Use the following free body diagram to answer questions 8 and 9:

8. Calculate the net force acting on the cart.
A 0N
B. +60N
C -60N
D. +240 N
9. Calculate the cart's rate of acceleration
A +60 m/s
B. +8 m/s
C.-2m/s
D. +2 m/s

Answer:

3A. This phenomenon can be seen in the discrete emission of the molecules.

3B. The emotion of the atoms is observed, from states high in energy to a state of minimum energy that is stable indefinitely.

3C. When an electron beam passes through an inhomogeneous magnetic field, it is divided into only two beams

3D. This is due to the stimulated emission

3E. The penetration of a potential barrier is observed in the radioactive emission of heavy atoms, where an alpha particle (Helium nucleus)

Explanation:

This problem asks for some experimental explanations of various quantum phenomena.

3A. This phenomenon can be seen in the discrete emission of the molecules.

In the classical explanation all states or energies are allowed, therefore when emitting energy (photons) there should be a continuum, this is not observed

In the correct quantum explanation only some states are allowed, therefore the emission must be discrete, which is observed in the emission or absorption of molecules and atoms

3B. The emotion of the atoms is observed, from states high in energy to a state of minimum energy that is stable indefinitely.

The incorrect classical explanation that if the minimum energy was zero the electrons cannot rotate around the nuclei and the atom collapses, this does not happen

3C. When an electron beam passes through an inhomogeneous magnetic field, it is divided into only two beams, which is evidence of the existence of two discrete states that we call spin, remember that a free electron beam has zero angular momentum.

3D. This is due to the stimulated emission that occurs when a photon passes through the emission zone, causing the atoms to have transitions and these emitted photons have the same initial photon location, the laser beam all photons are in phase and therefore it is coherent .

This is widely used for holographic and interference work

3E. The penetration of a potential barrier is observed in the radioactive emission of heavy atoms, where an alpha particle (Helium nucleus) leaves the atomic nucleus penetrating the barrier since its energy is lower than the nuclear barrier potential.

Answer:

Explanation:

Please provide some experimental demonstration or explanation of natural observations or characteristics of application which confirm(s) / display(s) the. Quantum mechanics is a fundamental theory in physics that provides a description of the physical properties of nature. Quantum mechanics arose gradually from theories to explain observations. By M Arndt · 2009 · Cited by 239 — Wave-particle duality of light: A paradigmatic example for this fact is the dual nature of light that manifests itself in Young's famous double-slit diffraction.

Answer:

The value  [tex]F  =  0.1396 \ N [/tex]

Explanation:

From the question we are told that

   The volume blood  ejected is  [tex]V = 65 \ cm^3 = 65*10^{-6} \ m^3[/tex]

    The velocity of the blood ejected is  [tex]v = 103 \ cm/s = \frac{103}{100} = 1.03 \ m/s[/tex]

    The density of blood is  [tex]\rho = 1060 \ kg/m^3[/tex]

     The heart beat is [tex]R = 59 \ bpm(beats \ per \ minute) = \frac{59}{60}= 0.9833\ bps[/tex]

The average force exerted by the blood on the wall of the aorta is mathematically represented as

      [tex]F = 2 * \rho * V * R * v[/tex]

=>    [tex]F  =  2 * 1060  *  65*10^{-6}  *  0.9833 *  1.03[/tex]

=>    [tex]F  =  0.1396 \ N [/tex]

Answer:

A. Join glycerol to fatty acids

Explanation:

I majored in Physics.

Answer:

1.04*10⁻⁵

Explanation:

light wave do showcase some behaviors whenever there is encounters with the end of the medium, some of the behaviors are - reflection, refraction, as well as diffraction. When visible light wave strikes a boundary that exist two different media, a portion of the energy will be transmitted into the new medium and some reflected.

Reflection of a light wave can be regarded as bouncing off of light wave from boundary. refraction on other hand is bending of the path of a light wave.

