Answer:
the envelope of gases surrounding the earth or another planet.
Answer:
An atmosphere is the layers of gases surrounding a planet or other celestial body. Earth's atmosphere is composed of about 78% nitrogen, 21% oxygen, and one percent other gases.
How does the suns energy affect the climate of an area
There wouldn't be any life on Earth and no climate if it weren't for the Sun.
The spherical form of the Earth results in changes in the amount of sunlight that reaches the surface at various latitudes, giving rise to various climatic zones. Seasons are caused by the tilt of the Earth's axis of rotation in relation to the ecliptic plane.
The surface of the sun is around 6000 C hot; at this temperature, the solar mostly produces visible light. The Earth radiates infrared energy back into space to maintain its energy balance after being warmed by incoming solar radiation. The Earth's surface has an average temperature of 15 C, and at this temperature, the planet emits infrared light.
Carbon dioxide, methane, nitrous oxide, and other atmospheric greenhouse gases let solar energy to pass through but absorb infrared radiation that the planet emits back into space. This feature is not shared by other atmospheric elements like oxygen and nitrogen. These greenhouse gases re-emit infrared radiation in all directions to maintain energy balance, some of which is aimed at the Earth's surface and warms it. More greenhouse gases will increase infrared absorption and raise surface temperatures. This is comparable to how much blanket you cover yourself with at night. The blanket's thickness affects how warm you feel. The result of heat brought on by rising levels of greenhouse gases in the atmosphere is climate change.
In the course of billions of years, the Sun will continue to get hotter and brighter, rendering the Earth uninhabitable.
Additionally, all fossil fuels (coal, petroleum, and natural gas) and renewable energy are directly and indirectly derived from the sun (wind, solar, biomass, and hydroelectric).
Thank you,
Eddie
(Please help 25 points)
Asteroids X, Y, and Z have equal mass of 6.0 kg each. They orbit around a planet with M=6.20E^24 kg. The orbits are in the plane of the paper and are drawn to scale.
In the statements below, TE is the total mechanical energy, KE is the kinetic energy, and PE is the potential energy.
The PE of X at r is .... that at s
The TE of Y is .... the TE of X
The PE of Y at u is .... the PE of Z at c
The speed of Y at i is .... that at u
The KE of Z at u is .... that at m
The PE of Y at s is .... the PE of X at c
The TE of Z is .... the TE of X
The PE of X at c is .... the PE of Z at c
(Options are: greater than, less than, equal to)
The potential energy (P.E) of X at r is less than P.E at s.
The total energy (T.E) of Y is less than total energy (T.E) of X.
The PE of Y at v is greater than the PE of Z at c.
The speed of Y at i is .greater than that at v.
The kinetic energy (KE) of z at u is less than that at n.
The PE of Y at s is less than the PE of X at c.
The PE of X at c is less than the PE of Z at c.
The PE of X at c is less than the PE of Z at c.
PE of X at r and at sThe potential energy (P.E) of X at r is less than P.E at s due to greater displacement at s.
TE of Y is .... the TE of XThe total energy (T.E) of Y is less than total energy (T.E) of X because X has greater displacement from the planet.
PE of Y at v is .... the PE of Z at cThe PE of Y at v is greater than the PE of Z at c due to greater displacement at v.
Speed of Y at i is .... that at uThe speed of Y at i is .greater than that at v because i is closer to the planet than v.
KE of Z at u is .... that at nThe kinetic energy (KE) of z at u is less than that at n because n is closer to the planet and will have greater velocity.
PE of Y at s is .... the PE of X at cThe PE of Y at s is less than the PE of X at c, because position c has greater displacement.
The TE of Z is .... the TE of XThe TE of Z is greater than the TE of X due to greater displacement of Z from the planet.
The PE of X at c is .... the PE of Z at cThe PE of X at c is less than the PE of Z at c due to greater displacement of Z from c.
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2. Two particles P and Q are shot vertically up. T
Particle P is first up with a velocity 40ms-¹. After
4s, particle Q is also shot up. Find 01 -
(a) where the two particles meet, if at the point of
meeting, P has velocity of 15ms™¹.
