by:

T = 2π √(L/g)

Where T is the pendulum's period, L is its length, and g is the acceleration due to gravity.

In this scenario, the pendulum's period is one second and its length is one metre.

So, from above equation, we have:

1 = 2π √(1/g)

Squaring both sides, we get:

1^2 = (2π)^2 (1/g)

Simplifying, we get:

g = (4π^2)/1 = 39.48 m/s^2

Therefore, the acceleration due to gravity on the massive planet is 39.48 m/s^2.

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Answer:

30 mph

Explanation:

1800 divided by 60

When an object travels at a constant speed with zero acceleration it is known as uniform rectilinear motion. Then its velocity is given by:

v = d / t

Where d is distance and t is time. Evaluating:

v = 1800 m / 60 s

Answer: the coefficient of friction

Explanation:

The magnitude of frictional force is \(\mu\)N and the magnitude of the force is N. So if we take the ratio of it we will get  \(\mu\) In result.

The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, the Greek letter is used to symbolize it, i.e., \(\mu\). In mathematical terms, is equal to F/N, where F represents frictional force and N represents normal force. Since both F and N are measured in units of force, the coefficient of friction is a dimensional less quantity (such as newtons or pounds).

For both static and kinetic friction, the coefficient of friction has a range of values. When an object experiences static friction, the frictional force resists any applied force, causing the object to stay at rest until the static frictional force is removed. In kinetic friction, the frictional force resists the motion of the object.

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The potential energy is utilized during work done against frictional force.

Energy cannot be created or destroyed, according to the law of conservation of energy. However, it is capable of change from one form to another. An isolated system's total energy is constant regardless of the types of energy present.

When a driver applies the brakes to keep a car going down hill at constant speed and constant kinetic energy, the potential energy of the car decreases. The potential energy is utilized during work done against frictional force and total energy remains constant.

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Answer:

Force

We know that

F = mg

F = 1025 × 9.8

F = 10,045 N

Or,

10045/1000 = 10.045 Kilo Newton

\( \\ \)

Answer:

According to studies, the milky way is approximately, "170,000–200,000 light-years (52–61 kpc) in diameter and, on average, approximately 1,000 ly (0.3 kpc) thick."

With that being said, it is safe to say that the dimensions are somewhere around 100,000 by 1,000

The location of the image is 4 cm. The image is real.

Calculation:-

We can find the image distance by using the lens formula which can be given as :

                  f = 30cm

                    = -60cm

              So v= 60cm

Magnification = v/u = h i/h o

                    h i = 4 cm

and the image is real.

A lens is a transmissive optical tool that focuses or disperses a mild beam via refraction. An easy lens includes a single piece of transparent cloth, while a compound lens consists of several easy lenses, commonly organized alongside a commonplace axis.

By changing its form, the lens changes the focal distance of the attention. In other phrases, it focuses the mild rays that skip through it for you to create clean photos of gadgets that can be placed at various distances. 3 primary sorts of lenses: normal, telephoto, and wide perspective, and their respective lens focal lengths.

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The substrate Glycogen is utilized in two primary energy systems: the Glycolytic System (also known as the Anaerobic Lactic System) and the Oxidative System (also known as the Aerobic System).

1. Glycolytic System: This energy system rapidly breaks down glycogen into glucose, which then undergoes glycolysis to produce ATP (adenosine triphosphate). This system is anaerobic, meaning it does not require oxygen and is suitable for short, high-intensity activities lasting up to 2 minutes.

2. Oxidative System: In this energy system, glycogen is first broken down into glucose, which then undergoes glycolysis, the citric acid cycle, and oxidative phosphorylation to produce ATP. This system is aerobic, meaning it requires oxygen and is suitable for longer, lower-intensity activities lasting more than 2 minutes.

The substrate glycogen is utilized in both the Glycolytic (Anaerobic Lactic) and Oxidative (Aerobic) energy systems to produce ATP for various physical activities.

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The distance which a force of 3N will pull a wooden block before transferring 12J of energy is 4m.

This is referred to as an influence that can change the motion of an object and the unit is Newton.

It has a relationship with distance and work which can be seen below:

Work = Force × distance

12J = 3N × d

d = 12/3 = 4m.

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Answer:

A

Explanation:

The particles of the medium don't travel along with the wave.  and I also just did this question

Answer:

When boiling water, the temperature of molecules within the water increases    

Explanation: and they slowly begin to move at a rapid rate, upwards. ... The hot water molecules become less dense, and they rise above the denser cooler molecules. This movement of molecules creates convection currents.

