Make a force diagram for the following scenarios:
A car is accelerating to the left at a rate of -3.5 m/s²

Option (C) is correct. \(T^{2}\) varies directly as \(R^{3}\) .

This is based on Kepler's law of Planetary motion.

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Answer:  C. R3

Explanation: plato :)

Answer: 1.Hotel/Motel Clerk 2.Hotel/Motel Supervisor or Manager  3.Housekeeper/Cleaning Worker 4.Concierge  5.Porter/Bellhop

A successfully professionalized field of employment is referred to as a profession.

It can be characterized as a disciplined team of people, professionals, who uphold ethical standards, present themselves as, and are recognized by the public as possessing, special knowledge and skills in a widely acknowledged body of learning derived from research, education, and training at a high level, and who are prepared to apply this knowledge and exercise these skills in the interest of others.

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Answer:

Net force is always m•a. In this case, the velocity is constant so the acceleration is zero and the net force is zero.

Explanation:

An example of a field force is an object that is falling freely.

In the most elementary sense of the word, we define force in physics as a push or a pull. There are basically two kinds of force that acts on an object and these two kinds of force are the field force and the contact force.

When we talk about the field force, we are talking about the kind of force that does not need to make any contact with the object upon which it does act. The contact force is the force that must have to make some kind of contact with the object upon which it must act.

Thus, an object that if falling is acted upon by the earth's gravitational force so it is a field force.

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Answer:

490 W

Explanation:

F = ma = (50 kg)(9.8 m/s²) = 490 N    (Weight = W = mg)

Work = W = Fd = (490 N)(5 m) = 2450 J

Power = W/t = 2450 J / 5 s = 490 J/s = 490 W  (watts)

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The units of supplementary quantities are dimensionless units.

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In the shells of the atom, we find the electrons while in the nucleus of the atom, we find the protons.

We define the electron as the smallest part of a substance that is able to take part in a chemical reaction. Thus the atoms of elements are quite different from each other but there are some of the feature of the atom that is general to all the atoms.

The atoms are made up of the sub atomic particles such as the electrons, the protons and the neutrons. Now we know that the protons and the protons can be found in the nucleus of the atom while the electrons are in the shells.

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The potential energy for a dust particle of mass 5.00×10−9kg and charge 2.00 nc at the position in part d is determined by the electric potential at that point and the charge of the particle.

The electric potential at a point in space is the amount of potential energy per unit charge that a particle would have if it were located at that point. It is measured in volts (V) and is a scalar quantity. The electric potential at a point due to a point charge q at a distance r from the charge is given by the equation: V = kq/r.

To find the potential energy, we first need to know the electric potential (V) at the position in part d. Unfortunately, you have not provided information about part d or the electric potential at that position. Once you have the value of V, you can proceed with the calculation. Assuming you have the electric potential value (V), you can now calculate the potential energy (U) using the formula U = qV. First, convert the charge of the dust particle from nC to C (Coulombs) by multiplying by 10^(-9), so 2.00 nC = 2.00 × 10^(-9) C. Then, plug the values of q and V into the formula to find the potential energy (U).
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The heat of fusion of lead is (a) 227°C.To solve this problem, we need to use the principle of conservation of energy, which states that the total energy in a closed system remains constant.

Option D is Correct answer. The ultimate temperature is well below the lead's (327°C) melting point.

To find the final temperature of the lead bullet, we need to use the equation for heat transfer
Q = mCΔT
where Q is the heat transferred, m is the mass of the bullet, C is the specific heat of lead, and ΔT is the change in temperature.
First, we need to find the initial temperature of the bullet. We are given that the air temperature is 20°C, but we don't know the initial temperature of the bullet. However, we can assume that the bullet is at the same temperature as the air before it strikes the steel plate. Therefore, the initial temperature of the bullet is also 20°C.
Next, we need to find the heat transferred when the bullet strikes the steel plate. We can assume that all of the kinetic energy of the bullet is converted into heat when it stops. Therefore:
Q = 0.5mv²
where m is the mass of the bullet and v is its velocity. Plugging in the values, we get:
Q = 0.5×(5 g)×(300 m/s)² = 2.25 J
Finally, we can use the equation for heat transfer to find the final temperature of the bullet:
ΔT = Q / (mC)
ΔT = 2.25 J / (5 g x 0.128 J/g °C) = 3.52°C
Therefore, the final temperature of the lead bullet is 20°C - 3.52°C = 16.48°C.
Note that this final temperature is well below the melting point of lead (327°C), so the bullet will remain solid. The heat of fusion of lead is not needed to solve this problem.

