Ddoes the F = C/r^2 relationship seem to hold?

Plant and animal cells have many similarities in terms of organelles, but there are a few organelles that are typically absent in both types of cells. These include:

Centrioles: Centrioles are cylindrical structures involved in cell division and the formation of spindle fibers. They are found in animal cells but are generally absent in plant cells.

Lysosomes (in some plant cells): While lysosomes are present in most animal cells, they are less common in plant cells. Plant cells usually have large central vacuoles instead, which serve similar functions such as storage and waste disposal.

It's important to note that there can be variations in cell structure and organelle presence among different plant and animal cell types. For example, certain specialized plant cells may contain centrioles, and some animal cells may have fewer or reduced lysosomes.

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The three following frequencies that resonate in the tube are: b) 250Hz, 500Hz, 750Hz. Option b is correct.

In an open tube, the lowest frequency that resonates is the fundamental frequency, which is given by:

where v is the speed of sound and L is the length of the tube. In this case, we are given that f1 = 250 Hz.

The next three resonant frequencies are the harmonics of the fundamental frequency, which are given by:

where n is an integer greater than 1.

Therefore, the next three resonant frequencies are:

Out of the given options, the frequencies that match these three resonant frequencies are option b) 250Hz, 500Hz, 750Hz. Hence Option b is correct.

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It will be False.

           The fictional dissipation term F (or energy loss coefficient) is not known for an orifice plate in a horizontal pipe, because it depends on several factors that may vary in different situations.

Even if F is known for a certain orifice plate in a horizontal pipe, it cannot be used in situations that are not horizontal, because the orientation of the pipe may affect the flow pattern and pressure distribution around the orifice plate.

To calculate F, or the energy loss coefficient of an orifice plate, you need to know the following parameters:

            The pressure difference between the section before and after the orifice plate (p₁- p₂)

The density of the fluid (ρ)

The diameter of the pipe (D₁)

The diameter of the orifice (D₂)

The discharge coefficient of the orifice plate (cd)

              The energy loss coefficient F can be calculated from the following equation12:

F = cd (π / 4) D₂2 [ 2 (p₁ - p₂) / ρ (1 - β₄) ]1/2

                                      This equation is derived from the Bernoulli equation and the continuity equation, which are based on the conservation of energy and mass for fluid flow.

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A periscope is an optical instrument that reflects images via a tube using a set of prisms, lenses, or mirrors.

Place rubber bands at the tube's two ends and move them so that they are parallel to one another. Leave the rubber bands in place or mark the angle with a pen or pencil.

Cut the tube at both ends. The mirrors are placed on the angled ends. However, you need viewing apertures on either end of the tube before you secure the mirrors. The tube's pointed sides have holes that descend from them. Cut them if you want to.

The mirrors should be set up and secured using tape or adhesive. Adjust the apertures' size and shape as necessary by looking through them.

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The tape will be attracted to the rod due to the electrostatic force generated by the charged rod and fur.

When a rod is rubbed with fur, it gains an electric charge due to the transfer of electrons between the materials.

This process is known as triboelectric charging.

As a result, the rod becomes charged, and the fur gains the opposite charge.

When you bring the charged rod close to the tape, the electrostatic force between the charges causes the tape to be attracted towards the rod.

This attraction occurs because the tape, being a neutral object, experiences an induced charge on its surface as the charges rearrange themselves in response to the charged rod's presence.

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A good comparison for an ultrasound beam is a flashlight. Just as a flashlight emits light in a focused beam, an ultrasound beam emits sound waves in a focused direction.

Both a flashlight and an ultrasound beam can be adjusted to control the direction and intensity of the emitted energy.
Another comparison could be to a radar system. Like ultrasound, radar also uses waves to detect and measure objects. However, radar waves are much longer and operate at higher frequencies than ultrasound waves.

The comparison to a flashlight is particularly useful in helping people understand how ultrasound imaging works. By imagining the ultrasound beam as a flashlight shining through the body, it becomes easier to visualize how the waves bounce off internal structures and create images.

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Yes, a light bulb is an ohmic device.

An ohmic device is one whose electrical resistance remains constant regardless of the voltage or current passing through it.

A light bulb operates on the principle of ohm's law, which states that the current flowing through a conductor is directly proportional to the voltage applied across it, provided its temperature and other physical conditions remain constant.

In other words, a light bulb's resistance remains constant, making it an ohmic device. As the voltage increases, the current flowing through the bulb also increases proportionally, resulting in the bulb's brightness increasing. Thus, a light bulb can be considered an ohmic device.