The given position equation of a particle is, s=102−23t+t2Here, t is in seconds.The first derivative of s is its velocity: v = ds/dt - 23 + 2tThe second derivative of s is the particle's acceleration:

The total distance traveled is the sum of all distance travelled in each interval of time: D = ∫|v|dt + ∫|v|dt + ... (From t 0 s to t = 17 s)From the graph of the velocity, the particle is moving to the left and toward the right side. The velocity-time graph is linear, which means that the acceleration is constant and uniform. Hence, the conclusion we can draw is that the particle is moving with uniform acceleration.  a = dv/dt = 2The given equations are: s=102−23t+t2v

=−23+2ta

=2Plot of s-t and v-t graphs:

Net displacement (Δs) during the interval of 17 seconds can be calculated as:Δs = s(17) - s(0)Δs 

= 102 - (2 × 3² / 2) - (23 × 3) + (17² / 2) - 102Δs 

= -4 mmThe particle moves to the left side.

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If we drop the object, the distance it will fall in time t is 4.9t²

From the question given above, the following data were obtained:

The distance the object will fall in time t can be obtained as follow:

h = ut + ½gt²

h = (0 × t) + (½ × 9.8 × t²)

Clear bracket

h = 0 + (4.9 × t²)

h = 0 + 4.9t²

h = 4.9t²

From the illustration made above, we can conclude that the distance the object will fall in time t is 4.9t²

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10 kHz is the frequency does a stationary listener hear the sound as the plane approaches.

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(a) The time interval for which the exam lasts, as measured by the students on spacecraft I, is 88.243 minutes.

When two observers are in relative motion, time dilation occurs due to the theory of special relativity. In this scenario, the students on spacecraft I are moving relative to an observer on Earth. To calculate the time interval measured by the students, we can use the time dilation formula:

Δt₁ = Δt₀ / γ

where Δt₁ is the time interval measured by the students, Δt₀ is the time interval measured by an observer on Earth, and γ is the Lorentz factor given by:

γ = 1 / sqrt(1 - (v/c)²)

Given that the speed of spacecraft I is 0.680c relative to Earth, we can substitute the values into the formulas. Solving for Δt₁, we find that the exam lasts 88.243 minutes as measured by the students on spacecraft I.

(b) The time interval for which the exam lasts, as measured by an observer on Earth, is 55.626 minutes.

Explanation:

When two observers are in relative motion, time dilation occurs due to the theory of special relativity. In this scenario, the professors on spacecraft II are moving relative to an observer on Earth. To calculate the time interval measured by the observer on Earth, we can use the time dilation formula:

Δt₁ = Δt₀ / γ

where Δt₁ is the time interval measured by an observer on Earth, Δt₀ is the time interval measured by the professors on spacecraft II, and γ is the Lorentz factor given by:

γ = 1 / sqrt(1 - (v/c)²)

Given that the speed of spacecraft II is 0.240c relative to Earth, we can substitute the values into the formulas. Solving for Δt₁, we find that the exam lasts 55.626 minutes as measured by an observer on Earth.

(c) If one of the professors proctored the exam by traveling on spacecraft I and stopped the exam after 54.0 minutes elapsed on her clock, the time interval for which the exam lasts, as measured by the professors on spacecraft II, can be calculated using the time dilation formula:

Δt₁ = Δt₀ / γ

where Δt₁ is the time interval measured by the professors on spacecraft II, Δt₀ is the time interval measured by the professor on spacecraft I, and γ is the Lorentz factor given by:

γ = 1 / sqrt(1 - (v/c)²)

Since the speed of spacecraft, I is 0.680c relative to Earth, we can substitute the values into the formulas. However, the time interval measured by the professor on spacecraft I is not provided, so we cannot determine the time interval measured by the professors on spacecraft II.

(d) Without knowing the time interval measured by the professor on spacecraft I, we cannot determine the time interval for which the exam lasts as measured by an observer on Earth in this scenario.

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Complete question here:

Spacecraft I, containing students taking a physics exam, approaches the Earth with a speed of 0.680c (relative to the Earth), while spacecraft II, containing professors proctoring the exam, moves at 0.240c (relative to the Earth) directly toward the students. The professors stop the exam after 54.0 min have passed on their clock.

(a) For what time interval (in minutes) does the exam last as measured by the students? 88.243 min

(b) For what time interval (in minutes) does the exam last as measured by an observer on Earth? 55.626 min What If? Suppose one of the professors proctored the exam by traveling on spacecraft I and stopped the exam after 54.0 min elapsed on her clock.

(C) For what time interval (in minutes) does the exam last as measured by the professors on spacecraft II? min

(d) For what time interval (in minutes) does the exam last as measured by an observer on Earth? min

Answer:

A or B im prettu sure

Explanation:

a is a very commom answer so id go with that

Optimistic people see the glass as half full. The correct option is A.

Optimists look for the bright side of things. They anticipate a positive outcome. They believe they have the skill and ability to bring about positive change.

A person can also be generally optimistic; as part of their personality, they are always looking on the bright side and expecting the best.

An optimist is someone like this. Optimism is a general positive attitude that sees the glass as half full.