We were to calculate the reflectivity at the boundary,

reflectivity at the boundary can be calculated using the expression below

Reflectivity= (n₂ - n₁)² /(n₂ + n₁ ) ²

where

n₁= Indices of refraction at first grain= 1.545

n₂= Indices of refraction at second grain=

1.555

(1.555 - 1.545)² / (1.555 - 1.545)²

=(0.01)²/ (3.1)²

= 0.0001/ 9.61

= 1.04*10⁻⁵

Hence, the reflectivity at the boundary if the indices of refraction for the two grains are 1.545 and 1.555 in the direction of light propagation is 1.04*10⁻⁵

Answer:

Explanation: The chemical action that occurs in the cell while the current is flowing causes hydrogen bubbles to form on the surface of the anode. This action is called POLARIZATION. Some hydrogen bubbles rise to the surface of the electrolyte and escape into the air, some remain on the surface of the anode. If enough bubbles remain around the anode, the bubbles form a barrier that increases internal resistance. When the internal resistance of the cell increases, the output current is decreased and the voltage of the cell also decreases.

   A cell that is heavily polarized has no useful output. There are several methods to prevent polarization or to depolarize the cell.

   One method uses a vent on the cell to permit the hydrogen to escape into the air. A disadvantage of this method is that hydrogen is not available to reform into the electrolyte during recharging. This problem is solved by adding water to the electrolyte, such as in an automobile battery. A second method is to use material that is rich in oxygen, such as manganese dioxide, which supplies free oxygen to combine with the hydrogen and form water.

   A third method is to use a material that will absorb the hydrogen, such as calcium. The calcium releases hydrogen during the charging process. All three methods remove enough hydrogen so that the cell is practically free from polarization.

LOCAL ACTION

   When the external circuit is removed, the current ceases to flow, and, theoretically, all chemical action within the cell stops. However, commercial zinc contains many impurities, such as iron, carbon, lead, and arsenic. These impurities form many small electrical cells within the zinc electrode in which current flows between the zinc and its impurities. Thus, the chemical action continues even though the cell itself is not connected to a load.

   Local action may be prevented by using pure zinc (which is not practical), by coating the zinc with mercury, or by adding a small percentage of mercury to the zinc during the manufacturing process. The treatment of the zinc with mercury is called amalgamating (mixing) the zinc. Since mercury is many times heavier than an equal volume of water, small particles of impurities weighing less than mercury will float to the surface of the mercury. The removal of these impurities from the zinc prevents local action. The mercury is not readily acted upon by the acid. When the cell is delivering current to a load, the mercury continues to act on the impurities in the zinc. This causes the impurities to leave the surface of the zinc electrode and float to the surface of the mercury. This process greatly increases the storage life of the cell.

Answer:

A

Explanation:

Explanation:

Given that,

Magnitude of vector A, |A| = 15

Magnitude of vector B, |B| = 25

We need to find the magnitude of this sum.

The maximum sum of the resultant vector,

[tex]R_{max}=|A_1|+|A_2|\\\\=15+25\\\\=45[/tex]

The minimum sum of the resultant vector,

[tex]R_{min}=|A_1|-|A_2|\\\\=15-25\\\\=-10[/tex]

So, the magnitude of this sum either 45 or -10.

Answer:

Valence electrons are always located in the outer most energy level.

Explanation:

Valence electrons are the ones that are involved in chemical bonds. In order to take part in a chemical bonding, the outermost/valence electron needs to be involved. Thus, the answer is Valence electrons are always located in the outer most energy level.

Answer:

1)    x_{cm} = 5 m ,  2)  I = 168.32 kg m²

Explanation:

1) An important concept of center of mass is

          [tex]x_{cm} = \frac{1}{M} \sum x_{i} m_{i}[/tex]

where M is the total mass of the system

Let's apply this equation to our case, suppose that all masses are equal and are worth 1 kg

         [tex]x_{cm}[/tex] = ⅓ (1 0 + 1 5 + 1 10)

         x_{cm} = 5 m

2) In this para indicates that there is an axis of rotation at the point xo = 3.8 m and they ask to calculate the moment inertia.

Let's use the parallel axes theorem

          I =    I_{cm} + M D

where    I_{cm}  is the moment of inertia with respect to the center of mass, D the distance between the two axes of rotation and M the total mass of the system

Let's look for the moment of inertia of the center of mass

         [tex]I_{cm}[/tex] = 1  0 + 1  5² + 1  10²

         I_{cm} = 125 kg m²

the total moment of inertia is

        I = 125 + 3 3.8²

        I = 168.32 kg m²

The moment of inertia of the 3-mass system about the new axis is 54.32 kgm/s².

We have three masses each of mass = 1kg such that they are in line with mass m at origin, m at 5m and m at 10m

(a) The center of mass:

[tex]X=\frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} \\ \\ X =\frac{ 1*0+1*5+1*10}{1+1+1}\\ \\ X = 5m[/tex]

Hence the center of mass of the system is at x = 5m.

