The electrical signals sent to the brain indicate the
loudness, pitch, and quality
pitch
of a sound wave.
loudness
quality

Given that a mass, [tex]\bold{m=0.275 \ kg}[/tex], swings in a circular path due to an attached string with a length, [tex]\bold{l=0.850 \ m}[/tex]. Refer to IMAGE #1 for a picture of the situation.

We are asked to answer the following...

(a) What forces always act on the mass throughout its swing?

(b) Draw force diagrams or free-body diagrams showing what forces are acting on the mass at the bottom of its swing vs the top.

(c) Given a velocity, [tex]\bold{\vec v = 5.20 \ m/s}[/tex], when the mass is at the top of its swing, find the tension in the string, [tex]\vec T[/tex].  

(d) Given the maximum tension,[tex]\bold{\vec T_{max}=22.5 N}[/tex], what is the fastest the mass can travel at the bottom of its swing, [tex]\vec v_{max}[/tex]?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

For part (a):

To answer part a lets list of all forces that can act on an object and then narrow them down.

List of forces:

1. Weight or the force of gravity which is the mass of an object multiplied by the magnitude of the acceleration from gravity

([tex]\vec w= m||\vec g|| \ where \ \vec g=-9.8 \ m/s^2[/tex]).

2. Tension force which can be descried as a pulling force.

3. Normal force which is the force an object applies to prevent an object from traveling through its surface.

4. Friction force is a force that resists an objects motion.

5. Spring force is a force provided by a spring.

6. Air resistance force which is caused by air, can be interpreted as a friction force.

7. Electrical force is a force provided by charges.

8. Magnetic force which is a force provided by a magnetic field.

9. Buoyant force which is an upward force a fluid exherts on an object, can be interpreted as a normal force.

Right away we can eliminate 5,7,8, and 9. As we are not dealing with any springs, electricity, magnets, or fluids. We are also going to assuming there is no resistive forces acting on the ball so we can eliminate 4 and 6. The mass is not sitting on something so we can eliminate 3.

That leaves us with the gravitational force (weight) and the tension force.  Which logically would make sense, gravity would act on the mass at all points in its path and the string will always provide a pulling force keeping the mass from unhooking and flying away in a straight line.

For part (b):

Refer to IMAGE #2

For part (c):

Refer to IMAGE #3

Add up forces acting in the centripetal axis. Note that the centripetal axis acts like the y axis and the tangential axis acts like the x axis.

[tex]\vec v = 5.20 \ m/s[/tex]

[tex]\Sigma \vec F_c: \vec Tsin90 \textdegree +\vec wsin90\textdegree=m\vec a_c[/tex] where [tex]\vec a_c[/tex] is centripetal acceleration. [tex]\vec a _c = \frac{\vec v^2}{r}[/tex]

[tex]\Longrightarrow \vec T(1) +\vec w(1)=m(\frac{\vec v^2}{r} ) \Longrightarrow \vec T+\vec w=m(\frac{\vec v^2}{r} ) \Longrightarrow \vec T=m(\frac{\vec v^2}{r} ) -\vec w[/tex]

[tex]\Longrightarrow \vec T=m(\frac{\vec v^2}{r} ) -m ||\vec g|| \Longrightarrow \vec T=m((\frac{\vec v^2}{r} ) - ||\vec g||) \Longrightarrow \vec T=(0.275)((\frac{ (5.20)^2}{0.850} ) - 9.8) \Longrightarrow \boxed{\boxed{\vec T = 6.05 N}} \therefore Sol.[/tex]

Thus, the tension in the string is found.

For part (d):

Refer to IMAGE #4

Add up forces acting in the centripetal axis. Note that the centripetal axis acts like the y axis and the tangential axis acts like the x axis.

