Recall that the blocks can only move along the x axis. the x components of their velocities at a certain moment are v1x and v2x. find the x component of the velocity of the center of mass (vcm)x at that moment. keep in mind that, in general: vx=dx/dt. express your answer in terms of m1, m2, v1x, and v2x.

The energy change associated with the transition from n = 3 to n = 2 in the hydrogen atom is about -3.03 × 10⁻¹⁹ J

[tex]\texttt{ }[/tex]

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 )

[tex]\texttt{ }[/tex]

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 !

[tex]\texttt{ }[/tex]

Given:

initial shell = n₁ = 3

final shell = n₂ = 2

Unknown:

ΔE = ?

Solution:

We will use this following formula to solve this problem:

[tex]\Delta E = R (\frac{1}{(n_2)^2} - \frac{1}{(n_1)^2})[/tex]

[tex]\Delta E = -2.18 \times 10^{-18} \times ( \frac{1}{2^2} - \frac{1}{3^2})[/tex]

[tex]\Delta E = -2.18 \times 10^{-18} \times ( \frac{1}{4} - \frac{1}{9} )[/tex]

[tex]\Delta E = -2.18 \times 10^{-18} \times \frac{5}{36}[/tex]

[tex]\Delta E \approx -3.03 \times 10^{-19} \texttt{ J}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: College

Subject: Physics

Chapter: Quantum Physics

the answer would be D

Answer:

D. billions of stars arranged in the circular pattern with a flattened core

Explanation:

The relationship between the force between two magnetic poles and the distance of separation is The magnetic force is inversely proportional to the square of the distance between them.

Coloumb's law of magnetism states the relationship between the magnetic strength of two magnets. The force of attraction or repulsion between two magnets is directly proportional to the strength of magnetic poles and inversely proportional to the square of the distance between them.

Coloumb's law of magnetism describes the electric field between two charged objects. The force of attraction or repulsion is inversely proportional to the absolute permeability of the surrounding medium.

The attractive and repulsive force, F∝ m₁×m₂ / μ₀r², where m₁, and m₂ are the magnetic strength or magnetic poles and r is the distance between two magnets.

The attractive or repulsive force, F = k (m₁×m₂) / μ₀r², where μ₀ is the permeability of the medium and k is the constant of proportionality. The unit is Newton/Tesla. The magnetic force is inversely proportional to the square of the distance of separation.

Hence, the ideal solution is option D.

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Answer:

Airbags:

" As the airbags comes in inbuilt form these days in most of the cars, as they fulfill the safety concerns of the passengers and make it sure that non of the individuals gets hurt in any accident faced during the travel."

Angle of Deployment:As the airbags will only deploy in the specific conditions provided, as the car must be speed over about the 25 km/hr and the accident must be more like a head-on-collision making the airbags to deploy. Or else if the car experiences an side wise crash with any other entity then it will never deploy in that case.

Answer:

SS and Ss

I did the test!!!!

By exerting the given constant pressure, work done on the system to decrease from initial volume to final volume is 2.84kJ.

Given the data in the question

The magnitude of the work done when a gas expands is equal to the product of the pressure of the gas and the change in the volume of the gas.

For work done by gas against a constant external pressure, we use the

expression:

[tex]Work\ done = -P\ *\ \delta V[/tex]

We substitute our given values into the equation

[tex]Work\ done = -P\ *\ ( v_2 - v_1)\\\\Work\ done = -4.0atm\ *\ ( 13.0L - 20.0L)\\\\Work\ done = -4.0atm\ *\ -7L\\\\Work\ done = 28L.atm\\\\Work\ done = 28L.atm\ *\ [ \frac{101.3J}{L.atm}] \\\\Work\ done = 2836.4J\\\\Work\ done = 2.84 kJ[/tex]

Therefore, by exerting the given constant pressure, work done on the system to decrease from initial volume to final volume is 2.84kJ.

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To decrease the volume of a system from 20.0 l to 13.0 l by exerting a constant pressure of 4.0 atm, a work of 2836.4 Joules must be done on the system.

In Physics, the concept of work done in a constant-pressure or isobaric process is defined as the product of pressure and the change in volume (W = PΔV). Since the pressure is given as 4.0 atm and must be converted to Joules using the conversion factor (1 atm = 101.3 J/L), the pressure in usable units becomes 4.0 atm * 101.3 J/L = 405.2 J/L. The change in volume is the initial volume subtracted from the final volume, so ΔV = 20.0L - 13.0L = 7.0L.

