Conventionally, the field strength around a charged object is the direction of the force acting on a .

Answer:

4500

Explanation:

Answer:

They are all radioactive

Explanation:

Radiopharmaceuticals, or medicinal radiocompounds, are a group of pharmaceuticals that contain radioactive isotopes. Radiopharmaceuticals are used as diagnostic and therapeutic agents.

A radioactive tracer, is a chemical compound in which one or more atoms have been replaced by a radionuclide so that as a result of of its radioactive decay the nuclide can be used to explore the mechanism of chemical reactions by carefully tracing the path that the radioisotope follows from reactants to products.

Radionuclides are species of atoms that emit radiation by undergoing radioactive decay leading to the emission of alpha particles (α), beta particles (β), or gamma rays (γ).

From the foregoing, radiopharmaceuticals, tracers, and radionuclides are all radioactive materials. Hence the answer.

Answer:

E = 400 V/m

Explanation:

It is given that, for a parallel plates :

Potential difference, V = 12 volts

Separation between the plates, d = 3 cm = 0.03 m

The relation between the electric field and the electric potential is given as :

[tex]E=\dfrac{V}{d}[/tex]

[tex]E=\dfrac{12\ V}{0.03\ m}[/tex]

E = 400 volts/m

So, the strength of the electric field is 400 V/m.                

The wave speed is calculated using the frequency and the wavelength. With the provided distance between crests (1.20 m) and the time for crests to pass (13 s), the wave speed is found to be approximately 0.8308 m/s.

The distance between two successive crests of a transverse wave is equal to one wavelength. If eight additional crests pass a given point in 13 seconds, it means that nine crests in total pass that point in 13 seconds because the first crest is observed at the beginning of the timing. The speed of the wave (v) can be calculated using the formula v = frequency x wavelength. Here, the wavelength (λ) is given as 1.20 meters.

To find the frequency, we use the number of waves passing a point divided by the time taken. Therefore, the frequency (f) is 9 crests / 13 seconds = 0.6923 Hz (where Hz represents Hertz or cycles per second).

Using the wave speed formula, we obtain the speed: v = 0.6923 Hz x 1.20 m = 0.8308 m/s. Thus, the wave speed is approximately 0.8308 meters per second.

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Sedimentary rock is formed from the compaction and cementation of sediment originating from other rocks or organic material. This process results in various types of sedimentary rocks, like sandstone and shale.

The type of rock that forms when particles from other rocks or the remains of plants and animals are pressed and cemented together is known as sedimentary rock. These rocks are created through a multi-step process starting with the deposition of sediment that comes from the weathering and erosion of pre-existing rocks (clasts), or from the accumulation of plant and animal material. Over time, these deposited sediments may harden into rock through two main processes:

Compaction: where sediments are squeezed together under the weight of sediments above them.

Cementation: where minerals fill in the spaces between the loose sediment particles.

This process creates various types of sedimentary rocks, such as clastic rocks which are fragments compacted and cemented together, and organic sedimentary rocks which are formed from the lithification of organic material. Examples of sedimentary rocks include sandstone, formed from cemented sand; shale, formed from compressed mud and silt; and conglomerate, composed of cemented gravel and pebbles.

The answer involves determining normal and shear stresses at point H, computing principal stresses using the relevant formula and plotting these values on a Mohr's circle.

To determine the principal planes and stresses, along with the maximum shear stress, we will need to work through a process of calculation and analysis. However, without numerical values for the forces applied to the pipe, we cannot perform accurate calculations. Theoretically, though, you would first find the normal and shear stresses on the element at point H. Subsequently, the formula for finding the principal stresses σ₁ and σ₂ would be σ₁/₂= (σx+σy) ÷ 2 ± sqrt ((σx-σy) ÷ 2)^2 + τxy^2. As for shear stress, τxy should be evaluated at the principal planes, where it reaches maximum and minimum. Lastly, to sketch Mohr's circle, you would plot the normal stress on the x-axis and shear stress on the y-axis.