[tex]\vec T_{max}=22.5 \ N[/tex]

[tex]\Sigma \vec F_c: \vec T_{max}sin90 \textdegree +\vec wsin90\textdegree=m\vec a_c[/tex]

[tex]\Longrightarrow \vec T_{max}sin90 \textdegree +\vec wsin90\textdegree=m(\frac{v^2_{max}}{r} ) \Longrightarrow \vec T_{max}(1) +\vec w(1)=m(\frac{v^2_{max}}{r} )[/tex]

[tex]\Longrightarrow \vec T_{max} +\vec m||\vec g||=m(\frac{v^2_{max}}{r} ) \Longrightarrow \vec v_{max}= \sqrt{r(\frac{\vec T_{max}+m||\vec g||}{m}) }[/tex]

[tex]\Longrightarrow \vec v_{max}= \sqrt{(0.850)(\frac{22.5+(0.275)(9.8)}{0.275}}) \Longrightarrow \boxed{\boxed{\vec v_{max}=8.82 \ m/s} } \therefore Sol.[/tex]

Thus, the max speed is found.

2. Hooke saw tiny openings in cork and named them "cells", 3. Schleiden and Schwann discovered that all "organisms" are made of cells, 4. Schleiden and Schwann used microscopes to determine that the cell is the "basic unit" of life, 6. Water - do not mix with lipids, 7. Nucleic acids - contain instructions, 8. Proteins - some help break down nutrients, 9. Lipids - do not mix with water, and 10. Carbohydrates - sugar is one.

Water is a polar molecule, meaning it has a positive and negative end. Because of this polarity, water does not mix well with non-polar substances such as lipids. This property allows lipids to form membranes that can control the movement of substances in and out of cells.

Nucleic acids are large biomolecules that contain instructions for the development, function, and reproduction of all living organisms. The two main types of nucleic acids are deoxyribonucleic acid (DNA) and ribonucleic acid (RNA), which store and transmit genetic information.

Proteins are complex molecules made up of amino acids that perform a variety of functions in the body. Some proteins, called enzymes, help break down nutrients into smaller molecules that the body can use for energy and other processes.

Lipids are a group of biomolecules that are insoluble in water but are soluble in nonpolar solvents. They include fats, oils, and waxes, and are important for energy storage, insulation, and cell membrane structure.

Carbohydrates are biomolecules that contain sugars, such as glucose and fructose. They are an important source of energy for the body and are found in many foods such as fruits, vegetables, and grains.

Therefore, The correct answers are 2. Hooke saw tiny openings in cork and named them "cells", 3. Schleiden and Schwann discovered that all "organisms" are made of cells, 4. Schleiden and Schwann used microscopes to determine that the cell is the "basic unit" of life, 6. Water - do not mix with lipids, 7. Nucleic acids - contain instructions, 8. Proteins - some help break down nutrients, 9. Lipids - do not mix with water, and 10. Carbohydrates - sugar is one.

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The banking angle is the angle that the surface makes with the horizontal, or the angle of inclination.

v = √rg tanθ

tanθ=v²/rg  

The relation gives the angle of banking of the cyclist going round the curve. Here v is the speed of the cyclist, r is the radius of the curve, and g is the acceleration due to gravity.

The banking angle is the angle that the surface makes with the horizontal, or the angle of inclination. The normal force acting on the car while travelling through such a curving road has a horizontal component. The centripetal force needed to prevent skidding is provided by this component.

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Answer:

The expansion of the universe has caused cosmic microwave background radiation to cool. As the universe expands, the radiation is stretched out, causing its wavelength to increase and its temperature to decrease. This phenomenon is known as cosmic microwave background radiation redshift.

the answer is D. It has caused it to red-shift.

Period, T [tex]\Longrightarrow T=\frac{60}{14} \Longrightarrow T \approx 4.286 \ s[/tex]

Using the equation, [tex]T=2 \pi \sqrt{\frac{l}{g} }[/tex], find l. ("g" is the acceleration from gravity, 9.8 m/s^2)

[tex]T=2 \pi \sqrt{\frac{l}{g} } \Longrightarrow 4.286=2 \pi \sqrt{\frac{l}{9.8} } \Longrightarrow \frac{4.286}{2 \pi} =\sqrt{\frac{l}{9.8} }[/tex]

[tex]\Longrightarrow (\frac{4.286}{2 \pi})^2 =\frac{l}{9.8} \Longrightarrow l =9.8(\frac{4.286}{2 \pi})^2 \Longrightarrow l \approx 4.560 \ m[/tex]

Thus, the radio booth is 4.560 m from the field.