(b) the velocity with which Q is shot. (g = 10ms-²)
(000)(9)
(a) The position where particle Q and P meet is 68.75 m.
(b) The velocity with which particle Q is shot is 15 m/s.
Time at which the two particles meet each otherThe time elapsed before the two particles meet is calculated as follows;
Distance Q - Distance P = Distance between them
Distance traveled by particle P:Trise = (V - V₀)/g
Trise = (0 - 40) / -10
Trise = 4.0 s
Hmax = V₀t + 0.5gt²
Hmax = 40 x 4 - (0.5)(10)4²
Hmax = 80 m.
P is falling when Q is moving up:h = Hmax - (V² - V₀²)/2g
h = 80 - ((15)²- 0) / 20
h = 68.75 m.
Thus, the position where particle Q and P meet is 68.75 m
Tfall = (V - V₀)/g = (15-0) / 10
Tfall = 1.5 s
Fall time. = Rise time for Q.
Distance traveled by particle Q:h = V₀t + 0.5gt²
h = 80 - 68.75
h = 11.25
V₀ x 1.5 - 5(1.5)² = 11.25
V₀ x 1.5 - 11.25 = 11.25
V₀ x 1.5 = 22.5
V₀ = 22.5 / 1.5
V₀ = 15 m/s
Thus, the velocity with which particle Q is shot is 15 m/s.
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what is relative density of a substance
Answer:
Relative density is the ratio of the density of a substance to the density of a standard, usually water for a liquid or solid, and air for a gas.
Describe at least four techniques used by astronomers to determine distances to stars and galaxies. Make sure you include the scientific theory behind each method and the range over which each method is considered valid.
Radar - measuring distances in our solar system
Parallax - measuring distances to nearby stars
Cepheids - measuring distances in our Galaxy and to nearby galaxies
Supernovae - measuring distances to other galaxies
Different techniques which can be used by the astronomers to determine the distances to the stars and galaxies include Radar, Parallax, Cepheids, and Supernova.
What are the techniques to determine distance of stars?
The distances to the nearby stars are precisely and accurately determined using the different techniques such as parallax. When a celestial body is seen from a different, widely separated viewing point, its position with respect to the more distant background stars or galaxies varies. This angular difference is known as parallax.
Radar is the measuring of the distances in the solar system.
Parallax is the measuring of distances to the nearby stars.
Cepheids is the measuring of distances in Galaxy and to nearby galaxies.
Supernovae is the measuring of distances to the other galaxies.
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[tex] \large{\rm{Question:}}[/tex]
Give one example where the displacement is zero but the distance traveled is not zero.
Let us assume a man travelled from a point A to a point B over a distance 'd'. After a while he travels the same distance back to point A.
Therefore, since his initial and final positions are the same, displacement is equal to zero, and the distance travelled is (d + d) = 2d, which is not zero.
Example:
Imagine a person walks 10 meters forward and then turns around and walks 10 meters back to the starting point. In this case, the displacement is zero (starting point to ending point), but the distance traveled is 20 meters (10 meters forward + 10 meters backward).
What is 2 chemical properties of acids and bases
⇒ Acids taste sour, react with metals, react with carbonates, and turn blue litmus paper red. Bases taste bitter, feel slippery, do not react with carbonates and turn red litmus paper blue.
The study of acids and bases is crucial to chemistry. The Lewis acid/base motif, which broadens the concept of an acid and base beyond H+ and OH- ions, is one of the most relevant theories.
Acids are ionic compounds, which means they have a positive or negative charge. In water, these ionic compounds separate to create hydrogen ions, or H+.
The quantity of H+ ions in the solution determines how strong an acid is. Acid is stronger the more H+ there is. Bases are ionic substances that separate in water to produce the negatively charged hydroxide ion (OH-). The quantity of Hydroxide ions in a base determines its strength (OH-). The strength of the base increases with OH- ion concentration.
Hope this helps,
- Eddie.
Stop and go traffic means a lot of opportunities for acceleration. This assignment looks at a typical experience of a car in typical
rush hour traffic. Calculate the different accelerations from the graph, interpret what your answers mean and show you really
understand acceleration.
