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Which accurately compares concave and convex lenses?
A. With both concave and convex lenses, the characteristics of the image do not depend on the placement of the object.
B. With both concave and convex lenses, the characteristics of the image depend on the placement of the object.
C. With concave lenses, the characteristics of the image do not depend on the placement of the object, but with convex lenses, they do.
D. With convex lenses, the characteristics of the image do not depend on the placement of the object, but with concave lenses, they do.

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.

D.        2.02 x [tex]10^{5}[/tex] Pa

The gauge pressure is the pressure measured relative to the atmospheric pressure.

Absolute pressure (or total pressure), is the sum of the gauge pressure and the atmospheric pressure. i.e

[tex]P_{ABS}[/tex] = [tex]P_{G}[/tex] + [tex]P_{ATM}[/tex]

Where;

[tex]P_{ABS}[/tex] = absolute pressure

[tex]P_{G}[/tex] = gauge pressure = 1.01 x [tex]10^{5}[/tex]Pa  (from the question)

[tex]P_{ATM}[/tex] =  atmospheric pressure = 1.01 x [tex]10^{5}[/tex]Pa

Substitute these values into the equation above;

[tex]P_{ABS}[/tex] = [tex]P_{G}[/tex] + [tex]P_{ATM}[/tex]

[tex]P_{ABS}[/tex] = 1.01 x [tex]10^{5}[/tex] + 1.01 x [tex]10^{5}[/tex]

[tex]P_{ABS}[/tex] = 2.02 x [tex]10^{5}[/tex] Pa

Therefore, the absolute pressure of the air inside the balloon is 2.02 x [tex]10^{5}[/tex] Pa

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.

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:

The answer would be 2 solar masses.

Answer:

its c

Explanation:

i took it

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.

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.

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).

the answer would be D

Answer:

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

Explanation:

Answer:

negatively charged electrons  

Explanation:

According to Dalton's Model, matter is made up of tiny particles known as atoms. Electrons were not known at that time. Thomson gave the plum pudding model which describe atom to be over all neutral with equal distribution of negatively charged electrons in the region of positive charges.

Thus, negatively charged electrons were a characteristic of the Thomson's model of the atom but not the Dalton's model.

The speed of an object when it reaches the center of the Earth depends on the conversion of gravitational potential energy into kinetic energy, and while exact calculation of this speed requires complex calculations considering Earth's non-uniform structure, it would be less than the Earth's escape velocity of approximately 11 km/s.

Speed of an Object Reaching the Center of the Earth

When an object is released in a shaft at the Earth's surface, to determine the speed it would have when it reaches the center, we must look at gravitational acceleration and potential energy. The acceleration due to gravity near the Earth's surface is 9.8 m/s². However, as the object falls towards the center, the gravitational force decreases because it is proportional to the distance from the center. By the time it reaches the center, the gravitational force is zero and the object would be in freefall.

Nevertheless, when an object initially falls from the surface it accelerates, gaining velocity until the gravitational force lessens. If the Earth were a uniform sphere, the problem could be simplified using the concept of simple harmonic motion where maximum speed would be reached at the center. For a non-uniform Earth, complex calculations involving Earth's density variation would be needed. Ignoring air resistance and Earth's rotation, an important value related to this concept is the escape velocity, which is about 11 km/s for Earth. This escape velocity gives us a frame of reference for the maximum potential speed that can be achieved by an object falling towards the center due to the conversation of potential energy into kinetic energy. In practical terms however, air resistance and the rotational speed of the Earth would affect the actual speed, making the calculation far more complex.

If we were to hypothesize, the object would convert gravitational potential energy into kinetic energy. Without precise calculations and factoring in Earth's interior structure, giving an exact velocity upon reaching the center is not straightforward. It is generally understood, however, that the speed would be significant, although less than the object's escape velocity.

Ignoring complicating factors, the speed of an object falling to the center of the Earth can be conceptualized like the parabolic velocity of 6.94 miles per second, adjusted for the varying gravitational pull as the object approaches the center. A detailed calculation would involve conservation of energy and Newton's law of gravitation.

To calculate the speed of an object as it reaches the center of the Earth after being released from the Earth's surface, we must consider the forces acting on the object. The acceleration due to gravity (9.8 m/s²) would act on the object as it falls. However, as the object moves closer to the center, the gravitational force acting on it decreases because the mass of the Earth underneath the object decreases. This means that the object will accelerate until it reaches the point where the gravitational forces from all parts of the Earth are equal.