Period, T [tex]\Longrightarrow T=\frac{60}{14} \Longrightarrow T \approx 4.286 \ s[/tex]

Using the equation, [tex]T=2 \pi \sqrt{\frac{l}{g} }[/tex], find l. ("g" is the acceleration from gravity, 9.8 m/s^2)

[tex]T=2 \pi \sqrt{\frac{l}{g} } \Longrightarrow 4.286=2 \pi \sqrt{\frac{l}{9.8} } \Longrightarrow \frac{4.286}{2 \pi} =\sqrt{\frac{l}{9.8} }[/tex]

[tex]\Longrightarrow (\frac{4.286}{2 \pi})^2 =\frac{l}{9.8} \Longrightarrow l =9.8(\frac{4.286}{2 \pi})^2 \Longrightarrow l \approx 4.560 \ m[/tex]

Thus, the radio booth is 4.560 m from the field.

Answer:

make exact measurements and follow the procedure exactly

Explanation:

accurate doesn't mean correct. It just means that the results from the experiment are the result of following the procedure

The ratio of the specific heat capacity of P to Q is 1:2, or cP/cQ = 1/2.

Specific heat is the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin). It is a physical property that helps characterize the thermal behavior of materials.

Let's use the formula for heat energy:

Q = mcΔT

where Q is the heat energy supplied, m is the mass of the object, c is the specific heat capacity, and ΔT is the change in temperature.

We can write two equations for P and Q using this formula:

For P: Q = (1/2)mP * cP * 2ΔT

For Q: Q = mQ * cQ * ΔT

We know that both P and Q receive the same amount of heat energy, so we can equate the two equations:

(1/2)mP * cP * 2ΔT = mQ * cQ * ΔT

Simplifying this expression, we get:

cP/cQ = mQ/(2mP)

We are given that P has half the mass of Q, so mP = (1/2)mQ. Substituting this into the expression above, we get:

cP/cQ = mQ/(2(1/2)mQ) = 1/2

Therefore, the ratio of the specific heat capacity of P to Q is 1:2, or cP/cQ = 1/2.

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If each galaxy is 150 kpc across, the event lasts for approximately 7.7 billion years.

To calculate the time for the event, we need to know the distance traveled by the galaxies during the collision. Since each galaxy is 150 kpc across and they are passing through each other, the total distance traveled is twice the diameter of one galaxy, or 300 kpc.

To find the time, we can use the formula:

time = distance / speed

where distance is 300 kpc and speed is 1200 km/s. However, we need to convert the distance to the same units as the speed, so let's convert kpc to km:

1 kpc = 3.086 × 10^16 meters

1 meter = 3.281 × 10^-6 miles

1 mile = 1.609 × 10^3 meters

1 kpc = 3.086 × 10^19 miles

So, 300 kpc is equal to:

300 kpc × 3.086 × 10^19 miles/kpc = 9.258 × 10^21 miles

Now we can plug in the values:

time = distance / speed

time = 9.258 × 10^21 miles / 1200 km/s

time ≈ 2.432 × 10^14 seconds

So the event lasts for approximately 2.432 × 10^14 seconds. To convert to a more meaningful unit, we can divide by the number of seconds in a year:

time ≈ 7.7 billion years

Therefore, the event lasts for approximately 7.7 billion years.

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Resistance of the resistors, R₁ and R₂ are 80Ω and 200Ω respectively.