A car traveled through rush hour traffic in a city. There was a lot of starting and stopping. The graph below shows the motion of
the car for a 16 second interval of time.
Explain why a Merry-Go-Round and a Ferris Wheel have a constant acceleration when they are moving?
You must answer using at least two complete sentences and vocabulary words from this lesson.
Q.1 Calculate the different accelerations from the graph
At initial point (let O) the velocity was 0 m/s to reach point A when the velocity was 5 m/s it took nearly 3 seconds as we all know that acceleration is nothing but the change in velocity per unit time
[tex]a \: = \frac{dv}{t} [/tex]
[tex]a = \frac{5 - 0}{3} = \frac{5}{3} \: m/ {s}^{2} [/tex]
Similarly from point A to point B the acceleration was
[tex]a = \frac{dv}{t} = \frac{7 - 5}{3} = \frac{2}{3} m/s²[/tex]
From Point B to C the velocity was constant so the acceleration at that point would be zero.
in the same way we can calculate acceleration for rest other points.
Q.2 Explain why a Merry-Go-Round and a Ferris Wheel have a constant acceleration when they are moving?
We all know that Velocity is nothing but the speed with direction. Merry Go Round and a Ferris Wheel has constant speed but dynamic direction due to which the magnitude of velocity at every point differs with different sign hence it have a constant acceleration.
define universal vibrations ?
Answer:
The second universal law, the law of vibration, posits that everything (every atom, object, and living thing) is in constant motion, vibrating at a specific frequency.
Explanation:
Please mark brainly
The second universal law defines this.
The second universal law, also known as the Law of Vibration, The Law of Vibration states that everything in the universe is in a constant state of movement. We refer to these movements as vibration, and the speed or rate at which something vibrates is called its frequency. The only difference between one object and another is its vibration rate.
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A thin spherical shell of mass M and radius r is allowed to roll from the edge of a hemispherical bowl of radius R = 80.0 cm. It rolls down with no slipping.
1) Find the speed of the center of mass of the spherical shell when it is at the bottom of the bowl, if r is very small.
2) Repeat part 1) if r = 10.0 cm. Moment of inertia of a thing spherical shell if 2/3Mr^2.
(a) The speed of the center of mass of the spherical shell when it is at the bottom of the bowl is 3.1 m/s when the radius is 80 cm.
(b) The speed of the center of mass of the spherical shell when it is at the bottom of the bowl is 1.1 m/s when the radius is 10 cm.
Speed of the shell at the bottom of the bowl
The speed of the shell at the bottom of the bowl is calculated by applying the principle of conservation of energy.
K.E(rot) + K.E(trans) = P.E
where;
P.E is the potential energy of the ball at the initial positionK.E(rot) is rotational kinetic energyK.E(trans) is translation kinetic energy¹/₂mv² + ¹/₂Iω² = mgh
where;
I is moment of inertia of the spherical shellh is the height of fallv is the speed at the bottomω is angular speed¹/₂mv² + ¹/₂(²/₃Mr²)(v/r)² = mgh
¹/₂v² + ¹/₂(²/₃r²)(v²/r²) = gh
¹/₂v² + ¹/₂(²/₃)(v²) = gh
¹/₂v² + ¹/₃v² = gh
⁵/₆v² = gh
v² = 6gh/5
v = √(6gh/5)
Let the vertical height from the edge of bowl to the bottom , h = R = 80 cm
v = √(6 x 9.8 x 0.8 /5)
v = 3.1 m/s
When the radius = 10 cmv = √(6gh/5)
v = √(6 x 9.8 x 0.1 /5)
v = 1.1 m/s
Thus, the speed of the center of mass of the spherical shell when it is at the bottom of the bowl is 3.1 m/s when the radius is 80 cm.
The speed of the center of mass of the spherical shell when it is at the bottom of the bowl is 1.1 m/s when the radius is 10 cm.
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When the current through a circular loop is 5.7 A, the magnetic field at its center is 3.9 ✕ 10−4 T. What is the radius (in m) of the loop?
Radius of the circular loop is 0.0091m.