(b) The moment of inertia about the axis passing through x = 3.8m

from the parallel axis theorem:

[tex]I = I_{cm} + Md^2[/tex]

where, [tex]I_{cm}[/tex] is the moment of inertial along an axis passing through the center of mass of the system, M is the total mass of the system and d is the distance of the given axis from center of mass.

M = 3kg

d = 5 - 3.8 = 1.2m

[tex]I_{cm}=1*5^2+1*0+1*5^2\\\\ I_{cm}=50 kgm/s^2[/tex]

Md² = 3×(1.2)²

Md² = 4.32 kgm/s²

I = 50 + 4.32

I = 54.32 kgm/s² is the moment of inertial about the given axis.  

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Answer:

The minimum coefficient of static friction required, µ = 0.10

Note. The question is incomplete. The complete question is given below:

While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll sideways, the police claim that the log could have slid out the back of the truck when accelerating from rest. The driver claims that the truck could not possibly accelerate at the level needed to achieve such an effect. Regardless, the police write a ticket anyway and now the driver court date is approaching.

The log has a mass of m = 929 kg; the truck has a mass of M = 8850 kg. According to the truck manufacturer, the truck can accelerate from 0 to 55 mph in 23.0 seconds, but this does not account for the additional mass of the log. Calculate the minimum coefficient of static friction μs needed to keep the log in the back of the truck.

Explanation:

First, velocity in mph is converted to m/s

1 mph = 0.447 m/s

55 mph ≈ 24.6 m/s

The acceleration of the empty truck is a = v/t = 24.6 / 23 = 1.07 m/s²

Force that can be generated by the truck, F = ma

F = 8850kg * 1.07 m/s² = 9469.5 N

However, with the added mass of the log on it, the acceleration of the truck will become;

a = F / m = 9469.5 N / (8550+929)kg = 0.97 m/s²

Frictional force between the log and the truck = 0.97 m/s² * 929 kg = 901.13 N

Normal reaction on the truck due to the weight of the log, R = mg

R = 929 kg * 9.8m/s² = 9104.2 N

Coefficient of static friction, µ = F/R

µ = 901.13/9104.2

µ = 0.098 ≈ 0.10

Therefore, the minimum static friction required is µ = 0.10

Answer:

3.1 m/s

Explanation:

The total distance she has to run is the addition of the three lengths:

47 + 63 + 76 = 186 meters.

She needs to cover it one minute (60 seconds). Therefore her speed must be:

186 m / 60 s = 3.1 m/s

Answer:

a

 [tex]\Delta P = 7558.6 \ Pa[/tex]

b

 [tex]h_1 = 10 \ m[/tex]

Explanation:

From the question we are told that

   The position of the column of mercury in the barometer is  [tex]h = 736 \ mm = 0.76 \ m[/tex]\

   The density of mercury is  [tex]\rho = 13,000 \ kg / m^3[/tex]

Generally the  pressure of the atmosphere at that column is mathematically represented as  

         [tex]P = \rho * g * h[/tex]

=>      [tex]P =13 000 * 9.8 * 0.736[/tex]

=>      [tex]P = 93766.4 \ Pa[/tex]

Generally the atmospheric pressure at sea level (Generally the pressure before the change in level of the mercury column)  is  [tex]P_a = 101325 \ Pa[/tex]

Generally the change in air pressure  is mathematically represented as

      [tex]\Delta P = P_a - P[/tex]

=>   [tex]\Delta P = 101325 - 93766.4[/tex]

=>   [tex]\Delta P = 7558.6 \ Pa[/tex]

Generally the height which the column will rise to is mathematically evaluated as  

         [tex]h_1 = \frac{P}{ \rho_w * g }[/tex]

Here [tex]\rho_w[/tex] is the density of water with value  [tex]\rho_w = 1000 \ kg/m^3[/tex]

So

       [tex]h_1 = \frac{ 93766.4}{ 1000 * 9.8 }[/tex]

=>   [tex]h_1 = 10 \ m[/tex]

Answer:

PLEASE MARK AS BRAINLIEST!!