Substituting the values into our equation, the work done to decrease the volume of the system becomes W = (405.2 J/L) * (7.0 L) = 2836.4 Joules. So, to decrease the volume of the system from 20.0 l to 13.0 l by exerting a constant pressure of 4.0 atm, a work of 2836.4 Joules must be done on the system.

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Final answer:

The wavelength of the monochromatic light in the double slit experiment is calculated to be 625 nm, using the provided measurements and the double slit interference formula.

Explanation:

The student is asking to determine the wavelength of monochromatic light based on observations from a double slit interference experiment. Given that the slit separation is 1.8 mm, the distance to the screen is 4.8 m, and there are 6.0 complete bright fringes per centimeter on the screen, we can calculate the wavelength using the formula for double slit interference, λ = Δy × d / D, where λ is the wavelength of light, Δy is the distance between adjacent bright fringes (the fringe spacing), d is the separation between the slits, and D is the distance from the slits to the screen.

First, we find the fringe spacing by noting that there are 6 bright fringes per centimeter, so Δy = 1 cm / 6 = 0.1667 cm = 1.667 mm. We can then use the given values to calculate the wavelength:

[tex]λ = (1.667 \times 10^{-3} m) \times (1.8 \times 10^{-3} m) / (4.8 m) = 6.25 \times 10^{-7} m = 625 nm.[/tex]

Therefore, the wavelength of the monochromatic light is 625 nm.

The rope is doing work against friction. The equation for work done by a force is:

W = F×d

W is the work done.

F is the force.

d is the distance covered by the object subjected to the force.

The force of friction is calculated using the following equation:

F = μN

F is the frictional force.

μ is the coefficient of friction between the object and the contact surface.

N is the normal force exerted on the object by the contact surface.

In this situation the normal force is equal to the force of gravity on the object. The normal force is then:

N = mg

m is the mass of the object and g is the acceleration due to gravity of the Earth (9.81m/s²).

Combine all of these equations to get the total work done:

W = μ×m×g×d

Given values:

μ = 0.06

m = 12kg

g = 9.81m/s²

d = 7.8m

Substitute the terms in the equation with the given values and solve for W:

W = 0.06×12×9.81×7.8

Final answer:

The work done by tension on a 12 kg block being pulled along a horizontal floor with a coefficient of kinetic friction of 0.06 for a distance of 7.8 m is 156 joules.

Explanation:

To calculate the work done by tension before the block reaches the incline, we must consider the effects of kinetic friction. Since the rope is parallel to the incline and the force is constant, we can use the formula Work (W) = Force (F) * distance (d) * cos(\theta), where \(\theta\) is the angle between the force and displacement. In this case, the force is the tension in the rope and is parallel to the displacement, so \(\theta = 0 degrees\) and \(\cos(0) = 1\). The work done by tension can be calculated as:

\[W = T \times d \times \cos(0 degrees)\]

\[W = 20 N \times 7.8 m \times 1\]

\[W = 156 J\]

Therefore, the work done by tension on the block before it gets to the incline is 156 joules.

When two objects have the same momentum, the one with the smaller mass will have larger kinetic energy, because it must have a higher velocity. Conversely, if they have the same kinetic energy, the object with larger mass will have a higher momentum due to its larger mass.

Understanding Momentum and Kinetic Energy.When considering two objects, one with a small mass and another with a large mass, that both have the same momentum, it's important to understand the relationship between momentum (p), mass (m), and velocity (v). Momentum is calculated as the product of mass and velocity (p = mv). To have the same momentum, a small mass must move with a larger velocity compared to a large mass.

When comparing kinetic energy (K), which is given by the formula K = (1/2)mv², the object with the smaller mass but higher velocity will have a larger kinetic energy. This is because kinetic energy is proportional to the square of velocity, and since the smaller mass has a higher velocity to maintain the same momentum, its kinetic energy will be greater. If the scenario is reversed and two objects have the same kinetic energy, then the large mass object must have a larger momentum. Since kinetic energy is the same, the larger mass object compensates for its lower velocity with its greater mass, leading to a greater momentum (p = mv).

Both athletes reach the same maximum height. This is because the kinetic energy they start with (due to their initial speed) is the same, which is fully transformed into potential energy when they reach their maximum heights.

The problem is solved by relating the kinetic energy and potential energy of the athletes.