What is magnetic field?Magnetic field is the area around a magnet where the magnetism influence is felt .
What is the magnetic field at the centre of a circular loop?The formula for magnetic field at the centre of a loop isB =(μ)I/2r
where B= Magnetic field at the centre of a circular loopμ= Magnetic permeability =4(π)*10^(-7)
I= current flowing through the loop
r= radius of the loop
Thus, radius of the loop =(4(π)×10^(-7)×5.7)/(2×3.9×10^(-4))=0.0091m
Thus, we can conclude that the radius of the loop is 0.0091m .
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A cell connected to two bulbs in series lasts longer than a cell connected to the same bulbs in parallel.Explain why?
Because the voltage stays the same while the amps grow, parallel batteries last longer. It will still be a 12 volt system if you connect two 12v 50ah batteries in parallel, but because the amps quadruple to 100ah, the batteries will last longer. Charges are spread among the bulbs in a parallel circuit because they are flowing in several directions over multiple paths. As a result, each bulb will have less charge since the charge is travelling in multiple directions, and the lights will shine less brightly. On the other hand, in a series circuit, there is only one channel and only one location for the charge to travel, meaning that all of the charge is directed at a single lightbulb. As a result, the lightbulb will shine more brilliantly since there is no need to spread the energy. The rule of conservation of energy is not broken because of the difference in power; instead, smaller resistance causes bigger current, which results in brighter lights. The energy traveling from the source to the bulbs remains constant. Because higher power means that more energy in a given time goes quicker from the source to the bulb, brighter lights require more power rather than energy. Because the energy traveling from the source to the bulbs remains constant, the rule of conservation of energy is not being broken; rather, there is merely a difference in power as a result of smaller resistance leading to bigger current, which will result in brighter lights. This means that brighter lights require more power rather than energy since higher power enables more energy to go from the source to the bulb more quickly.
Thank you,
Eddie
If the gravitational potential energy of an object 10 m above the ground is 50 J, what is its Ep, if it moves to 30 m above the ground?
Answer:
150 J
Explanation:
Moving 3 times higher will increase the P E x 3 = 150 J
Complete the following:
When light is incident parallel to the principal axis and then strikes a lens, ___
the light will remain parallel after refracting through the lens
the light will refract through the focal point on the opposite side of the lens
the light will not refract at all
When light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.
To find the answer, we have to know about the rules followed by drawing ray-diagram.
What are the rules obeyed by light rays?If the incident ray is parallel to the principal axis, the refracted ray will pass through the opposite side's focus.The refracted ray becomes parallel to the major axis if the incident ray passes through the focus.The refracted ray follows the same path if the incident light passes through the center of the curve.Thus, we can conclude that, when light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.
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A solid, homogeneous sphere with a mass of m0, a radius of r0 and a density of ρ0 is placed in a container of water. Initially the sphere floats and the water level is marked on the side of the container. What happens to the water level, when the original sphere is replaced with a new sphere which has different physical parameters? Notation: r means the water level rises in the container, f means falls, s means stays the same. Combination answers like 'f or s' are possible answers in some of the cases.
The new sphere has a density of ρ > ρ0 and a mass of m = m0.
The new sphere has a mass of m > m0 and a radius of r = r0.
The new sphere has a radius of r = r0 and a density of ρ > ρ0.
Based on the given data about, volume, mass, and density of the spheres;
the water level will fall; fthe water level may rise or fall; r or fthe water level will rise; rWhat determines whether an object floats are sinks in water?The density, mass, volume, and shape of an object determines whether it will float or sink in water.
Density = mass/volumeFor an object to float, it will displace its own weight of the fluid in which it floats. This is according to Archimedes' principle which states that the upward force acting on a body immersed fully or partially in a fluid, is equal to the weight of the fluid displaced.
Considering the given scenarios:
1. The new sphere has a density of ρ > ρ0 and a mass of m = m0.
This means that the volume of new sphere is smaller than the volume of the original sphere, therefore, the water level will fall,
2. The new sphere has a mass of m > m0 and a radius of r = r0.
This means that the new sphere will have a greater volume and mass. However, the density may be greater or less.. Therefore, the water level may rise or fall.