V(A) = 12 V

ΔV(B) = 2 V

ΔV(C) = 5 V

R₃ = 200Ω

Since, the total voltage-drop along the upper branch must be 12 V, according to loop rule, the voltage-drop across resistor 3 is 5.0V

So, current through the loop,

I = V(C)/R₃

I = 5/200

I = 25 x 10⁻³A

a) Therefore, the resistance,

R₁ = V(B)/I

R₁ = 2/25 x 10⁻³

R₁ = 80Ω

b) Resistor 2 has the same voltage-drop as resistor 3. So, its resistance, R₂ is 200Ω.

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Answer: 125 m

Explanation:

[tex]V_{i}[/tex] = 0

[tex]V_{f}[/tex] = no se conoce

a = 2,5 [tex]\frac{m}{s^{2} }[/tex]

t = 10 s.

d= x no los piden

Formula:

[tex]d=V_{i} .t + \frac{a.t^{2} }{2}[/tex]

Remplazamos:

[tex]d=0.(10) + \frac{2,5.10^{2} }{2}[/tex]  =  0 + [tex]\frac{2,5.(100)}{2}[/tex]  = 2,5.(50) = 125 m

C. It would red-shift. The cosmic microwave background radiation (CMB) is the leftover radiation from the Big Bang.

Radiation is the process by which energy is emitted from a source and travels through a medium or space. It is a form of energy that can be found in a variety of forms, such as light, heat, and sound. Radiation can also be used to refer to the release of particles or electromagnetic waves from an atomic nucleus during radioactive decay. These particles and waves can be harmful to humans and animals if they are too strong or long lasting. Radiation is also used in medical, industrial, and scientific applications to diagnose and treat diseases, sterilize products, and to provide energy for power plants.

It is composed of a low-energy form of light called microwaves. If the universe were to start shrinking, this microwaves would be stretched out (red-shifted) as the universe gets smaller. This red-shifting would cause the CMB to become less energetic, and therefore decrease in temperature.

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Josh was blowing bubbles through a straw into a glass of tap water. As Josh exhaled into the straw, he blew carbon dioxide gas into the water. The harder he blew, the more bubbles entered the water. The more bubbles, the more saturated the solution gas/water solution.

This is because the bubbles increase surface area of gas-water Interface, allowing for more gas molecules to come into contact with  water and dissolve. As more gas dissolves,  concentration of the gas in the water increases, and the solution becomes more saturated . This effect is known as Henry's law, which states that amount of gas that dissolves in a liquid is directly proportional to the partial pressure of the gas above liquid.

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A simplified version of Rutherford’s calculation of the size of the gold nucleus. A piece of gold foil that is 0.010 cm thick and whose area is 1 cm x 1 cm is used in the experiment. It is observed that 99.93% of all particles go through undeflected. The density of gold is 19,300 kg/m3.

Rutherford's experiment involved firing alpha particles (helium nuclei) at a thin sheet of gold foil to study the structure of atoms. Based on the results of this experiment, Rutherford was able to deduce that atoms have a small, dense nucleus at their center.

In this problem, we will go through a simplified version of Rutherford's calculation of the size of the gold nucleus.

First, we need to calculate the total number of gold atoms in the foil. We know that the foil is 0.010 cm thick and has an area of 1 cm x 1 cm, so its volume is

V = thickness x area = 0.010 cm x (1 cm x 1 cm) = 0.010 [tex]cm^{3}[/tex]

The density of gold is 19,300 kg/[tex]m^{3}[/tex], which is equivalent to 19.3 g/[tex]cm^{3}[/tex]Therefore, the mass of the gold foil is

m = density x volume =  19.3 g/[tex]cm^{3}[/tex] x 0.010 [tex]cm^{3}[/tex] = 0.193 g.

The molar mass of gold is 197 g/mol, so the number of gold atoms in the foil is

N = (0.193 g) / (197 g/mol) x (6.022 x [tex]10^{23}[/tex] atoms/mol) = 1.86 x [tex]10^{21}[/tex] atoms

Next, we need to determine the fraction of alpha particles that are deflected by the gold nucleus. We are told that 99.93% of all alpha particles go through undeflected, which means that only 0.07% of the alpha particles are deflected. This is a very small fraction, which suggests that the size of the gold nucleus must be very small compared to the size of the atom.

