The gauge pressure for the air in a balloon equals 1.01 x 10^5 Pa. If atmospheric pressure is equal to 1.01 x 10^5 Pa, what is the absolute pressure of the air inside the balloon?
a.
0 Pa
b.
5.05 x 10^4 Pa
c.
1.01 x 10^5 Pa
d.
2.02 x 10^5 Pa
D. 2.02 x [tex]10^{5}[/tex] Pa
Explanation:
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
A person pulls a toboggan for a distance of 37.9 m along the snow with a rope directed 32.8 ° above the snow. the tension in the rope is 105 n. (a) how much work is done on the toboggan by the tension force? (b) how much work is done if the same tension is directed parallel to the snow?
if a vector a - b = 2c, a + b = 4c and c = 3i + 4j,find a and b
The rock in the picture is pushed over the edge and it begins to falls. Which type of energy conversion is taking place?
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.
FILL IN TUE BLANK WITH ANSWER DOWN BELOW!!! When storm clouds produce lightning and thunder, 1)_________ energy changes to2)______ energy and 3)_______ energy
1) electric potential, gravitational potential, or sound?
2) Chemical, electric potential, or radiant?
3) Chemical, electric potential, or sound?
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.
What is the current when a typical static charge of 0.234~\mu\text{c}0.234 μc moves from your finger to a metal doorknob in 0.595~\mu\text{s}0.595 μs?
A soccer ball is kicked with an initial horizontal velocity of 17 m/s and an initial vertical velocity of 20 m/s. 1) what is the initial speed of the ball?
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.
A 1 ev electron has a 10-4 "probability of tunneling through a" 2.0 ev "potential barrier. what is the probability of a" 1 ev "proton tunneling through the same barrier
The air inside the building is easily pressed out the windows because the air is composed of widely spaced molecules in the _____ phase.
solid
liquid
gas
plasma
Monochromatic light of a given wavelength is incident on a metal surface. however, no photoelectrons are emitted. if electrons are to be ejected from the surface, how should the incident light be adjusted? monochromatic light of a given wavelength is incident on a metal surface. however, no photoelectrons are emitted. if electrons are to be ejected from the surface, how should the incident light be adjusted? light of a longer wavelength should be used. light of a shorter wavelength should be used. the intensity of the light should be decreased. the intensity of the light should be increased.
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.
The allowed energies of a quantum system are 1.0 ev, 2.0 ev, 4.0 ev, and 7.0 ev. what wavelengths appear in the system's emission spectrum
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:
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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
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.
what causes the end of a star?
A culture is a group of people that share the same __________.
A. development goals
B. professions and work
C. beliefs and behaviors
D. philosophies and religion
An ebonite (very hard, black rubber) rod is rubbed with the four different substances below. Choose the one that is MOST likely to produce the largest static charge by friction.
A 3.00 kg mud ball has a perfectly inelastic collision with a second mud ball that is initially at rest. the composite system moves with a speed equal to one- third the original speed of the 3.00 kg mud ball. what is the mass of the second mud ball?
A 1200-kg ore cart is rolling at 10.8 m/s across a flat friction-free surface. a crane suddenly drops of ore vertically into the cart. how fast does the cart move just after being loaded with the ore?
Only moving things have energy
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.
Which statements accurately describe elements? Check all that apply.
Elements are made up of two or more types of atoms.
Elements are made up of only one type of atom.
Each element has a unique chemical symbol.
Elements can be identified by their atomic number.
One element cannot be combined with another element.
Answer: The correct statements are:
- Elements are made up of only one type of atom.
- Each element has a unique chemical symbol.
- Elements can be identified by their atomic number.
Explanation: and just for a little bit more information about elements.
An element is a pure substance that is made from a single type of atom. Elements are the building blocks for all the rest of the matter in the world. Examples of elements include iron, oxygen, hydrogen, gold, and helium. An important number in an element is the atomic number.
A solid object heated to 120°C is placed inside a container of water at 33°C. What is the BEST explanation of the resulting change in temperature?
A) The temperature of both the water and solid will decrease.
B) The temperature of the solid will decrease and the water will decrease.
C) The temperature of the water will increase; the temperature of the solid will decrease.
D) The temperature of the water will remain constant; the temperature of the solid will decrease.
Answer:
its c
Explanation:
i took it
Select all 3 of the reactants (things needed) necessary for photosynthesis to occur. A. oxygen B. carbon dioxide C. sunlight D. water E. sugar (glucose)
A 2.90-kg ball, moving to the right at a velocity of 4.69 m/s on a frictionless table, collides head-on with a stationary 8.80-kg ball. find the final velocities of (a) the 2.90-kg ball and of (b) the 8.80-kg ball if the collision is elastic. (c) find the magnitude and direction of the final velocity of the two balls if the collision is completely inelastic.
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.
__________ helps us adapt to our environment. It also generally __________ with age. A. Plasticity . . . decreases B. Development . . . increases C. Plasticity . . . increases D. Development . . . decreases
Answer: A, Plasticity ... decreases
Explanation:
Which of these was a characteristic of the Thomson model of the atom, but not the Dalton model?
negatively charged electrons
positive charge concentrated at the center of the atom
unequal positive and negative charges
indivisible atoms
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.
Which image editing technique allows you to reverse an image (horizontally or vertically)?
cropping
rotating
resizing
flipping
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.
What do you mean by image editing technique?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:
Crop your images and clean them up.Adjust the white balance.Adjust exposure and contrast.Adjust color vibrancy and saturation.Sharpen images.Finalize and share.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.
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A block of mass 2.5 kg is attached to a spring with spring constant 150 n/m and slides on a rough, horizontal surface (coefficient of kinetic friction 0.25). the spring is initially extended by a distance of 45 cm from its equilibrium position and released from rest. (a) what is the speed of the block when it is at the equilibrium position of the spring? (b) where does the block stop?
Burning magnesium gives us magnesium oxide. This is a example of
A mass m = 12 kg is pulled along a horizontal floor, with a coefficient of kinetic friction μk = 0.06, for a distance d = 7.8 m. then the mass is continued to be pulled up a frictionless incline that makes an angle θ = 25° with the horizontal. the entire time they massless rope used to pull the block is pulled parallel to the incline at an angle of θ = 25° (thus on the incline it is parallel to the surface) and has a tension t = 20 n. 1) what is the work done by tension before the block gets to the incline
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
W = 55.1JFinal 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.
If vx = 7.90 units and vy = -6.50 units, determine (a)the magnitude and (b)direction of v
An element which has strong intermolecular forces is most likely to have
a. a boiling point below room temperature
b. a melting point below room temperature
c. a boiling point very close to its melting point
d. a very high melting point
According to the forces of attraction, an element which has strong inter molecular forces is most likely to have a very high melting point.
What are forces of attraction?
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.
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If an object is released in the shaft at the earth's surface, what speed will it have when it reaches the center of the earth?
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.