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It is a longitudinal wave. In a longitudinal wave the particles of the medium vibrate in a direction parallel to the direction of energy transport. Sound waves are mechanical waves as well as longitudinal waves that require a medium. They do not travel through a vacuum.

I₁ = initial moment of inertia before pulling in the arms

I₂ = final moment of inertia after pulling in the arms = I₁ /2

w₁ = initial angular velocity before pulling in the arms

w₂ = final angular velocity after pulling in the arms

using conservation of angular momentum

I₁ w₁ = I₂ w₂

I₁ w₁ = (I₁/2 ) w₂

w₂ = 2 w₁

KE₁ = initial rotational kinetic energy before pulling in the arms = (0.5) I₁ w²₁

KE₂ = final rotational kinetic energy after pulling in the arms = (0.5) I₂ w²₂

Ratio of final rotational kinetic energy to initial rotational kinetic energy is given as

KE₂ /KE₁ = (0.5) I₂ w²₂/((0.5) I₁ w²₁ )

KE₂ /KE₁ = ((I₁/2 ) (2 w₁)²)/(I₁ w²₁)

KE₂ /KE₁ = 2

KE₂ = 2 KE₁

hence the kinetic energy becomes twice

Answer:

f = 3×10¹¹ Hz

Explanation:

Relation ship between frequency and wavelength

"The wave speed (v) is defined as the distance traveled by a wave per unit time. If considered that the wave travels a distance of one wavelength in one period,

ν=λ/T

As we know that T = 1/f, hence we can express the above equation as,

V = f λ

The wave speed is equal to the product of its frequency and wavelength, and this implies the relationship between frequency and wavelength."

The relation between frequency and wavelength is  

λ×f = c

c = speed of light = 3×10⁸ m/s

λ = 1.00 mm = 10⁻³ m

f=c/λ

f=(3×10⁸ m/s)/   10⁻³ m

f = 3×10¹¹ Hz

Answer:

B.) Each piece will still have a north pole and a south pole

Explanation:

I just did it

When a bar magnet is cut into two pieces, each half becomes a smaller magnet with its own north and south pole. Magnetic domains within the magnet ensure that these poles always exist in pairs, a fundamental characteristic of magnets called magnetic dipoles, and cannot be isolated. Even at the smallest scale, no magnet exists with only a single pole.

If you cut a bar magnet into two pieces, you will not end up with one piece having two north poles and another piece having two south poles. Instead, each piece will have its own north and south pole, making each piece a smaller, complete magnet. This property is consistent even down to the smallest particles with magnetic properties.

All magnets have a north and a south pole, and they always occur in pairs—this is why they are called magnetic dipoles ('di' meaning two). The presence of both poles in any fragment of a magnet implies that if you were to continue cutting the magnet, no matter how small, you would still get pieces with both a north and a south pole. No matter the size, from subatomic particles to stars, a magnet cannot have a single isolated pole.

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The answer is B i know this because i just did it on usatestprep

Answer:

A. Energy from various energy sources, such as wind or from burning fossil fuels, is used to spin the blades of the turbine.

Explanation:

The answer above mine was EXTREMELY HELPFUL because I had a similar question needing to be answered, and that helped me figure out the correct answer to my question, so THANK YOU!!!

Hope that this was helpful information for anyone needing it! <3

Unpaired electron is the electron, that occupies the a place in orbital without the pair of electron. The number of unpaired electron in 3d sub shell of octahedral coordination complex [tex][FeX_6]^{2-}[/tex] is 3.

The given octahedral coordination complex in the problem is [tex]{FeX_6]^{2-}[/tex]

Here, [tex]X[/tex] is the halogen.

Unpaired electron is the electron, that occupies the a place in orbital without the pair of electron.

In the given complex ion +3 oxidation state [tex]F[/tex] (iron) represents in,

[tex]_{26}F^{3+}=1s^2, 2s^2 2p^6, 3s^23p^63d^6,4s^2\\[/tex]

[tex]F^{3+}=1s^2, 2s^2 2p^6, 3s^23p^63d^6[/tex]

As the unpaired electron in 3d sub shell is 3.