3. The new sphere has a radius of r = r0 and a density of ρ > ρ0.
The new sphere has equal volume, greater density and mass than the original sphere. Therefore, the water level will rise.
In conclusion, the level of the water in the container depends on the volume, mass and density of the object placed in it.
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The diameters of the main rotor and tail rotor of a single-engine helicopter are 7.70 m and 1.03 m, respectively. The respective rotational speeds are 449 rev/min and 4,150 rev/min. Calculate the speeds of the tips of both rotors.
main rotor ______m/s
tail rotor _______m/s
Compare these speeds with the speed of sound, 343 m/s.
vmain rotor = _______ vsound
vtail rotor = _______ vsound
(a) The speeds of the tips of both rotors; main rotor 181.02 m/s and tail rotor 223.8 m/s.
(b) The speed of the main rotor is 52.8 % speed of sound, and the speed of the tail rotor is 65.2 % speed of sound.
Linear speed of main motor and tail rotor
v = ωr
where;
ω is the angular speed (rad/s)r is radius (m)v(main rotor) = (449 rev/min x 2π rad x 1 min/60s) x (0.5 x 7.7 m)
v(main rotor) = 181.02 m/s
v(tail rotor) = (4,150 rev/min x 2π rad x 1 min/60s) x (0.5 x 1.03 m)
v(tail rotor) = 223.8 m/s
Speed of the rotors with respect to speed of sound% speed (main motor) = 181.02/343 = 0.528 = 52.8 %
% speed (tail motor) = 223.8/343 = 0.652 = 65.2 %
Thus, the speed of the main rotor is 52.8 % speed of sound, and the speed of the tail rotor is 65.2 % speed of sound.
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Why are enthalpy changes always lower than their expected values?
A. They aren't. They are always greater than expected.
B. Heat losses during the experiment
C. Heat gains during the experiment
D. Faulty measurements
Enthalpy changes are always lower than their expected values due to faulty measurements; option D.
What are enthalpy changes?Enthalpy changes refers to the changes in the heat content of substances in the course of a reaction.
Enthalpy changes occurs as a result of bond breaking and bond formation in the reactant molecules and product molecules respectively.
Enthalpy change = Energy of bond formation - Energy of bond breakage.Enthalpy changes are measured from the changes in temperature that are observed in the course of a given reaction.
Due to faulty measurements in a particular experiment measuring enthalpy changes, enthalpy changes are always lower than their expected values.
Therefore, accurate measurements are required if results as close as possible to the actual enthalpy changes are to be obtained.
In conclusion, enthalpy changes measure the heat changes that occur during a particular chemical reaction.
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A 750 kg car is stalled on an icy road during a snowstorm. A 1000 kg car traveling eastbound at 13 m/s collides with the rear of the stalled car. After being hit, the 750 kg car slides on the ice at 4 m/s in a direction 30 ∘ north of east.
A) What is the magnitude of the velocity of the 1000 kg car after the collision?
B) What is the direction of the velocity of the 1000 kg car after the collision?
C) Calculate the ratio of the kinetic energy of the two cars just after the collision to that just before the collision. (You may ignore the effects of friction during the collision.)
(a) The magnitude of the velocity of the 1000 kg car after the collision is 10.5 m/s.
(b) The direction of the velocity of the 1000 kg car after the collision is 8.2 ⁰ north west.
(c) The ratio of the kinetic energy of the two cars just after the collision to that just before the collision is 0.72.