Explanation:

ANSWER IS IN THE IMG BELOW

Answer: fossil fuels are using more in the existing world because fossil fuels are of great importance because they can burned producing significant amount of energy prr unit mass the use of coal as a fuel prefates recorded history

Explanation:

Answer:

A fossil fuel is a fuel formed by natural processes, such as anaerobic decomposition of buried dead organisms, containing organic molecules originating in ancient photosynthesis[1] that release energy in combustion.[2] Such organisms and their resulting fossil fuels typically have an age of millions of years, and sometimes more than 650 million years.[3] Fossil fuels contain high percentages of carbon and include petroleum, coal, and natural gas.[4] Peat is also sometimes considered a fossil fuel.[5] Commonly used derivatives of fossil fuels include kerosene and propane. Fossil fuels range from volatile materials with low carbon-to-hydrogen ratios (like methane), to liquids (like petroleum), to nonvolatile materials composed of almost pure carbon, like anthracite coal. Methane can be found in hydrocarbon fields alone, associated with oil, or in the form of methane clathrates.

Coal, a fossil fuel

As of 2018, the world's main primary energy sources consisted of petroleum (34%), coal (27%), and natural gas (24%), amounting to an 85% share for fossil fuels in primary energy consumption in the world. Non-fossil sources included nuclear (4.4%), hydroelectric (6.8%), and other renewables (4.0%, including geothermal, solar, tidal, wind, wood, and waste).[6] The share of renewables (including traditional biomass) in the world's total final energy consumption was 18% in 2018.[7] Compared with 2017, world energy-consumption grew at a rate of 2.9%, almost double its 10-year average of 1.5% per year, and the fastest since 2010.[8]

Although fossil fuels are continually formed by natural processes, they are generally classified as non-renewable resources because they take millions of years to form and known viable reserves are being depleted much faster than new ones are generated.[9][10]

Most air pollution deaths are due to fossil fuel combustion products, it is estimated to cost over 3% of global GDP,[11] and fossil fuel phase-out would save 3.6 million lives each year.[12]

The use of fossil fuels raises serious environmental concerns. The burning of fossil fuels produces around 35 billion tonnes (35 gigatonnes) of carbon dioxide (CO2) per year.[13] It is estimated that natural processes can only absorb a small part of that amount, so there is a net increase of many billion tonnes of atmospheric carbon dioxide per year.[14] CO2 is a greenhouse gas that increases radiative forcing and contributes to global warming and ocean acidification. A global movement towards the generation of low-carbon renewable energy is underway to help reduce global greenhouse-gas emissions.

Explanation:

To derive an equation you must indicate the variable you want to solve for.

Here we have tension of an object A and Tension 1.

Two variables or unknown are given hence we cannot derive any other equations.

Answer:

The value is  [tex]m = 69.24 \ kg[/tex]

Explanation:

From the question we are told that

   The value of the external force is  [tex]F = 59.0 \ N[/tex]

    The magnitude of the astronaut's acceleration is  [tex]a = 0.852 \ m/s[/tex]

Generally Newton's Second Law of Motion from the mass of the astronauts is mathematically represented as

            [tex]m = \frac{F}{a}[/tex]

=>         [tex]m = \frac{59 }{0.852 }[/tex]

=>         [tex]m = 69.24 \ kg[/tex]  

Answer:

you can show them a vid

Explanation:

Explanation:

wavelength =200cm=2m

frequency is reciprocal of wavelength

f=1/2

f=0.5Hz or 1/s

The frequency of the wave with a Wavelength of 200 cm should be considered as the f=0.5Hz or 1/s.

Since the wavelength is 200 cm or 2m

Also we know that

frequency is reciprocal of wavelength

So,

f=1/2

f=0.5Hz

Hence, The frequency of the wave with a Wavelength of 200 cm should be considered as the f=0.5Hz or 1/s.

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Answer:

When an object moves with a constant speed in a circular path, then its motion is called as Circular motion.

Examples of Uniform Circular Motion :--

The moon moves in uniform circular motion around the earth.

A stoned tied to a thread and rotated in circular motion.

An athlete running on a circular track.

A satellite revolving around Earth.

Answer:

Explanation:Capillary action is the ability of a liquid to flow in narrow spaces without the assistance of, ... This article is about the physical phenomenon. ... If the diameter of the tube is sufficiently small, then the combination of surface tension (which is caused by cohesion ... They derived the Young–Laplace equation of capillary action.

Answer:

62.5m

Explanation:

Given parameters:

Initial velocity  = 0m/s

Time  = 5s

Acceleration  = 5m/s²

Unknown:

Distance traveled  = ?