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Answer: No

Explanation:

Whenever light travelling on a straight line encounters obstruction, it diffracts and scatter.

Scattering of light occurs when light passes through a rough path or a diffused surface.

But in case of spectral diffusion, which is the fluctuation in spectroscopy as a result of time dependent frequency shifts.

Spectral diffusion occurs in particular molecules initiated by excessive excitation energy.

Fluctuation in frequency does not mean diffraction of light or particles

Therefore, spectral diffusion does not cause light to scatter.

Answer:

Most minerals can be characterized and classified by their unique physical properties: hardness, luster, color, streak, specific gravity, cleavage, fracture, and tenacity.

The interaction of an attracting or repulsive force between two charged bodies is referred to as an electric force.

This force is comparable to other forces in that it strikes and affects a specific item, and it is simple to illustrate using Newton's law of motion. One of the forces acting on other bodies is the electric force.

Electric force is the interaction of any two charged bodies that is either repellant or attracting. Similar to any force, Newton's laws of motion describe how it affects the target body and how it does so. The list of additional forces that apply to objects includes the electric force.

The electrostatic force between the two electrons is one of repulsion since electrons are negatively charged and items with like charges repel one another.

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Answer of the all following horizontal force question as follows.

Horizontal force is a type of force that acts parallel to the ground, or in a horizontal direction. It is also known as a lateral force. This type of force is often encountered in everyday life and in physics.

a) The height of Aaron's kick can be calculated using the equation: h = (v^2sin2θ)/2g, where v is the initial velocity, θ is the angle of projection and g is the acceleration due to gravity. Plugging in the given values, we get h = (67^2sin(43))/ (2*9.8) = 36.3 m.

b) The time of travel can be found using the equation: t = (2v sinθ)/g. Plugging in the given values, we get t = (2*67sin(43))/9.8 = 7.72 s.

c) The distance of the kick can be found using the equation: d = (v^2sin2θ)/g. Plugging in the given values, we get d = (67^2sin(2*43))/9.8 = 130.3 m.

18-

a. Hailey kicked the ball to a height of 21.5 m.

b. The time of travel in the air was 2.5 s.

c. The ball hit the ground 91 m away from the goal. Since she was standing 100 m away, she will not score a goal.

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Answer:

Earth’s gravitational force is stronger

Explanation:

Earth’s gravitational force is stronger than the moon because of the presence of more massive objects in earth; on the flip side, the moon’s gravity is weaker than the earth due to smaller in size as compared to the earth. Earth gravity is denoted by g; conversely, moon gravity is denoted by gm

Answer: The greater an object's mass, the more gravitational force it exerts. So, to begin answering your question, Earth has a greater gravitational pull than the moon simply because the Earth is more massive. That's why the moon isn't pulled out of Earth's orbit by the gravity of larger planets or by the sun.

Explanation:

The speed of the airplane at the top point during the circular motion is 44.27 m/s.

The passengers feel weightless, because they are experiencing a free fall at that point.

Radius of the circular path, r = 200 m

Acceleration due to gravity, g = 9.8 m/s²

The passengers feel weightless just above the floor when the plane is at the top of the circular path, because they are experiencing a free fall at that point.

So, the centripetal force is given by,

mv²/r = mg

v²/r = g

Therefore, the speed at the top point is,

v = √rg

v = √(200 x 9.8)

v = √1960

v = 44.27 m/s

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The figure (4) is said to be in equilibrium with the force 2F in the centre and F on the top and bottom of the hockey puck.