Velocity of the 1000 kg after the collisionApply the principle of conservation of linear momentum as follows;
Final velocity in x directionm₁u₁ + m₂u₂ = m₁v₁x + m₂v₂x
where;
m₁ is mass of 750 kg caru₁ is initial velocity of 750 kg massm₂ is mass of 1000 kg caru₂ is initial velocity of 1000 kg massv₁ is final velocity of 750 kg massv₂ is final velocity of 1000 kg mass750(0) + 1000(13) = 750(4 cos 30) + 1000v₂x
13000 = 2,598.1 + 1000v₂x
10,401.9 = 1000v₂x
v₂x = 10.4 m/s
Final velocity in y directionm₁u₁ + m₂u₂ = m₁v₁y + m₂v₂y
750(0) + 1000(0) = 750(4 sin 30) + 1000v₂y
0 = 1500 + 1000v₂y
v₂y = -1500/1000
v₂y = -1.5 m/s
Resultant final velocityv = √(v₂ₓ² + v₂y²)
v = √[(10.4)² + (-1.5)²]
v = 10.5 m/s
Direction of the final velocity of 1000 kg cartanθ = v₂y/v₂ₓ
tanθ = -1.5/10.4
tanθ = -0.144
θ = arc tan(-0.144)
θ = 8.2 ⁰ north west
Kinetic energy of the cars before the collisionK.Ei = 0.5m₁u₁² + 0.5m₂u₂²
K.Ei = 0.5(750)(0)² + 0.5(1000)(13)²
K.Ei = 84,500 J
Kinetic energy of the cars after the collisionK.Ef = 0.5(750)(4)² + 0.5(1000)(10.5)²
K.Ef = 61,125 J
Ratio of the kinetic energyK.Ef/K.Ei = 61,125/84,500
K.Ef/K.Ei = 0.72
Thus, the magnitude of the velocity of the 1000 kg car after the collision is 10.5 m/s.
The direction of the velocity of the 1000 kg car after the collision is 8.2 ⁰ north west.
The ratio of the kinetic energy of the two cars just after the collision to that just before the collision is 0.72.
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10. Complete each of the following radioactive decay equations.
235U → Th+a
92
90
87
37Rb Sr +
6C
→
F
e
1¹B ++ie
1. Alpha
Number next to Th is 231
(because 231 + 4 = 235)
The other number below the 4 is 2 - helium
(because 90 + 2 = 92)
2. Beta
Missing numbers are zero and one. Same as the one below it.
3. Beta
11 is the number above the C
(because 11 + 0 = 11)
5 is the number next to the B
(because 5 + 1 = 6)
Basically, the missing numbers must balance on both side of the equation; creating the elements you've started of with, meaning if you where to add the right side of the equation up - you should end up with the same protons and mass numbers you began with.
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It has been suggested that rotating cylinders about 19.5 mi long and 5.66 mi in diameter be placed in space and used as colonies. What angular speed must such a cylinder have so that the centripetal acceleration at its surface equals the free-fall acceleration on Earth?
_____rad/s
The angular speed such a cylinder must have so that the centripetal acceleration at its surface equals the free-fall acceleration on Earth is 0.046 rad/s.
Angular speed of the cylinder
The angular speed such a cylinder must have so that the centripetal acceleration at its surface equals the free-fall acceleration on Earth is calculated as follows;
a = v²/r
v² = ar
(ωr)² = ar
ω²r² = ar
ω²r = a
ω² = a/r
ω = √(a/r)
where;
a is centripetal acceleration = acceleration due to gravity, gr is radius of the cylinder = 5.66 m / 2 = 2.83 mi = 4554.4 mω is angular speedω = √(g/r)
ω = √(9.8/4554.4)
ω = 0.046 rad/s
Thus, the angular speed such a cylinder must have so that the centripetal acceleration at its surface equals the free-fall acceleration on Earth is 0.046 rad/s.
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A container with 3 kg of pure water at a temperature of 12 degrees Celsius is placed in a refrigerator where the air temperature is kept at 4 degrees Celsius. how much heat (in joules) is gained by the air when the water cools to 4 degrees Celsius?
Answer:
100 368 J
Explanation:
Heat lost by the water = heat gained by the air ( which the 'frige works to expel and keep at 4 degrees)
specific heat for water ( in joules) : 4182 J / kg-C
3 kg * 4182 J / kg-C * (12-4 C) = 100 368 J
A tree limb is blown loose from a tree during a storm. As it falls, it gains
speed. Which type of energy is the tree limb gaining as it falls?
O A. Kinetic energy
B. Gravitational potential energy
O C. Nuclear energy
OD. Light energy
Answer:
B Gravitational potential energy
Explanation:
How much heat is required to raise the temperature of 0.210 g of water from 19.2 ∘C to 32.0 ∘C?