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Static charges build up due to friction between objects, causing electrons to transfer and create an imbalance of charges.

The theory behind static charges building up when objects rub against each other is called the triboelectric effect.

As objects come into contact, friction causes electrons to be transferred between their surfaces.

This creates an imbalance in the electric charges of the objects.

One object becomes negatively charged by gaining electrons, while the other becomes positively charged by losing electrons.

When these objects are separated, the imbalance in charge creates a static electric field.

This can result in static electricity discharges, such as sparks, when the objects come into contact with other materials or objects.

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During a collision, a city bus exerts a greater force than a moped due to its significantly larger mass. Both vehicles experience the same acceleration, but the force exerted is directly proportional to their masses, making the force from the city bus much larger.

A collision in physics happens when two or more objects come into contact with each other and their velocities are altered. Whether or not the kinetic energy is preserved determines whether the collision is categorised as elastic or inelastic. The entire kinetic energy of the objects before and after a collision is preserved in an elastic collision. In other words, the objects collide and bounce off one another without transferring any energy to heat or sound. The kinetic energy is not conserved in an inelastic collision, on the other hand, and part of it is lost to other forms, causing the total kinetic energy of the objects to drop after the impact.

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For both supports to exert the same force on the ladder while keeping it in equilibrium, the ladder must be placed symmetrically between the supports.

When a ladder is in equilibrium, the forces acting on it (i.e., the force exerted by each support and the ladder's weight) must balance each other out.

To achieve this balance and have both supports exert the same force, the ladder must be positioned symmetrically between the supports, with its center of gravity aligned with the midpoint between the supports.

In this scenario, the forces from each support will be equal, and the sum of their torques around the center of gravity will be zero, satisfying the equilibrium conditions.

Summary: To keep a ladder in equilibrium with both supports exerting the same force, it must be positioned symmetrically between the supports, ensuring a balance of forces and torques.

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A phase shift is a shift of an entire wave with respect to an identical wave along the x-axis.

A waveform's displacement in time is referred to as a phase shift. It happens when the entire waveform is phase-shifted away from a reference waveform by a predetermined amount.

The amplitude of the waveform against time can be shown in a graph as a horizontal shift along the x-axis.

Phase shifts can result from a number of things, including modifications to the waveform's speed or frequency.

A sine wave with a frequency of 10 Hz, for instance, will be 90 degrees out of phase with the original wave if its wavelength is shifted by a quarter.

This indicates that the peaks and troughs of the two waves will occur at various times.

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The distance between m = 0 and m = 1 bright fringes is approximately 0.44 meters. Approximately 3 bright fringes will be observed on the screen.

To find the number of bright fringes that can be observed on the screen, we can use the equation:

nλ = d sinθ

where n is the order of the bright fringe, λ is the wavelength of the laser beam, d is the distance between the lines of the diffraction grating, and θ is the angle between the incident beam and the diffracted beam.

Rearranging the equation, we get: θ = sin⁻¹(nλ/d)

Plugging in the values given, we get: θ = sin⁻¹(1 × 750 × 10⁻⁹ / (1/750 × 10⁻³))

θ ≈ 19.5°

This means that the first bright fringe will be observed at an angle of 19.5° with respect to the incident beam.

To find the distance between m = 0 and m = 1 bright fringes, we can use the equation:

Δy = L tanθ

where Δy is the distance between adjacent bright fringes, L is the distance between the diffraction grating and the screen, and θ is the angle of the first bright fringe.

Plugging in the values given, we get: Δy = 1.4 × tan(19.5°)

Δy ≈ 0.44 m

Therefore, the distance between m = 0 and m = 1 bright fringes is approximately 0.44 meters.

Since we know that bright fringes occur at regular intervals, we can divide the distance between the grating and the screen by the distance between adjacent fringes to find the total number of fringes that will be observed:

Total number of bright fringes = (1.4 m) / (0.44 m) ≈ 3.2

So, approximately 3 bright fringes will be observed on the screen.

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The spacing between the two slits is approximately 1.5 micrometers.

To calculate the spacing between the two slits, we need to use the formula for the spacing of fringes in the interference pattern of a double-slit experiment, which is:

d sin θ = mλ

where d is the spacing between the two slits, θ is the angle between the line connecting the slit and the center of the screen and the line perpendicular to the screen, and m is the order of the bright fringe (starting from 0 for the central fringe), and λ is the wavelength of the light.