For a body to be in equilibrium, the net force acting on the body and the net moment acting on the body must be zero.

Let us consider the net force acting on the body of pucks (force toward positive direction is positive and force towards negative direction is negative).

For puck 1, net force = 3F - F -2F = 0

For puck 2, net force = 2F + F - F = 2F

For puck 3, net force = F + 2F - F = 2F

For puck 4, net force = F + F - 2F = 0

For puck 5, net force = 3F - F - 2F = 0

In puck 1 and puck 5, net torque about the centre mass is not zero.

In puck 4, net torque about the centre mass is zero.

Thus, puck 4 is said to be in equilibrium.

The question is incomplete. The figure is missing in the question. It is attached in the below attachment.

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False. Slope can be expressed as either a fraction or a decimal. Both forms are valid and commonly used in mathematics. The choice of representation depends on the context and personal preference.

When slope is expressed as a fraction, it is in the form of rise over run. For example, a slope of 2/3 means that for every 2 units of vertical change (rise), there is 3 units of horizontal change (run). This fraction form can provide a clear visual understanding of the slope. On the other hand, slope can also be expressed as a decimal, which is the result of dividing the rise by the run. Decimal slope values are commonly used in calculations and equations. They can be more precise and convenient for certain mathematical operations. In summary, whether to express slope as a fraction or a decimal is a matter of choice and can vary based on the specific mathematical context or personal preference. Both forms are valid and serve different purposes in different situations.

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Answer:

All the planets would be remain in place and orbit in darkness

Explanation:

I'm assuming the sun didn't go nova, but it would sure kill off all life on Earth!

Have a great weekend!

Answer:

A- did this test :P

Explanation:

Answer:

what exactly are the following?

Answer:

a low protein diet helps you lose weight because lack of protein can make you lose muscle mass

Explanation:

It cuts your strength it makes it harder to keep your balance and slows your metabolism and later you get tired easily it leads to muscle wasting overtime and sheds your fat

With the use of K.E formula, the correct answers are:

a.) At point B, velocity = 2m/s

b.) At point C, velocity = 4m/s

C.) At point D, velocity = 3.5 m/s

POTENTION ENERGY

At maximum energy, total energy will be equal to maximum kinetic energy.

Given that a 2.5 kg mass starts from rest at point A and moves along the x-axis subject to the potential energy shown in the given figure.

Let us assume the that their maximum potential energies are given.

Total energy = maximum P.E = Maximum K.E

According to the graph, the P.E at point B,C, and D are:

Since Maximum P.E = Maximum K.E = 1/2m\(v^{2}\)

At point B

5 = 1/2 x 2.5 x \(v^{2}\)

\(v^{2}\) = 10/2.5

\(v^{2}\) = 4

v = \(\sqrt{4}\)

v = 2 m/s

At point C

20 = 1/2 x 2.5 x \(v^{2}\)

\(v^{2}\) = 40/2.5

\(v^{2}\) = 16

v = \(\sqrt{16}\)

v = 4 m/s

At point D

15 = 1/2 x 2.5 x \(v^{2}\)

\(v^{2}\) = 30/2.5

\(v^{2}\) = 12

v = \(\sqrt{12}\)

v = 3.5 m/s

b.) The turning point should be any point where velocity is negative. Since there is no point where velocity is negative, let us consider the points where potential energy is increasing.

The turning points of the mass will be at point B and D

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Cameras that use adaptive optics provide higher spatial resolution images primarily because c. Deformable mirrors are used to correct the blurring due to Earth’s atmosphere.

Adaptive optics is a technique used in imaging systems to compensate for the distortion and blurring caused by Earth’s atmosphere. The atmosphere introduces turbulence and variations in the refractive index, which degrade the image quality of astronomical observations or even ground-based photography. To overcome this, adaptive optics systems use deformable mirrors that can rapidly adjust their shape to counteract the atmospheric turbulence. These mirrors have numerous actuators that change their surface shape in real-time, based on the feedback received from wavefront sensors.