Using the principle of conservation of energy, where potential energy is converted into kinetic energy, and assuming a uniform density of the Earth, one could calculate this speed; however, the exact calculation would require advanced knowledge of differential equations and concepts from Newton's law of gravitation.

It is crucial to note that this is a simplification, as several factors like the Earth's rotation, the changing density of Earth's layers, and air resistance are not considered here. If we ignore these factors and assume constant acceleration, an object starting from rest will be at its maximum velocity when it reaches the center. This velocity can be conceptualized similarly to the parabolic velocity of 6.94 miles per second, which represents the speed of an object affected only by Earth's gravitational pull from space without atmospheric resistance.

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] .

Learn More:

1. Calculate the wavelength of an electron (m = 9.11 × 10-28 g) moving at 3.66 × 106 m/s https://brainly.com/question/1979815

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

3. What is the frequency of light for which the wavelength is 7.1 × 102 nm https://brainly.com/question/9559140

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.

Final answer:

The problem requires applying the conservation of momentum and kinetic energy to find final velocities for elastic collisions, and only momentum conservation for inelastic collisions.

Explanation:

The question involves a 2.90-kg ball colliding with an 8.80-kg stationary ball. For elastic collisions, we use conservation of momentum and conservation of kinetic energy. For inelastic collisions, we only use the conservation of momentum.

(a) For the 2.90-kg ball, we first apply the law of conservation of momentum:

where m1 = 2.90 kg, v1_initial = 4.69 m/s, m2 = 8.80 kg, and v2_initial = 0 m/s.

(b) The final velocity of 8.80-kg ball is also determined by applying conservation of momentum and energy for an elastic collision.

c) If the collision is completely inelastic, the balls stick together and move with the same final velocity. This is found using:

We can solve these equations to find the final velocities of both balls for both scenarios.

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: 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.

Answer:

1) electric potential, Radiant Energy(Lightening) , or sound Energy(Thundering)

Explanation:

During lightening we know that clouds are at high electric potential and earth surface is taken at low potential.

Due to this potential difference the charge particles in the atmosphere has tendency to move from high potential to low potential.

Also the moisture present and few dipole gases present in the atmosphere will have tendency to show breakdown and due to this breakdown and motion of charge the phenomenon of lightening occurs.

So here the electric potential energy of charge particles will change and it will convert into sound energy as it produce sound as well as in Radiant energy in form of lightening occurs in the sky.

According to the forces of attraction, an element which has strong inter molecular  forces is most likely to have a very high melting point.

Forces of attraction  is a force by which atoms in a molecule  combine. it is basically an attractive force in nature.  It can act between an ion  and an atom as well.It varies for different  states  of matter that is solids, liquids and gases.

The forces of attraction are maximum in solids as  the molecules present in solid are tightly held while it is minimum in gases  as the molecules are far apart . The forces of attraction in liquids is intermediate of solids and gases.

The physical properties such as melting point, boiling point, density  are all dependent on forces of attraction which exists in the substances.

Learn more about forces of attraction,here:

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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?

To eject electrons from a metal surface, monochromatic light of a shorter wavelength should be used, as this will increase the energy of photons to overcome the metal's work function.

When monochromatic light is incident on a metal surface and no photoelectrons are emitted, it indicates that the energy of the photons is not sufficient to overcome the work function of the metal. To eject electrons from the surface, one should use monochromatic light of a shorter wavelength. According to the photoelectric effect, there is a minimum frequency (threshold frequency) required for electrons to be emitted, and since frequency and wavelength are inversely related (the shorter the wavelength, the higher the frequency), a shorter wavelength would increase the energy of the photons. It is important to note that increasing the intensity of light only increases the number of electrons ejected but does not affect their kinetic energy or the threshold frequency.

Answer:

Flipping

Explanation:

The flip tool reverses an image on the horizontal or vertical axis.

The image editing technique that allows you to reverse an image (horizontally or vertically) is known as flipping. Thus, the correct option for this question is D.

Image editing technique may be defined as a type of technique that modifies or improves digital or traditional photographic images using different techniques, tools or software. Images produced by scanners, digital cameras, or other image-capturing devices may be good, but not perfect.

It encompasses the processes of altering images, whether they are digital photographs, traditional photo-chemical photographs, or illustrations. Some basic image editing process is as follows:

Therefore, the image editing technique that allows you to reverse an image (horizontally or vertically) is known as flipping. Thus, the correct option for this question is D.

To learn more about Image editing techniques, refer to the link:

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