Hence the number of unpaired electron in 3d sub shell of octahedral coordination complex [tex][FeX_6]^{2-}[/tex] is 3.

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Final answer:

The [FeX₆]³− octahedral complex, with iron in a d5 configuration and halides as weak field ligands, will have five unpaired electrons due to the high-spin configuration created by the ligands.

Explanation:

To determine the number of unpaired electrons in the octahedral coordination complex [FeX₆]³−, where X represents any halide, we will apply crystal field theory (CFT). First, we note that iron in this complex exists in a +3 oxidation state, which gives it a d5 electron configuration since the neutral iron atom has 8 valence electrons (3d6 4s2).

For any halide as a weak field ligand in an octahedral complex, the crystal field splitting is not sufficient to overcome the electron pairing energy. This results in a high-spin complex for iron (III), where all the d-orbitals initially receive one electron each before any pairing occurs.

Thus, the [FeX₆]³− complex will have five unpaired electrons, one in each of the 3d orbitals, as halides create a high-spin configuration for a d5 metal ion like iron (III).

your weight on the surface of a neutron star with the same mass as the Sun and a diameter of 25.0 km would be approximately [tex]\( 2.8 \times 10^{12} \, \text{N} \).[/tex]

To calculate your weight on the surface of a neutron star with the given mass and diameter, we can use the formula for gravitational force ( F ) and the definition of weight ( W ).

The formula for gravitational force between two objects is given by:

[tex]\[ F = \frac{{G \cdot m_1 \cdot m_2}}{{r^2}} \][/tex]

Where:

- ( F ) is the gravitational force,

- ( G ) is the gravitational constant [tex](\( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))[/tex],

- [tex]\( m_1 \) and \( m_2 \)[/tex] are the masses of the objects,

- ( r ) is the distance between the centers of the objects.

On the Earth's surface, your weight ( W ) is calculated using your mass \( m \) and the acceleration due to gravity ( g ):

[tex]\[ W_{\text{Earth}} = m \cdot g_{\text{Earth}} \][/tex]

Given:

- Your weight on Earth[tex]\( W_{\text{Earth}} = 670 \, \text{N} \)[/tex],

- Mass of the Sun [tex]\( m_{\text{Sun}} = 1.99 \times 10^{30} \, \text{kg} \)[/tex],

- Diameter of the neutron star[tex]\( r = 25.0 \, \text{km} = 25,000 \, \text{m} \)[/tex].

First, let's calculate the gravitational force between you and the Earth using your weight:

[tex]\[ W_{\text{Earth}} = \frac{{G \cdot m \cdot m_{\text{Earth}}}}{{r_{\text{Earth}}^2}} \][/tex]

Solve for your mass \( m \):

[tex]\[ m = \frac{{W_{\text{Earth}} \cdot r_{\text{Earth}}^2}}{{G \cdot m_{\text{Earth}}}} \][/tex]

Substitute the given values:

[tex]\[ m = \frac{{670 \, \text{N} \cdot (6.371 \times 10^6 \, \text{m})^2}}{{6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \cdot 5.97 \times 10^{24} \, \text{kg}}} \][/tex]

Calculate the mass ( m ) on Earth:

[tex]\[ m \approx 69 \, \text{kg} \][/tex]

Now, use this mass to calculate your weight[tex]\( W_{\text{Neutron Star}} \)[/tex] on the surface of the neutron star:

[tex]\[ W_{\text{Neutron Star}} = \frac{{G \cdot m \cdot m_{\text{Sun}}}}{{r_{\text{Neutron Star}}^2}} \][/tex]

Substitute the given values for the neutron star:

[tex]\[ W_{\text{Neutron Star}} = \frac{{6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \cdot 69 \, \text{kg} \cdot 1.99 \times 10^{30} \, \text{kg}}}{{(25,000 \, \text{m})^2}} \][/tex]

Calculate[tex]\( W_{\text{Neutron Star}} \)[/tex]:

[tex]\[ W_{\text{Neutron Star}} \approx 2.8 \times 10^{12} \, \text{N} \][/tex]

So, your weight on the surface of a neutron star with the same mass as the Sun and a diameter of 25.0 km would be approximately [tex]\( 2.8 \times 10^{12} \, \text{N} \).[/tex]

Your weight on the neutron star would be approximately 1.45 × 10¹³ N due to the extremely high gravitational acceleration of 2.13 × 10¹¹ m/s².