Before the athletes grab the rope, they have kinetic energy due to their motion (1/2mv2). Once they reach the maximum height, this kinetic energy is fully transformed into gravitational potential energy (mgh).

In the first case, the kinetic energy is (1/2)m(v2) and this is equal to the potential energy h at the maximum height (mgh), thereby, h = (v2)/2g.

In the second case, the kinetic energy is (1/2)(2m)(v2), and this is equal to the potential energy at the maximum height (2m)(g)(2h). In this scenario, the maximum height reached is 2h = (v2)/2g.

So, comparing the two scenarios, we clearly see that the maximum height reached by both athletes is identical. Although the second athlete is twice as heavy, they both start with the same speed and so the initial kinetic energy is the same.

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Final answer:

The weight of a student with a mass of 65 kg on Earth is 637 N (newtons).

Explanation:

According to Newton's second law of motion, force (F) is the product of an object's mass (m) and its acceleration (a). The formula for this relationship is F = ma. When talking about weight, we refer to the force of gravity acting on an object's mass. The acceleration due to gravity (g) on Earth is approximately 9.8 m/s2. Therefore, to calculate the weight of an object on Earth, we use the aforementioned relationship but with g as the acceleration, leading to W = mg. Substituting the given mass (65 kg) and the acceleration due to gravity, we get W = 65 kg × 9.8 m/s2 = 637 N, which is the student's weight on Earth.

Answer:

Angular acceleration, α = -2βt

Explanation:

Angular velocity of fan is [tex]\omega_{z(t)}=\gamma -\beta t^2[/tex]

[tex]\gamma=4.90\ rad/s[/tex]

[tex]\beta=0.750\ rad/s^3[/tex]

Angular acceleration is given by :

[tex]\alpha=\dfrac{d\omega}{dt}[/tex]

[tex]\alpha=\dfrac{d(\gamma -\beta t^2)}{dt}[/tex]

[tex]\alpha=-2\beta t[/tex]

Hence, the above equation is the angular acceleration as a function of time.   

Most people perceive the least intense sounds at a frequency around 2000 Hz with an intensity level at the threshold of hearing, which is about 0 dB or 10^-12 W/m².

The frequency at which most people perceive the least intense sounds is typically around the threshold of human hearing. This threshold is at about 0 dB, which corresponds to an intensity level of approximately 10-12 W/m². However, frequency and intensity are not the only factors that contribute to the perception of sound, as the sensitivity of the human ear varies across different frequencies. The human ear has the maximum sensitivity to frequencies ranging from 2000 to 5000 Hz. Sounds at these frequencies are perceived as louder compared to those at lower or higher frequencies, even if they have the same intensity level. Therefore, the least intense sounds that humans can perceive are at the lower end of this range, around 2000 Hz, at the absolute threshold of hearing of 0 dB.

Final answer:

The initial speed of the soccer ball can be calculated using the Pythagorean theorem and is found to be approximately 26.24 meters per second.

Explanation:

The initial speed of a soccer ball that is kicked with an initial horizontal velocity of 17 m/s and an initial vertical velocity of 20 m/s can be calculated using the Pythagorean theorem. The theorem states that the square of the hypotenuse (the initial speed, in this case) is equal to the sum of the squares of the other two sides (the horizontal and vertical velocities).

To find the initial speed, we use the following equation: [tex]initial speed = \sqrt{(horizontal velocity)^2 } +\sqrt{(vertical velocity)^2[/tex]. Substituting the values, we get initial speed = [tex]\sqrt{(17 m/s)^2} + \sqrt{(20 m/s)^2}[/tex]

≈[tex]\sqrt{289 + 400} m^2/s^2[/tex] ≈ 26.24 m/s. Therefore, the initial speed of the ball is approximately 26.24 meters per second.

Third option is correct. The mass of an object remains same when an unbalanced force is applied to it.

the mass on an object doesn't depend on the applied force. it is constant.When an unbalanced force is applied to an object, it will get accelerated. The velocity of the object will increase or decrease depending on the direction of the force but the mass of the object remains same. That doesn't change with the applied force.

0.35 m is the maximum distance that the block will compress the spring after the collision.

Given :

Mass, [tex]\rm m_b = 15 \;Kg\;and\; m_s = 3\;Kg[/tex]

Spring constant, K = 500 N/m

Solution :

Before the collision the total momentum is,

[tex]\rm p_i = m_s \times v_s[/tex]

Whare [tex]m_s[/tex] is mass of stone and [tex]\rm v_s[/tex] is the velocity of stone beforethe collision.