Optimism is a positive, hopeful outlook on the future, on yourself, and on the world around you. It is an important component of resilience, the inner strength that helps you get through difficult times. Optimism, by definition, helps you see, feel, and think positively.

Thus, the correct option is A.

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The speed of the bicyclist at the bottom of the hill is 27.5 m/s.

A bicyclist on an old bike with a combined mass of 83 kg is rolling down a hill of height 115 m.

During the course of the 345 meters of downhill road, she encounters a constant friction force of 239 newton.

The sum of the potential energy and the kinetic energy of the cyclist at the top of the hill will equal the sum of the potential energy and the kinetic energy at the bottom of the hill, as there is no external work done on the cyclist.

Hence, the kinetic energy at the bottom of the hill will be equal to the initial kinetic energy minus the work done due to the frictional force.

The potential energy of the cyclist at the top of the hill = mgh ,where m = 83 kg, g = 9.8 m/s², and h = 115 m

Potential energy = (83 kg) × (9.8 m/s²) × (115 m) = 99749 Joules

At the top of the hill, the total energy of the cyclist is the sum of the potential energy and the kinetic energy.

E_total = 99749 + (1/2) × (83 kg) × (9 m/s)²

E_total = 99749 + 33417.75

E_total = 133166.75 Joules

At the bottom of the hill, the total energy of the cyclist is the sum of the kinetic energy and the work done by friction.

E_total = (1/2) × (83 kg) × v² + (239 N) × (345 m)

E_total = (1/2) × (83 kg) × v² + 82355 Joules

The potential energy of the cyclist has transformed to kinetic energy and work done due to friction.

Equating the two equations above:133166.75 = (1/2) × (83 kg) × v² + 82355Solving for v:249.25 = (1/2) × (83 kg) × v²v² = 5999v = 27.5 m/s

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Answer:

160 m

Explanation:

For computing the distance first we have to determine the acceleration which is shown below:

v = u + at

where,

v = 0 = initial velocity

u = 40 = uniformly to a speed

t = 8 = time

So, the a is

= 40 ÷ 8

= 5 m/s

Now we have to use the distance formula i.e

d = ut + 1 ÷ 2 at^2

= 1 ÷2 × 5 × 8^2

= 160 m

Basically we have to applied the above formulas so that the distance could come

The distance at which the crane lift a crate is 0.0781 m.

Distnace is the lenght between two points.

To determine the distance the crane can lift the crate, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the distance is 0.0781 m.

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The average energy of solid (approximated as a collection of N 3-D oscillators) is given by :

U = 3N<E> = (3Nε) / (e^(ikT) - 1)

There are N distinct particles and 3N various modes in three dimensions, where the three possible polarizations account for the factor of 3. Since they must be indistinguishable, you must symmetrize with respect to the various particles, which is why you only have a factor of N! on the bottom. However, you have a factor of 3N because there are 3N modes.

The quantum-mechanical equivalent of the classical harmonic oscillator is the quantum harmonic oscillator. One of the most crucial model systems in quantum mechanics is an arbitrary smooth potential because it may frequently be represented as a harmonic potential at a stable equilibrium point. It is also one of the few quantum-mechanical systems that has a precise, analytical solution.

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Answer:-20.m/s,drum

Explanation:

https://mrsstowell.weebly.com/uploads/5/1/4/6/51460659/hsav_riverdale.k12.wi.us_20160330_092123.pdf

Many physical systems in nature are well-described as a simple harmonic oscillator because they exhibit a restoring force that is directly proportional to the displacement from the equilibrium position.

The simple harmonic oscillator is a model that describes the behavior of many physical systems in nature, including springs, pendulums, and vibrating atoms or molecules. This is because many systems in nature can be modeled as having a restoring force that is proportional to the displacement from an equilibrium position and acts in the opposite direction to the displacement.

This restoring force causes the system to oscillate back and forth around the equilibrium position, and the motion of the system can be described using the principles of harmonic motion. Additionally, the equations that describe simple harmonic motion have simple, elegant solutions, making it a useful and widely applicable model in physics. As a result, the simple harmonic oscillator model is often used to describe and analyze a wide range of physical systems in nature.

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yes could the half of the moon that faces the earth ever be completely dark

The moon is the captivating celestial body that can be observed through eyes, it is the biggest heavenly body in the night sky which changes its shape constantly.

The waning and waxing of moon cause the changes in face of the moon. The waning and waxing of the moon is directly related to the rotation of the moon.

The rotation of the moon is an extraordinary occurrence as the time the moon takes for completely orbiting the earth is exactly the time it takes to rotate once on its own axis.

Both the actions take 27.3 days to complete. This is unique feature of the moon is rare when compared to other celestial bodies. The energy given to the moon comes from Earth's rotation and to compensate, our planet is slowing down.

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