Assuming that the alpha particles are deflected only by the gold nucleus and not by the electrons, we can use the principle of conservation of momentum to estimate the size of the gold nucleus. When an alpha particle approaches the gold nucleus, it experiences a repulsive electrostatic force that causes it to change direction. The magnitude of this force is given by Coulomb's law

F = k[tex]q_{1}[/tex][tex]q_{2[/tex] / [tex]r^{2}[/tex]

Where k is Coulomb's constant, [tex]q_{1}[/tex] and [tex]q_{2}[/tex] are the charges of the alpha particle and gold nucleus, respectively, and r is the distance between them. Since the alpha particle has a positive charge and the gold nucleus has a positive charge, the force is repulsive.

If we assume that the alpha particle is initially moving directly toward the center of the gold nucleus, then at the point of closest approach, the alpha particle will have a velocity v that is perpendicular to the direction from the alpha particle to the gold nucleus. At this point, the force on the alpha particle will be perpendicular to its velocity, which means that it will change only the direction of the alpha particle's velocity, not its magnitude.

Using conservation of momentum, we can relate the angle of deflection θ to the distance of closest approach r.

m[tex]v^{2}[/tex] / r = k[tex]q_{1}[/tex][tex]q_{2[/tex] / [tex]r^{2}[/tex]

Where m is the mass of the alpha particle. Solving for r, we get

r = k[tex]q_{1}[/tex][tex]q_{2[/tex] / m[tex]v^{2}[/tex]

To estimate the size of the gold nucleus, we assume that the alpha particles are deflected by a single, stationary gold nucleus at the center of the atom. In reality, the gold nucleus is not stationary, but this assumption gives us a rough estimate of its size.

Hence, the alpha particles are undeflected with a probability of 0.9993, we can assume that they do not interact with the gold nucleus and that their path is a straight line through the foil.

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B) Both employ the triple point of water and absolute zero.

Absolute zero and the triple point of water serve as the foundation for both the Celsius and Kelvin scales. Absolute zero, or 0 K on the Kelvin scale and -273.15 °C on the Celsius scale, is the temperature at which all matter has no thermal energy.

The triple point of water is determined to be 273.16 K on the Kelvin scale and 0.01 °C on the Celsius scale. It is the temperature and pressure at which water can coexist in all three phases (solid, liquid, and gas) in equilibrium.

Absolute zero and the triple point of water serve as identical fiduciary points for both the Celsius and Kelvin temperature systems. The temperature ranges for each scale are determined by the fiduciary points, which are essential to temperature scales.

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A 6.47 mm high firefly sits on the axis of, and 12.3 cm in front of, the thin lens A, whose focal length is 5.77 cm, the height of the image is 2.61 mm.

We can use the thin lens equation to find the image distance and then use the magnification equation to find the height of the image.

Let's call the distance between the firefly and lens A "d1", the distance between the lenses "d2", the image distance from lens B "d3", and the height of the firefly "h1".

Using the thin lens equation for lens A:

1/fA = 1/d1 + 1/d3

Since the firefly is very small, we can assume that the rays of light from it are parallel to the axis of the lenses, so d1 = 12.3 cm.

Solving for d3, we get:

1/d3 = 1/fA - 1/d1

1/d3 = 1/5.77 cm - 1/12.3 cm

d3 = -23.46 cm

The negative value for d3 indicates that the image is formed on the same side of lens B as the firefly, which means it is a virtual image.

Now we can use the magnification equation:

m = -d3/d2

where m is the magnification of the image. The negative sign indicates that the image is inverted.

Using the distance between the lenses, d2 = 58.1 cm, we get:

m = -(-23.46 cm) / 58.1 cm

m = 0.403

This tells us that the image is smaller than the firefly, and its height is:

h2 = m * h1

h2 = 0.403 * 6.47 mm

h2 = 2.61 mm

Therefore, the height of the image is 2.61 mm.

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If |Qh| is the total energy that enters the system by heat in one cycle.