In this case, we know that λ = 663 nm, m = 11, and the distance from the slits to the screen is 3 m. To find θ, we need to use the small angle approximation:

θ ≈ tan θ = y/L

where y is the distance between the central fringe and the 11th bright fringe, which is 52 mm, and L is the distance from the slits to the screen, which is 3 m.

θ = tan⁻¹ (y/L) = tan⁻¹ (0.052/3) ≈ 0.97°

Now we can solve for d:

d sin θ = mλ

d = mλ/sin θ = (11)(663 nm)/(sin 0.97°) ≈ 1.5 μm

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The statement "If a wave travels from a medium in which its speed is faster to a medium in which its speed is slower, the reflected wave has the same orientation as the original wave" is FALSE because when a wave travels from one medium to another medium in which the speed of the wave is slower, the wave changes direction.

This change in direction is called refraction. The amount of bending of the wave depends on the difference in the speed of the wave between the two media, as well as the angle of incidence (the angle at which the wave approaches the boundary between the two media).

When a wave is incident on a boundary between two media, some of the wave is reflected back into the original medium. The angle of reflection is equal to the angle of incidence, but the reflected wave is flipped with respect to the normal line (a line perpendicular to the boundary).

Therefore, the reflected wave has the same amplitude and frequency as the incident wave, but it is flipped with respect to the normal line. The orientation of the reflected wave is opposite to that of the incident wave.

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The sum of the areas of all the patches on the Gaussian sphere is equal to the surface area of the sphere, which is 4πr².

The sum of the areas of all the patches on a Gaussian sphere with radius r can be written as:

4πr²

This is due to the surface area of a sphere with a radius r is 4πr², and the Gaussian sphere is also a sphere which has the same radius r.

Therefore, the sum of the areas of all the patches on the Gaussian sphere is equal to the surface area of the sphere, which is 4πr².

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The average power developed by the shot-putter is approximately 961 W when a shot-putter accelerates a 7.3-kg shot from rest to 14 m/s in 1.5 s.

To find the average power developed by the shot-putter, we need to use the equation for power:
Power = Work/Time
We can calculate the work done by the shot-putter using the equation for kinetic energy:
[tex]KE = (1/2)mv^2[/tex]
Where m is the mass of the shot and v is the final velocity. Substituting the given values, we get:
[tex]KE = (1/2)(7.3 kg)(14 m/s)^2[/tex]
KE = 1441.4 J
The work done by the shot-putter is equal to the kinetic energy gained by the shot. Now we need to divide this by the time taken to achieve this velocity:
Power = Work/Time
Power = 1441.4 J / 1.5 s
Power = 960.93 W
Therefore, the average power developed by the shot-putter is approximately 961 W.

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The magnetic field at the center of a circular current in the xy plane rotating clockwise when viewed from the positive z-axis is directed along the positive z-axis.

When a circular current flows, it creates a magnetic field around it. The direction of the magnetic field at any point is perpendicular to the plane of the circular current and follows the right-hand rule.

The right-hand rule states that if the thumb of the right hand points in the direction of the current, then the curled fingers will point in the direction of the magnetic field. In the case of a circular current rotating clockwise, the magnetic field at the center will be directed along the positive z-axis.

This is because the magnetic field lines will form a vertical loop around the current, with the direction of the field lines being perpendicular to the plane of the current and pointing upwards.

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When a neutral object is charged by induction, the polarity of the charge acquired by the neutral object is opposite to that of the charged object that polarized it.

Charging by induction occurs when a charged object is brought near a neutral object, causing the charges in the neutral object to redistribute.

The charged object induces a separation of charges in the neutral object, with like charges repelling and opposite charges attracting.

When the neutral object is grounded (or connected to a conducting path), the like charges are free to leave, leaving the neutral object with an overall charge opposite to that of the polarizing charged object.

Summary: In the process of charging by induction, a neutral object acquires a charge with a polarity opposite to the charged object that induced the polarization.

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The north pole of a compass is the end of the needle that points towards the geographic North Pole.

This is the point located at the northernmost part of the Earth's axis.

On average, all compasses have the same north pole, which is the end of the needle that points towards the geographic North Pole.

To determine which end of the needle is the north pole, you can use a map or a globe to find the location of the geographic North Pole and then hold the compass near it.

The end of the needle that points towards the geographic North Pole is the north pole of the compass.

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The calculated value of the electric force between the two protons is 2.3 × [tex]10^{(-28)[/tex] N.