After the collision stone and block both are moving therefore total momentum is,

[tex]\rm p_f = m_sv_s' + m_bv_b[/tex]

where, [tex]\rm v_s'[/tex] is the velocity of stone after collision, [tex]\rm m_b[/tex] is the mass of block and [tex]\rm v_b[/tex] is the speed of the block after collision.

Now, momentum should be conserve

[tex]\rm p_i = p_f[/tex]

[tex]\rm m_sv_s= m_sv_s'+ m_bv_b[/tex]

[tex]\rm v_b = \dfrac{m_s}{m_b}(v_s - v_s')[/tex]  ------- (1)

[tex]\rm v_s = 8\;m/sec[/tex]      (Given)

[tex]\rm v_s' = 2 \;m/sec[/tex]      (Given)

Now putting the values in equation (1) we get,

[tex]\rm v_b = \dfrac{3}{15}(8-2)[/tex]

[tex]\rm v_b = 2\;m/sec[/tex]

Therefore, block slides with the speed of 2 m/s to the right after the collision.

We know that the block starts to move with speed 2 m/s, so its kinetic energy is

[tex]\rm K = \dfrac{1}{2}m_bv_b^2[/tex]

As the spring is compress, block speed decreases and its kinetic energy is converted into elastic potential energy of the spring and the elastic potential energy of the spring is,

[tex]\rm U = \dfrac{1}{2} Kx^2_m_a_x[/tex]

Now through conservation of energy we have,

[tex]\rm K = U[/tex]

[tex]\rm m_bv_b^2 = Kx^2_m_a_x[/tex]

[tex]\rm x_m_a_x = \sqrt{\dfrac{m_bv_b^2}{K}}[/tex]----- (2)

Putting the values of  [tex]\rm m_b,\;v_b\;and\; K[/tex]  in equation (2),

[tex]\rm x_m_a_x= \sqrt{\dfrac{15\times2^2}{500}}[/tex]

[tex]\rm x_m_a_x=0.35\;m[/tex]

0.35 m is the maximum distance that the block will compress the spring after the collision.

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Answer:

The brightness of the beam of light increases.

Explanation:

As per Einstein's experimental verification he proves that light is of particle nature and his equation for energy balance is given as

[tex]h\nu = \phi + KE[/tex]

so incident photons on a photosensitive plate will eject the electrons from the plate with some sufficient KE.

So here light is considered as particles which means one photon will eject one electron. So here intensity of light is considered as

Intensity = (number of photons per second) (energy of one photon)

so it is given as

[tex]I = \frac{N}{t}(\frac{hc}{\lambda})[/tex]

now if the intensity of light is increased for same wavelength of light then it means the number of photon must have to be increase.

so correct answer will be

The brightness of the beam of light increases.

Answer: Potential energy into kinetic energy.

Explanation: When the rock is in the edge, it is not actually moving, so it only has potential energy U = m*g*h (where m is the mass of the rock, g is the gravity's acceleration and h is the height at which the rock is, in this case, the height of the edge with respect to the ground).

Now, when the rock starts to fall, now the rock has velocity, so it has kinetic energy, which is written as K = (m/2)*v^2 where v is the velocity.

You can see that as the rock starts to fall, the height decreases, and also the potential energy. This is because the potential energy is being converted into kinetic energy, and this is why the velocity of the rock is increasing.

The marine life zones that are warm, well-lit, and have an abundance of marine life are known as the photic zone of the ocean. This zone includes both the Sunlight Zone and the Twilight Zone, which extends to a depth of about 200 meters below the sea surface. The photic zone is characterized by sufficient sunlight to support photosynthesis, which enables the growth of phytoplankton, the primary producers in marine ecosystems. Consequently, this zone supports a diverse array of marine life.

The photic zone is the uppermost layer of the marine biome where sunlight penetrates and enables photosynthesis. Phytoplankton, tiny photosynthetic organisms, form the base of the marine food web in this area, supporting a complex and diverse ecosystem. Other marine organisms, such as fish, marine mammals, and various invertebrates, depend on the photic zone directly or indirectly for sustenance. The availability of sunlight, warmth, and nutrients makes this zone the most lively and productive part of the ocean, contrasting with the aphotic zone, where the lack of light limits life. The health of the photic zone is vital for marine biodiversity and for global ecological processes such as oxygen production and carbon dioxide absorption.