Solution:

To solve this problem, we use the motion equation given below:

S = ut + [tex]\frac{1}{2}[/tex] at²

S is the distance traveled

u is the initial velocity

a is the acceleration

t is the time taken

Now, insert the parameters and solve;

 S =( 0 x 5) +( [tex]\frac{1}{2}[/tex] x 5 x 5²) = 62.5m

Answer:

Explanation:

The average kinetic energy can be found by using the formula

[tex]k = \frac{1}{2} m {v}^{2} \\ [/tex]

m is the mass

v is the velocity

From the question we have

[tex]k = \frac{1}{2} \times 200 \times {5}^{2} \\ = 100 \times 25[/tex]

We have the final answer is

Hope this helps you

Answer:

20 m/s.

Explanation:

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

Distance travelled = 600 m

Time taken = 30 s

Speed =?

Speed can be defined as the distance travelled per unit time. Mathematically, it is expressed as:

Speed = Distance /time

With the above formula, we can easily calculate the speed of the bus as follow:

Distance travelled = 600 m

Time taken = 30 s

Speed =?

Speed = Distance /time

Speed = 600 / 30

Speed = 20 m/s

Therefore, the speed of the bus is 20 m/s.

Answer:

same but from what i know is the newtons thingy

Explanation:

hope that helps!

The Pelican  must aim differently than where the fish appears to be because light is refracted from air to water.

Refraction is the change in the direction of a wave at the interface between two media. The effects of refraction are easily seen when light is travelling from air to water hence objects are not really where they appear to be when viewed above the water surface.

Hence, the Pelican  must aim differently than where the fish appears to be because light is refracted from air to water.

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Answer:

PART A: Maximum speed = 0.314 m/s

PART B: Maximum acceleration = 1.23 m/s²

Explanation:

A simple Harmonic motion is a repetitive motion through an equilibrium point.

Amplitude = 8.0cm = 8/100 = 0.08m (highest displacement)

period (T) = 0.80s

A) maximum speed [tex](V_{max)[/tex]

[tex]V_{max} = 2\pi fA\\where:\\A = Amplitude = 0.08m\\f = frequency = \frac{1}{period(T)} = \frac{1}{0.8} = 1.25 Hz\\\therefore 2\pi fA = 2\pi \times 1.25 \times 0.08\\= 0.314\ m/s[/tex]

B) maximum acceleration [tex](a_{max})[/tex]

[tex]a_{max} = (2\pi f)^2A\\where:\\f = 1.25Hz\\A = 0.08m\\a_{max} = (2\pi \times 1.25)^2 \times 0.08\\= 1.23\ m/s^2[/tex]

Answer:

Informal education is important because it can help individual to learn how to react and control situations.

It help individual to improve on its existing knowledge, new skills or ideas. This kind of education can happen any where and it can add values to the learner.

Explanation:

Informal education is a type of education that is learned from different life experiences, happenings outside a structured curriculum.

Informal education is important because it can help individual to learn how to react and control situations.

It help individual to improve on its existing knowledge, new skills or ideas. This kind of education can happen any where and it can add values to the learner.

Answer:

The minimum speed needed at the top of the circle so that the cord remains tensioned and the ball's path remains circular is approximately is 9.903 meters per second.

Explanation:

By the Principle of Energy Conservation we understand that the minimum speed needed by the ball is that speed such that maximum height reached is equal to the diameter of the vertical circle, that is:

[tex]K =U_{g}[/tex] (1)

Where:

[tex]K[/tex] - Translational kinetic energy, measured in joules.

[tex]U_{g}[/tex] - Gravitational potential energy, measured in joules.

By definitions of translational kinetic and gravitational potential energies, we expand the equation above and clear the initial speed of the ball:

[tex]\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g\cdot h[/tex]

[tex]v = \sqrt{2\cdot g\cdot h}[/tex] (2)

Where:

[tex]m[/tex] - Mass, measured in kilograms.

[tex]v[/tex] - Initial speed, measured in meters per second.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]h[/tex] - Maximum height of the ball, measured in meters.

If we know that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]h = 5\,m[/tex], then the initial speed of the ball is:

[tex]v = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (5\,m)}[/tex]

[tex]v\approx 9.903\,\frac{m}{s}[/tex]

The minimum speed needed at the top of the circle so that the cord remains tensioned and the ball's path remains circular is approximately is 9.903 meters per second.