To find the magnitude of the electric force between the two protons in a helium nucleus, we can use Coulomb's Law:
F = (k * q1 * q2) / r²
where F is the force, k is Coulomb's constant (8.99 × [tex]10^9[/tex] N·m²/C²), q1 and q2 are the charges of the protons (1.6 × [tex]10^{(-19)[/tex] C), and r is the distance between them (1 × [tex]10^{(-15)[/tex] m).
F = (8.99 × 10^9 N·m²/C² * [tex](1.6 *10^{(-19)} C)^2[/tex]) / [tex](1 * 10^{(-15)}m)^2[/tex]
F ≈
None of the given options matches the calculated force. There might be a typographical error in the options provided. However, the calculated value of the electric force between the two protons is 2.3 × [tex]10^{(-28)[/tex] N.

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The maximum displacement of a wave from its equilibrium point is known as its amplitude.

Amplitude is defined as maximum distance that a point on a wave or vibrating body can travel before returning to equilibrium.

Amplitude is the same as the vibration path's half-length.

A sound wave's height can be determined by looking at its amplitude. The loudness or the maximum displacement of the medium's vibrating particles from their mean position at the time the sound is produced are two ways to describe a sound wave's amplitude.

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The air temperature is approximately 17°C.

The speed of sound in the air depends on the temperature of the air. At a given temperature, the speed of sound can be calculated using the formula:

v = fλ

where v is the speed of sound, f is the frequency, and λ is the wavelength.

In this case, we are given the frequency and wavelength of the sound wave, so we can calculate the speed of sound:

v = fλ = (500 Hz)(0.68 m) = 340 m/s

Now we can use the speed of sound formula to solve for the air temperature. The formula for the speed of sound in air is:

[tex]v = (\gamma RT)^{0.5[/tex]

where γ is the ratio of specific heat (1.4 for air at standard temperature and pressure), R is the gas constant (287 J/(kg·K) for air), and T is the temperature in Kelvin.

We can rearrange this formula to solve for the temperature T:

[tex]T =\dfrac{ v^2}{\gamma R^{\frac{1}{2}}}[/tex]

Substituting the known values, we get:

[tex]T = (\dfrac{340^2}{(1.4 \times 287)})^\frac{1}{2} = 290 K[/tex]

Converting to Celsius, we get:

T = 290 - 273 = 17°C

Therefore, the air temperature is approximately 17°C.

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Electric fields can be visualized by drawing force lines, which give an indication of the direction and strength of the electric field at different points in space.

Force lines, also known as electric field lines or lines of force, are imaginary lines that represent the direction in which a positive test charge would move if placed in the electric field. The lines always point away from positively charged objects and towards negatively charged objects.

The closer the lines are to each other, the stronger the electric field is at that point. The spacing of the lines indicates the strength of the electric field. Therefore, by visualizing the force lines, one can gain an understanding of the nature of the electric field, including its direction, strength, and how it varies in space around charged objects.

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When a patient sees an out-of-network physician, the payment rules depend on the type of insurance plan the patient has.

In general, out-of-network care is not covered or is only partially covered by insurance plans, and the patient may be responsible for paying the difference between the amount charged by the out-of-network physician and the amount that the insurance plan covers.

For example, in a preferred provider organization (PPO) plan, patients may have some coverage for out-of-network care, but they will generally have to pay higher copays, coinsurance, and deductibles.

In a health maintenance organization (HMO) plan, out-of-network care may not be covered at all, except in emergency situations.

In some cases, out-of-network physicians may be willing to accept the insurance plan's payment as payment in full, but this is not guaranteed, and patients should check with their insurance plan and the physician's office to understand their financial responsibility.

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The magnitude of the electric field decreases as the distance from the particle increases, following the inverse square law.

The electric potential at a point due to a negatively charged particle is negative. The electric potential is defined as the amount of work that must be done to bring a unit charge from infinity to a particular point in an electric field.

Since the negatively charged particle repels other negative charges and attracts positive charges, the work done in bringing a positive charge from infinity to that point would be negative. Therefore, the electric potential due to a negatively charged particle is negative.

On the other hand, the electric field at a point due to a negatively charged particle is directed radially outward from the particle and is proportional to the inverse square of the distance from the particle.

The electric field at a point is defined as the force per unit charge experienced by a small positive test charge placed at that point.

Since the negatively charged particle repels positive charges and attracts negative charges, a positive test charge placed at that point would experience a repulsive force away from the particle, giving rise to an electric field directed radially outward.