As per the question only moving objects have energy.

The statement given here is not always true.

we know that every moving object has  kinetic energy.But in the universe there are other kinds of energy also.

We may take a simple  example of potential energy which is the energy possessed  by a body due to its configuration and position.

A stone present on the top of a mountain  has  zero kinetic energy but here the total mechanical energy is in the form of potential energy.

The electrostatic potential energy is not due to the motion of objects also.

There are also various other kinds of energy existing in the nature which are not  due to the motion of objects.

Answer:

The answer would be 2 solar masses.

The amplitude of the block's oscillation is 0.103 A.

The speed of the block at the given position is 0.34 m/s.

How to calculate the amplitude of the oscillation?

The amplitude of the block's oscillation is calculated by applying the principle of conservation of energy as follows.

K.E = U

¹/₂mv² = ¹/₂kA²

mv² = kA²

A² = mv²/k

A = √ (mv² / k)

where;

A = √ (1.1 x 0.36² / 13.5)

A = 0.103 A

The speed of the block at the given position is calculated as follows;

¹/₂kA² = ¹/₂mv² + ¹/₂kx²

kA² = mv² + kx²

mv² = kA² - kx²

v² = (kA² - kx²) / m

v = √ (13.5 x 0.103²  -  13.5(0.3 x 0.103)² ) / 1.1

v = 0.34 m/s

The complete question is below:

A 1.10 kg block is attached to a spring with spring constant 13.5 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 36.0 cm/s .

What are the amplitude of the subsequent oscillations and the block's speed at the point where x = 0.3 A?

The possible wavelengths that appear in the emission spectrum of the system are [tex]\boxed{210\,{\text{nm,}}\,{\text{250}}\,{\text{nm,}}\,{\text{410}}\,{\text{nm,}}\,{\text{620}}\,{\text{nm,}}\,{\text{1200}}\,{\text{nm}}}[/tex] .

Further Explanation:

Given:

The quantum energy levels allowed in the system are:

[tex]\begin{aligned}{E_1}&=1.0\,{\text{eV}}\hfill\\{E_2}&=2.0\,{\text{eV}}\hfill\\{E_3}&=4.0\,{\text{eV}}\hfill\\{E_4}&=7.0\,{\text{eV}}\hfill\\\end{aligned}[/tex]

Concept:

As the transition of electron takes from one energy level to another, there is an emission of particular wavelength from the transition. The relation between the wavelength of the emission and the energy of the energy level is expressed as:

[tex]\boxed{{E_f}-{E_i}=\frac{{hc}}{\lambda }}[/tex]

Here, [tex]{E_f}[/tex]  is the final energy level, [tex]{E_i}[/tex]  is the initial energy level and [tex]\lambda[/tex]  is the wavelength of emission.

(1). Transition of electron from [tex]{E_1}[/tex]  to [tex]{E_2}[/tex]  energy level:

[tex]\begin{aligned}{E_2}-{E_1}&=\frac{{hc}}{\lambda }\\\lambda &=\frac{{\left({6.63\times{{10}^{-34}}}\right)\times\left({3\times{{10}^8}}\right)}}{{\left({2-1}\right)\left({1.6\times{{10}^{-19}}}\right)}}\,{\text{m}}\\&{\text{=1}}{\text{.244}}\times{\text{1}}{{\text{0}}^{-6}}\,{\text{m}}\\&\approx{\text{1200}}\,{\text{nm}}\\\end{aligned}[/tex]

(2). Transition of electron from [tex]{E_1}[/tex]  to [tex]{E_3}[/tex]  energy level:

[tex]\begin{aligned}{E_3} - {E_1}&=\frac{{hc}}{\lambda }\\\lambda &=\frac{{\left({6.63\times{{10}^{-34}}}\right)\times\left({3\times{{10}^8}}\right)}}{{\left({4-1}\right)\left({1.6\times{{10}^{-19}}}\right)}}\,{\text{m}}\\&{\text{=0}}{\text{.414}}\times{\text{1}}{{\text{0}}^{-6}}\,{\text{m}}\\&\approx410\,{\text{nm}}\\\end{aligned}[/tex]

(3). Transition of electron from [tex]{E_1}[/tex]  to [tex]{E_4}[/tex]  energy level:

[tex]\begin{aligned}{E_4} - {E_1}&=\frac{{hc}}{\lambda }\\\lambda &=\frac{{\left({6.63\times{{10}^{-34}}}\right)\times\left({3\times{{10}^8}}\right)}}{{\left({7-1}\right)\left({1.6\times{{10}^{-19}}}\right)}}\,{\text{m}}\\&{\text{=0}}{\text{.207}}\times{\text{1}}{{\text{0}}^{-6}}\,{\text{m}}\\&\approx210\,{\text{nm}}\\\end{aligned}[/tex]

(4). Transition of electron from [tex]{E_2}[/tex]  to [tex]{E_3}[/tex]  energy level:

[tex]\begin{aligned}{E_3} - {E_2}&=\frac{{hc}}{\lambda }\\\lambda &=\frac{{\left({6.63\times{{10}^{-34}}}\right)\times\left({3\times{{10}^8}}\right)}}{{\left({4-2}\right)\left({1.6\times{{10}^{-19}}}\right)}}\,{\text{m}}\\&{\text{=0}}{\text{.621}}\times{\text{1}}{{\text{0}}^{-6}}\,{\text{m}}\\&\approx620\,{\text{nm}}\\\end{aligned}[/tex]

(5). Transition of electron from [tex]{E_2}[/tex]  to  [tex]{E_4}[/tex] energy level:

[tex]\begin{aligned}{E_4} - {E_2}&=\frac{{hc}}{\lambda }\\\lambda &=\frac{{\left({6.63\times{{10}^{-34}}}\right)\times\left({3\times{{10}^8}}\right)}}{{\left({7-2}\right)\left({1.6\times{{10}^{-19}}}\right)}}\,{\text{m}}\\&{\text{=0}}{\text{.248}}\times{\text{1}}{{\text{0}}^{-6}}\,{\text{m}}\\&\approx250\,{\text{nm}}\\\end{aligned}[/tex]

(6). Transition of electron from  [tex]{E_3}[/tex] to  [tex]{E_4}[/tex] energy level:

[tex]\begin{aligned}{E_4} - {E_3}&=\frac{{hc}}{\lambda }\\\lambda &=\frac{{\left({6.63\times{{10}^{-34}}}\right)\times\left({3\times{{10}^8}}\right)}}{{\left({7-4}\right)\left({1.6\times{{10}^{-19}}}\right)}}\,{\text{m}}\\&{\text{=0}}{\text{.414}}\times{\text{1}}{{\text{0}}^{-6}}\,{\text{m}}\\&\approx410\,{\text{nm}}\\\end{aligned}[/tex]

Thus, The possible wavelengths that appear in the emission spectrum of the system are [tex]\boxed{210\,{\text{nm,}}\,{\text{250}}\,{\text{nm,}}\,{\text{410}}\,{\text{nm,}}\,{\text{620}}\,{\text{nm,}}\,{\text{1200}}\,{\text{nm}}}[/tex] .

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2. Microwave ovens emit microwave energy with a wavelength of 12.5 cm. What is the energy of exactly one photon of this microwave radiation https://brainly.com/question/2385939

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Answer Details:

Grade: College

Subject: Physics

Chapter: Modern Physics

Keywords:

Allowed energies, quantum system, energy levels, wavelengths appear, emission spectrum, 1.0eV, 2.0eV, 4.0eV, 7.0eV, transition of electrons.

The rover sends photographs of Mars back to Earth. These pictures show scientists the landscape of the planet. The rover also collects specimens and looks for evidence of water so we can determine whether life could ever exist on Mars.

Answer:

Option-(D):  "A cat jumps from a tree limb to the ground."

Explanation:

Negative displacement(Assume positive position is measured vertically upward along a y-axis):

The displacement can be visualized going from the initial point been supposed as the ground level. While, the limb is considered more elevated as compared to the ground level. So, going from the lower level to the higher position is considered as positive displacement while the movement from the higher position to the lower point is considered as negative displacement.

Answer:

Need to know THEIR TEMPERATURES

Explanation:

Answer:

if value of k is high then it results less stretchy spring

Explanation:

As we know by formula of spring force

F = kx

here we know

k = spring constant

x = stretch in the spring

Now we know that when same force is applied on two types of spring then the stretch of the spring is given as

[tex]x = \frac{F}{k}[/tex]

so if the value of k is higher then the stretch in the spring will be less as it is inversely depends on the spring constant.

So here we can say that if value of k is high then it results less stretchy spring