By continuously measuring the distortion caused by the atmosphere and adjusting the deformable mirror accordingly, adaptive optics systems can effectively correct the blurring and improve the spatial resolution of the images. This allows for sharper and more detailed observations, enabling higher spatial resolution. Options a, b, d, and e are not the primary reasons why adaptive optics provides higher spatial resolution images. While operating above Earth’s atmosphere (a) can indeed improve image quality, it is not specific to adaptive optics. Infrared light (b) and correcting for chromatic aberration with composite lenses (d) are separate aspects of imaging technology. Simulating a larger telescope € is not the primary purpose of adaptive optics but rather achieving better image quality by compensating for atmospheric distortions.

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The Paschen series is a spectral series of hydrogen-like atoms that is similar to the Balmer series, but with a different value of the principal quantum number.

Specifically, the Paschen series corresponds to transitions between an excited state with principal quantum number n greater than or equal to 3, and the n=3 energy level in hydrogen-like atoms.

Like the Balmer series, the Paschen series also produces a series of spectral lines, which are the result of the emission of electromagnetic radiation as the electrons in the atom transition from higher energy levels to lower energy levels.

However, the wavelengths of the spectral lines in the Paschen series are longer than those in the Balmer series, due to the higher energy levels involved.

Overall, the Paschen series is an important aspect of the study of atomic physics, and provides valuable insights into the behavior of hydrogen-like atoms and their spectral emissions.

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Answer:

0.1 amps or A

Explanation:

I believe to calculate that you would have to use the formula V = IR and change it to:

I = V/R

I = 1.5/15

I = 0.1 amps or A

Answer:

Rigorous exercise

Explanation:

For your heart rate is been around 100 bpm you should be doing a rigorous exercise . but if ain't doing any rigorous exercise and your heart rate is constantly above 100 bpm you should consult your doctor.

An average heart is between 60 bpm to 100 bpm.

This statement is true. If the rock contained 6 grams of radium-226 at the closure temperature and 0.375 grams at the time of discovery, the remaining mass difference (6 - 0.375 = 5.625 grams) is attributed to the decay of radium-226 into radon-222. Therefore, the rock had 3.375 grams of radon-222 when it was discovered. Based on the given information, statement 5 is the only true statement.

To determine the statements that are true, let's consider the information provided.

The rock reached its closure temperature 6,400 years ago.

This statement is not true. The closure temperature is the temperature at which a mineral retains its daughter isotopes without any further loss or gain. However, the closure temperature is not directly related to the half-life of the parent isotope.

The rock reached its closure temperature 4,800 years ago.

This statement is not true. Similar to the previous statement, the closure temperature is not determined solely based on the half-life of the parent isotope.

The rock had 2.625 grams of radon-222 1,600 years ago.

This statement is not true. The amount of radon-222 present 1,600 years ago cannot be determined directly from the information provided.

When the rock was discovered, it had 5.625 grams of radon-222.

This statement is not true. The amount of radon-222 when the rock was discovered is given as 0.375 grams, not 5.625 grams.

When the rock was discovered, it had 3.375 grams of radon-222.

This statement is true. If the rock contained 6 grams of radium-226 at the closure temperature and 0.375 grams at the time of discovery, the remaining mass difference (6 - 0.375 = 5.625 grams) is attributed to the decay of radium-226 into radon-222. Therefore, the rock had 3.375 grams of radon-222 when it was discovered.

Based on the given information, statement 5 is the only true statement.

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Answer:

welll

Explanation:

i honestly dk

Answer:

b,d,f

Explanation:

Answer:

Aluminium

Explanation:

If we calculate the heat transfer for all the substances, we have

Mc(∆T)

Where M is mass

C is heat capacity

∆T is temperature change.

For iron

Mc(∆T) = 500 ×0.440×100= 22000J

For copper

Mc(∆T) = 500 ×0.385×100= 19250J

For aluminium

Mc(∆T) = 500 ×0.897×100= 44850J

Aluminium has the highest heat capacity and would take longest for this heat capacity to be dissipated under similar condition.