To determine your weight on a neutron star, we first need to calculate the gravitational acceleration on its surface. Given:

The formula for the gravitational acceleration gstar on the surface of a spherical object is:

gstar = G M / R²

Plugging in the values:

[tex]g^* = \frac{(6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2) \times (1.99 \times 10^{30} \, \text{kg})}{(2.5 \times 10^4 \, \text{m})^2}[/tex]

gstar ≈ 2.13 × 10¹¹ m/s²

This is the gravitational acceleration on the neutron star. To find your weight, we use:

Weightstar = Mass * gstar

Your mass (from Earth weight):

Mass = WeightEarth / gEarth = 670 N / 9.81 m/s² ≈ 68.3 kg

Therefore, your weight on the neutron star:

Weightstar = 68.3 kg * 2.13 × 10¹¹ m/s² ≈ 1.45 × 10¹³ N

Therefore, your weight on the neutron star would be approximately 1.45 × 10¹³ N

Answer:

The change in momentum increases because the impact time increases.

Explanation:

When a baseball player catches a baseball, the glove "gives" or moves backward. Momentum of a body is equal to the product of mass and velocity. Also, the change of momentum of an object when it is acted by a force is called impulse. It is represented by J. Mathematically, it can be written as :

[tex]J=F.\Delta t=m\Delta v[/tex]

F is applied force

[tex]\Delta t[/tex] is the time taken by an object to change its momentum

The momentum depends on the force and the time of impact. In this case, as the a baseball player catches a baseball, the gloves move in backward direction. By doing so, the ball comes to rest in larger time.

So, the change in momentum increases because the impact time increases.

Final answer:

The wavelength corresponding to the most intense light emitted by a giant star with a surface temperature of 4400 K is approximately 681.8 nm.

Explanation:

According to Wien's law, the wavelength at which the most intense light is emitted by a giant star can be determined using the formula:
wavelength = 3 x 10^6 / temperature

In this case, the surface temperature of the giant star is 4400 K. Plugging in this value into the formula, we get:
wavelength = 3 x 10^6 / 4400 = 681.8 nm

Therefore, the wavelength corresponding to the most intense light emitted by the giant star is approximately 681.8 nm.

The energy of exactly one photon of this microwave radiation is 1.53 × 10⁻²⁴Joule

The term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is:

[tex]\large {\boxed {E = h \times f}}[/tex]

E = Energi of A Photon ( Joule )

h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )

f = Frequency of Eletromagnetic Wave ( Hz )

The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.

[tex]\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}[/tex]

[tex]\large {\boxed {E = qV + \Phi}}[/tex]

E = Energi of A Photon ( Joule )

m = Mass of an Electron ( kg )

v = Electron Release Speed ( m/s )

Ф = Work Function of Metal ( Joule )

q = Charge of an Electron ( Coulomb )

V = Stopping Potential ( Volt )

Let us now tackle the problem!

Given:

λ = 13.0 cm = 13.0 × 10⁻² m

Unknown:

E = ?

Solution:

[tex]E = h f[/tex]

[tex]E = h \frac{c}{\lambda}[/tex]

[tex]E = 6.63 \times 10^{-34} \frac{3 \times 10^8}{13.0 \times 10^{-2}}[/tex]

[tex]\large {\boxed {E = 1.53 \times 10^{-24} ~ Joule } }[/tex]

Grade: High School

Subject: Physics

Chapter: Quantum Physics

Keywords: Quantum , Photoelectric , Effect , Threshold , Frequency , Electronvolt

The wavelength of infrared radiation with a frequency of 9.76 x 10^13 Hz is calculated to be 3070 nanometers by using the relationship between wavelength, frequency, and the speed of light.