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Tripling the distance between the sound source and the listener will change the intensity, as detected by the listener, by a factor of 1/9. The answer is (a) 1/9.

The intensity of sound is inversely proportional to the square of the distance from the sound source. This relationship is described by the inverse square law. Mathematically, the equation for the intensity of sound is:

I ∝ 1/distance²

where I is the intensity of sound and "distance" refers to the distance between the listener and the sound source.

If we triple the distance between the sound source and the listener, the new distance will be 3 times the original distance. Therefore, the new intensity of sound can be calculated as follows:

New intensity = Original intensity x (Original distance/New distance)²

New intensity = Original intensity x (1/3)²

New intensity = Original intensity x 1/9

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The final angular velocity of the merry-go-round depends on the mass and velocity of the child, as well as the radius and rotational inertia of the merry-go-round : mRv/I.

To understand why, let's first define the terms. The radius, R, is the distance from the center of the merry-go-round to the edge. The rotational inertia, I, is a measure of how difficult it is to change the merry-go-round's rotation, and depends on the mass distribution and shape of the object. When the child jumps on, they add their mass, m, to the merry-go-round. This increases the total rotational inertia to I + [tex]mR^{2}[/tex] (where [tex]R^{2}[/tex] is the square of the radius). To find the final angular velocity, we can use the principle of conservation of angular momentum, which states that the total angular momentum of a system remains constant unless acted upon by an external torque.

Before the child jumps on, the merry-go-round has zero angular momentum, since it is at rest. After the child jumps on, the angular momentum of the system is: ([tex]L_{initial}[/tex] + [tex]L_{child}[/tex]) = Iω[tex]_{final}[/tex]. where [tex]L_{initial}[/tex] is the initial angular momentum (zero), [tex]L_{child}[/tex] is the angular momentum of the child, and ω[tex]_{final}[/tex] is the final angular velocity. The angular momentum of the child is mRv, since they are moving with velocity v and their mass is m. Plugging this into the equation and solving for ω[tex]_{final}[/tex], we get: (0 + mRv) = (I + [tex]mR^{2}[/tex])ω[tex]_{final}[/tex], ω[tex]_{final}[/tex] = mRv / (I + [tex]mR^{2}[/tex]). Simplifying this expression, we get the answer C, mRv/I.

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The required, as an astronaut in a sealed spacesuit in the vacuum of space, the only way to transfer heat to the environment is through radiation. Optio C is correct.

Conduction and convection both require a medium such as a gas, liquid, or solid to transfer heat, but in space, there is no medium to transfer heat to or from. Evaporation requires the presence of a liquid, which is not available in the vacuum of space.

Radiation is the transfer of heat through the emission of electromagnetic waves, and it can occur even in a vacuum. Objects in space, including the astronaut's spacesuit, will radiate heat away as long as they are at a higher temperature than their surroundings. Therefore, the astronaut can only transfer heat to the environment by radiation.

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The answers are based on the laws of reflection and Bob can see just his father from the mirror.

a) After reflection we can see that Bob can see his father from the mirror and for this answer pls refer to the image attached.

b) Bob can see only one person.

c) Bob can not see all the people from the other side of the screen because when the ray is reflected from the mirror in and the angle of reflection is equal to the angle of incidence the ray doesn't cover all the people from the other side thus preventing Bob from seeing everyone there.

d) Yes, the person whom Bob sees from this side can also see Bob from the other side as the ray can be traced back on the same line from the father to Bob.

e) Bob would have to sit in seat C in order to prevent from being seen by everyone in the family sitting on the other side as the angle which that ray would have made would be very big and as a result the ray will pass from behind of the brother.

f) No, there is nowhere in the place where Bob can sit in order to be able to see his sister. As finding a suitable angle on the seats will not be possible.

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The force constant of the spring with a mass of 0.250 kg stone attached to the ideal spring is 24.1 N/m. Hence, option B is correct.

From the given,

mass of the stone = 0.250 kg

time period = 0.64 s

Force constant (k) =?

To find the force constant,

T = 2π(√m/k), where T is the time period of simple harmonic oscillations. k is the force constant and m is the mass of the stone.

0.64 = 2π×(√0.25/k)

On squaring on both sides

(0.64×0.64) = (2×π)²×(m/k)

k = (4×3.14×3.14×0.25) / (0.4096)

  = 9.8596 / 0.4096

 = 24.07 N/m

Thus the force constant k, is 24.1 N/m. Thus, the ideal solution is Option  B.

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