The ratio of |Qh|/(PiVi) is a measure of the efficiency of a heat engine, where |Qh| is the total energy that enters the system by heat in one cycle, Pi is the initial pressure, and Vi is the initial volume of the system.

In a heat engine, energy is transferred from a high-temperature reservoir (|Qh|) to a low-temperature reservoir (|Qc|) to perform work. The efficiency of the heat engine is given by the ratio of the work output (W) to the heat input from the high-temperature reservoir (|Qh|), which can be written as

Efficiency = W/|Qh|

Using the first law of thermodynamics, we can relate the work output to the difference between the heat input and the heat output.

W = |Qh| - |Qc|

Substituting this into the efficiency equation, we get

Efficiency = (|Qh| - |Qc|)/|Qh|

Rearranging this expression, we get

Efficiency = 1 - |Qc|/|Qh|

The quantity |Qc|/|Qh| is known as the heat rejection ratio, which is the ratio of the heat output to the heat input. Since energy cannot be created or destroyed, the total energy entering the system by heat (|Qh|) is equal to the sum of the work done by the system (W) and the energy rejected as heat (|Qc|) we get

|Qh| = W + |Qc|

Substituting this into the efficiency equation, we get

Efficiency = W/(W + |Qc|)

We can also write the work output as the product of the pressure and volume change.

W = PiVi - PfVf

Where Pi and Vi are the initial pressure and volume, and Pf and Vf are the final pressure and volume. Substituting this into the efficiency equation and simplifying, we get

Efficiency = (PiVi - PfVf)/(PiVi)

Rearranging this expression, we get

(PiVi - PfVf)/(PiVi) = |Qh|/(PiVi + |Qc|)

Since the heat engine operates in a cycle, the final volume and pressure are the same as the initial volume and pressure, and |Qc| is equal to zero. Therefore, we can simplify the expression to

(PiVi - PiVi)/(PiVi) = |Qh|/(PiVi)

Which simplifies further to

Efficiency = 1 - (|Qc|/|Qh|) = 1 - 0 = 1

Hence, the maximum efficiency of a heat engine is 1, which is achieved when all the energy transferred by heat is converted into work. In other words, the ratio of |Qh|/(PiVi) represents the maximum theoretical efficiency of a heat engine, which is known as the Carnot efficiency.

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Answer:

Displacement S= 17mHere the velocity will increase Displacement S= 17mHere the velocity will increase so it will be C or B

Explanation:

It is a true statement that the sensory hair cells of the semicircular canals convert vibrations into electrical impulses.

The sensory hair cells located in the semicircular canals of the inner ear are responsible for converting mechanical vibrations caused by head movements into electrical impulses that are then transmitted to the brain for processing. This process helps the body maintain balance and spatial orientation.

The semicircular canals are part of the vestibular system in the inner ear, which plays a crucial role in detecting changes in head position and movement. The sensory hair cells in the semicircular canals are embedded in a gelatinous structure called the cupula, which moves in response to the flow of endolymphatic fluid inside the canals.

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When a current of 1.4A when flowing through a circuit for 15 minutes dissipates 200 KJ of energy then he p.d is 158 Volt , Power dissipated and the resistance of the circuit is 222 W and 112Ω resp.

Electric circuit, a channel for carrying electric current. An electric circuit consists of a device that provides energy to the charged particles that make up current, such as a battery or a generator, as well as equipment that consume current, such as lights, electric motors, or computers, and the connecting wires or transmission lines. Ohm's law and Kirchhoff's rules are two fundamental laws that quantitatively define the behaviour of electric circuits.

Given,

Current I = 1.4 A

Energy Dissipated E = 200 kJ

time t = 15 minute = 900s

Power dissipated in the circuit,

P = E/t = 200000/900 = 222 W

Power is a voltage times current,

P =VI

222 = 1.4 × V

V = 158 Volt

according to ohms law

V = RI

R = V/I = 158/1.4 =  112Ω.