The question asks us to calculate the wavelength of infrared radiation with a given frequency of 9.76 x 1013 hertz (Hz). To find the wavelength, we will use the formula that relates wavelength (λ), frequency (ƒ), and the speed of light (c):

λν = c

Where λ is the wavelength, ν is the frequency, and c is the speed of light (c = 3 x 108 meters per second). Solving for λ, we get:

λ = c / ν

Now we can plug in the given values:

λ = (3 x 108 m/s) / (9.76 x 1013 Hz)

λ = 3.07 x 10-6 meters

Since the question asks for the wavelength in nanometers, we convert meters to nanometers by multiplying by 109:

λ = 3.07 x 10-6 meters x 109 nm/meter

λ = 3070 nm

Therefore, the wavelength of the infrared radiation is 3070 nanometers.

If a charged particle is projected with its initial velocity parallel to a uniform magnetic field, then the resulting path would be a straight line parallel to the field, therefore the correct answer is option 3.

A magnetic field could be understood as an area around a magnet, magnetic material, or an electric charge in which magnetic force is exerted .

As given in the problem if a  charged particle is projected with its initial velocity parallel to a uniform magnetic field then we have to find out what would be the resulting path of the particle .

Thus , the right response is option 3, as the path of a charged particle projected with its starting velocity parallel to a uniform magnetic field would be a straight line parallel to the field .

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Final answer:

When the total energy of a system consisting of a block and a spring is doubled, the new maximum speed of the block will be sqrt(2) times its original maximum speed. If the original speed was 20 cm/s, the new max speed will be about 28.28 cm/s.

Explanation:

The maximum speed of a block oscillating on a spring is determined by its total mechanical energy, which is the sum of its kinetic and potential energy. In a frictionless system, when the kinetic energy is maximum, the potential energy is zero, and vice versa. When the total energy of the system is doubled, both the kinetic and potential energy will double when the block is at their maximum values. Since the maximum speed (v_max) is directly related to the maximum kinetic energy (which is 1/2 m v_max^2, where m is the mass of the block), if the total energy doubles and mass remains the same, the new maximum speed will be the square root of 2 times the original maximum speed. Hence, if the original maximum speed is 20 cm/s, the new maximum speed will be 20 cm/s * sqrt(2), approximately equal to 28.28 cm/s.

Given the initial maximum speed of 20 cm/s, the new maximum speed will be approximately 28.3 cm/s.

To determine the block's maximum speed when its total energy is doubled, we need to understand how the energy in a spring-mass system is related to its speed.

In a spring-mass system, the total mechanical energy (E) is the sum of its potential energy (U) and kinetic energy (K). The equation for kinetic energy (K) at maximum speed (V(max)) is given by:

where

Since the problem states that the initial maximum speed (V(max)) is 20 cm/s (or 0.2 m/s), we need to find the new maximum speed when the total energy is doubled.

Knowing that the total energy is proportional to the square of the maximum speed, we can set up the following relationship:

Considering the kinetic energy component:

Canceling out the common factors, we get:

Given the initial maximum speed (V(max)) is 20 cm/s:

Thus, the block's maximum speed will be approximately 28.3 cm/s if its total energy is doubled.

A photon of violet light with a wavelength of 4.5×10⁻⁷ meters carries approximately 2.75 electron-volts (eV) of energy.

To find the energy of a photon of violet light with a wavelength of 4.5×10⁻⁷ meters, we can use the formula for energy in terms of wavelength:

Energy (E) = h * c / λ

Here,

Substituting the values into the formula:

E = (6.626 × 10⁻³⁴ J·s) * (3 × 10⁸ m/s) / (4.5 × 10⁻⁷ m)

E = 4.414 × 10⁻¹⁹ J

Since we want the energy in electron-volts (eV), we convert from joules using the conversion factor 1 eV = 1.602 × 10⁻¹⁹ J:

E = (4.414 × 10⁻¹⁹ J) / (1.602 × 10⁻¹⁹ J/eV)

E ≈ 2.75 eV

Therefore, a photon of violet light with a wavelength of 4.5×10⁻⁷ meters carries approximately 2.75 eV of energy.