Data:
m = 0.210 g T₀ = 19.2 °C + 273 = 292. 2 KT = 32.0 °C + 273 = 305 KCe = 4.18 J / GKWe apply the following formula
Q = mcₑΔT
Q = mcₑ (T - T₀)
We substitute Our data in the formula and solve:
Q = 0.210 g * 4.18 J / g K (305 k - 292.2 k)
Q = 11.23 J
Which one of the following statements concerning weight and energy balance is most accurate?
A. People generally need the same amount of physical activity to maintain weight stability.
B. Regular physical activity doesn’t impact the percentage of body fat in children and adolescents.
C. It’s possible to achieve weight stability by doing the equivalent of 60–120 minutes a week of moderate-intensity walking.
D. The optimal amount of physical activity needed to maintain weight is unclear.
Answer: D. I took the test and got it right
The correct answer choice concerning weight and energy balance which is most accurate is the optimal amount of physical activity needed to maintain weight is unclear.
What is energy balance?Energy balance refers to the way in balance is achieved when intake of energy is equal to energy expended.
Energy refers to the impetus behind all motion and all activity. If is also the capacity to do work. Energy is measured in a unit dimensioned in mass × distance²/time² (ML²/T²) or the equivalent.
So therefore, the correct answer choice concerning weight and energy balance which is most accurate is the optimal amount of physical activity needed to maintain weight is unclear.
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A ball of mass 0.200 kg has a velocity of 150m/s; a ball of mass 0.300 kg has a velocity of - 0.4m/s. They meet in a head-on elastic collision. (a) Find their velocities after the collision. (b) Find the velocity of their center of mass before and after the collision.
(a) The velocity of 0.2 kg after the collision is 30.48 m/s and the velocity of the 0.3 kg mass is 119.92 m/s.
(b) The velocity of their center of mass before collision is 60.24 m/s and after the collision is 84.14 m/s.
Velocity of the balls after collision
Apply the principle of conservation of linear momentum to determine the velocity of the balls;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.2(150) + (0.3)(-0.4) = 0.2v₁ + 0.3v₂
29.88 = 0.2v₁ + 0.3v₂
Apply one directional linear velocity
u₁ + v₁ = u₂ + v₂
v₁ = u₂ + v₂ - u₁
v₁ = -0.4 + v₂ - 150
v₁ = v₂ - 150.4
Substitute the value of v₁ into the first equation;
29.88 = 0.2(v₂ - 150.4) + 0.3v₂
29.88 = 0.2v₂ - 30.08 + 0.3v₂
59.96 = 0.5v₂
v₂ = 59.96/0.5
v₂ = 119.92 m/s
v₁ = 119.92 - 150.4
v₁ = -30.48 m/s
Velocity of their center mass before collisionV(cm) = (0.2 x 150 + 0.3 x 0.4) / (0.2 + 0.3)
V(cm) = 60.24 m/s
Velocity of their center mass after collisionV(cm) = (0.2 x 30.48 + 0.3 x 119.92) / (0.2 + 0.3)
V(cm) = 84.14 m/s
Thus, the velocity of 0.2 kg after the collision is 30.48 m/s and the velocity of the 0.3 kg mass is 119.92 m/s.
The velocity of their center of mass before collision is 60.24 m/s and after the collision is 84.14 m/s.
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How do "inner planets" such as: Mercury, Venus, Earth and Mars compare with the "outer planets" such as: Jupiter, Saturn, Uranus and Neptune? Identify three characteristics that differ between these groups of planets.
Answer:
The four inner planets have shorter orbits, slower spin, no rings, and they are made of rock and metal. The four outer planets have longer orbits and spins, a composition of gases and liquids, numerous moons, and rings. The outer planets are made of hydrogen and helium, so they are called gas giants.
Compute your average velocity in the following two cases: (a) You
walk 73.2 m at a speed of 1.22 m/s and then run 13.2 m at a speed
of 3.02 m/s along a straight track. (b) You walk for 1.00 min at a
speed of 1.22 m/s and then run for 1.00 min at 3.05 m/s along a
straight track. (c) Graph x versus t for both cases and indicate how
the average velocity is found on the graph.