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Answer:

3.33×10^-5 meters above the origin along the positive y-axis

Explanation:

The weight of an object is given by its mass multiplied by the acceleration due to gravity, which is 9.81 m/s² near the surface of the earth. For an electron with mass me, its weight is:

W = me * g

= 9.11×10^-31 kg * 9.81 m/s²

= 8.93×10^-30 N

Since the electric force acting on the electron is opposite to its weight, the electric force must have the same magnitude as its weight but in the opposite direction:

|F_E| = |W| = 8.93×10^-30 N

The electric force between two point charges is given by Coulomb's law:

F_E = k * |q1| * |q2| / r²

where k is the Coulomb constant, q1 and q2 are the charges of the two particles, and r is the distance between them.

In this problem, q1 is the fixed charge of -0.55 nC at the origin, and q2 is the charge of the electron, which we want to find. Since we know the magnitude of the electric force between the two charges, we can solve for the distance between them:

r² = k * |q1| * |q2| / |F_E|

= 8.99×10^9 N⋅m²/C² * 0.55×10^-9 C * |q2| / (8.93×10^-30 N)

= 6.95×10^6 * |q2|

Taking the square root of both sides, we get:

r = 2.64×10^-3 * sqrt(|q2|)

Now, we need to find the distance r at which the electric force between the two charges is equal in magnitude but opposite in direction to the weight of the electron. Equating the expression for r above with the distance y along the y-axis where the electron is placed, we get:

2.64×10^-3 * sqrt(|q2|) = y

Since the electron is placed on the y-axis, its x and z coordinates are zero, and the distance between the electron and the fixed charge is simply the y-coordinate. The electric force between the charges will be attractive (i.e., in the negative y direction), so the direction of the force vector will be opposite to the positive y direction. Therefore, we can write the electric force on the electron as:

F_E = - k * |q1| * |q2| / y²

Setting this equal to the weight of the electron, we have:

k * |q1| * |q2| / y² = |W|

|q2| = |W| * y² / (k * |q1|)

= 8.93×10^-30 N * y² / (8.99×10^9 N⋅m²/C² * 0.55×10^-9 C)

= 1.56×10^-20 * y²

Substituting this expression for |q2| into the expression for r above, we get:

r = 2.64×10^-3 * sqrt(1.56×10^-20 * y²)

= 1.35×10^-11 * y

Equating this expression for r with the expression for y above, we have:

2.64×10^-3 * sqrt(1.56×10^-20 * y²) = y

Squaring both sides and simplifying, we get:

y = 3.33×10^-5 m

Therefore, the electron must be placed at a distance of 3.33×10^-5 meters above the origin, along the positive y-axis, in order for the electric force acting on it to be exactly opposite to its weight.

To summarize, we used Coulomb's law to relate the electric force between the electron and the fixed charge at the origin to the distance between them, and equated this force with the weight of the electron. We then solved for the distance at which the two forces are equal in magnitude but opposite in direction, and found that the electron must be placed 3.33×10^-5 meters above the origin along the positive y-axis.

Hope this helps!

Answer:

Given:

Water content at field capacity = 0.15 kg kg-1

Water content at wilting point = 0.06 kg kg-1

PAWC = Water content at field capacity - Water content at wilting point

PAWC = 0.15 kg kg-1 - 0.06 kg kg-1

PAWC = 0.09 kg kg-1

To convert kg kg-1 to depth units, we need to multiply by the bulk density of the soil, and divide by the density of water.

Bulk density of soil = 1250 kg m-3

Density of water = 1000 kg m-3

PAWC in depth units = (PAWC * Bulk density of soil) / Density of water

PAWC in depth units = (0.09 kg kg-1 * 1250 kg m-3) / 1000 kg m-3

PAWC in depth units = 0.1125 m

So, the Plant Available Water Capacity (PAWC) is 0.1125 m or 11.25 cm.

2.2 Plant Available Water (PAW) for the crop with rooting depth of 40 cm can be calculated as:

PAW = PAWC * Rooting depth

PAW = 0.1125 m * 0.40 m

PAW = 0.045 m or 4.5 cm

So, the Plant Available Water (PAW) for the crop is 0.045 m or 4.5 cm.

2.3 To raise the soil water content from the depletion level back to field capacity for the rooting depth of the crop, the farmer needs to apply water equivalent to the difference between PAW at depletion level and PAWC, per hectare of land.

Given:

PAW at depletion level = 65% of PAWC = 0.65 * 0.1125 m = 0.073125 m

PAWC = 0.1125 m

Volume of water to be applied = (PAWC - PAW at depletion level) * Area of land

Let's assume the area of land is 1 hectare, which is equivalent to 10,000 m^2.

Volume of water to be applied = (0.1125 m - 0.073125 m) * 10,000 m^2

Volume of water to be applied = 0.039375 m * 10,000 m^2

Volume of water to be applied = 393.75 m^3

So, the farmer must apply 393.75 m^3 of water per hectare of land to raise the soil water content from the depletion level back to field capacity for the rooting depth of the crop.

Explanation:

For electromagnetic waves    c = wavelength * frequency

 c = speed of light = 3 x 10^8 m/ s

                                    3 x 10^8 m/s = wl * 33 x 10^9  Hz

                                        wl = .009 m      ( or  9 mm)

If a radar signal that has a frequency of 33 GHz, Then the wavelength of the radar signal is  9 mm.

Wavelength is a fundamental concept in the study of waves, which are disturbances that propagate through space or a medium. It is defined as the distance between two consecutive points in a wave that are in phase, meaning that they have the same position in their respective cycles.

In other words, the wavelength is the spatial period of a wave, which is the distance over which the wave repeats itself. It is commonly represented by the Greek letter lambda (λ) and is measured in meters (m), although it can also be expressed in other units such as nanometers (nm) or micrometers (μm).

Wavelength is a key property of waves, as it determines many of their characteristics and behavior. For example, the wavelength of an electromagnetic wave (such as light or radio waves) determines its color or frequency, and thus its energy and ability to interact with matter. Similarly, the wavelength of a sound wave determines its pitch, and thus its perceived tone and musical quality.

The relationship between wavelength, frequency, and velocity is described by the wave equation, which states that the velocity of a wave is equal to the product of its wavelength and frequency. This relationship is important for understanding how waves behave and interact with their environment, such as when they are reflected, refracted, or diffracted.

So, the wavelength is a crucial concept in the study of waves, as it defines their properties and behavior. It is the distance between two consecutive points in a wave that are in phase, and is measured in meters or other units. The relationship between wavelength, frequency, and velocity is described by the wave equation, which is fundamental to the study of waves in various fields such as physics, engineering, and communication.

Here in the Question,

The wavelength of a radar signal can be calculated using the formula:

wavelength = speed of light/frequency

where the speed of light is approximately 3 x 10^8 meters per second.

Plugging in the given frequency of 33 GHz (33 x 10^9 Hz), we get:

wavelength = 3 x 10^8 / (33 x 10^9)

wavelength = 0.009090909... meters

Therefore, By rounding to the nearest millimeter, the wavelength of the radar signal is approximately 9 mm.

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Answer:

True

Explanation:

"Passive sonar uses specialized transducers called hydrophones (or underwater microphones) to listen to sounds in the ocean. These hydrophones convert received sounds into electrical signals that are then sent to a computer for a sonar operator to look at, listen to, and analyze."

Acoustic amplification is not an application of ultrasonic waves.

Ultrasonic waves are used in many different applications such as medical imaging, where they are used to create images of the internal structures of the human body.

Echolocation, which is used by animals such as bats and dolphins to navigate their environment, also relies on the use of ultrasonic waves. Additionally, ultrasonic waves are used in nondestructive testing to detect flaws or defects in materials without damaging them.

Acoustic amplification, on the other hand, involves the use of sound waves to amplify or enhance the sound produced by a musical instrument or a speaker. It does not involve the use of ultrasonic waves.

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