(a) Walking 73.2 m at 1.22 m/s would take
[tex]\dfrac{73.2\,\rm m}{1.22 \frac{\rm m}{\rm s}} = 60 \,\rm s[/tex]
and running 13.2 m at 3.02 m/s would take
[tex]\dfrac{13.2\,\rm m}{3.02\frac{\rm m}{\rm s}} \approx 4.37\,\rm s[/tex]
You've undergone a total displacement of 73.2 + 13.2 = 86.4 m in a matter of approximatly 64.37 s, so your average velocity is
[tex]\dfrac{\Delta x}{\Delta t} = \dfrac{86.4\,\mathrm m}{64.37\,\rm s} \approx \boxed{1.34\dfrac{\rm m}{\rm s}}[/tex]
(b) In the first 1.00 min = 60 s, you undergo a displacement of
[tex](60\,\mathrm s) \left(1.22 \dfrac{\rm m}{\rm s}\right) = 73.2 \,\rm m[/tex]
and in the second minute, you undergo a displacement of
[tex](60\,\mathrm s) \left(3.05\dfrac{\rm m}{\rm s}\right) = 183 \,\rm m[/tex]
Your total displacement is then 73.2 + 183 = 256.2 m in a matter of 2.00 min = 120 s, so your average velocity is
[tex]\dfrac{\Delta x}{\Delta t} = \dfrac{256.2\,\mathrm m}{120\,\rm s} \approx \boxed{2.14\dfrac{\rm m}{\rm s}}[/tex]
(c) For part (a), your displacement [tex]x(t)[/tex] (in m) at time [tex]t[/tex] (in s) is given by
[tex]x(t) = \begin{cases}1.22t & \text{for } 0 \le t \le 60 \\ 73.2 + 3.02 (t-60) & \text{for } t > 60\end{cases}[/tex]
and for part (b), your displacement is given by the very similar
[tex]x(t) = \begin{cases}1.22 t & \text{for } 0 \le t \le 60 \\ 73.2 + 3.05(t-60) & \text{for } t > 60 \end{cases}[/tex]
See the attached plots. The average velocity for the given situation is the slope of the dotted line.
A directional loudspeaker aims a sound wave of Frequency 200 H₂ at wall. At what distance from the wall would you stand and hear no sound at all? (v= 332 m/s).
The distance from the wall you would stand and hear no sound at all is 0.83 m.
What is speed of wave?The speed of a wave is the rate of change of distance traveled by a wave with time.
Distance of the waveThe distance at which the wave will have zero amplitude, there will be no sound at all since amplitude of a sound is proportional to intensity of the sound.
The point of zero amplitude, L = λ/₂
Where;
λ is wavelength of the wave
The wavelength of the wave is calculated as follows;
λ = V/f
where;
V is speed of sound wavef is frequency of the waveλ = 332/200
λ = 1.66 m
Distance from the wall a zero soundL = 1.66/2
L = 0.83 m
Thus, the distance from the wall you would stand and hear no sound at all is 0.83 m.
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If something is a good conductor, what type of insulator is it?
1. also a good insulator
2. a metal
3. it depends
4. a poor insulator
4. a poor insulator
If rest other things are kept constant or unchanged then a good conductor can be termed as a poor insulator.
Answer:
4 conductors and insulators are opposites of each other.
Explanation:
Consider the f(x) = Acos(x) function shown in the figure in blue color. What is the value of amplitude A for this function?
As a reference the g(x) = cos(x) function is shown in red color, and green tick marks are drawn at integer multiples of π.
The amplitude of the red colored wave is 1 unit and the amplitude of the red colored wave is 2.1 unit.
What is amplitude of a wave?
The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.
Amplitude of the red colored waveFrom the graph, the amplitude of the red colored wave is 1 unit.
Amplitude of the blue colored waveFrom the graph, the amplitude of the red colored wave is 2.1 unit.
Thus, the amplitude of the red colored wave is 1 unit and the amplitude of the red colored wave is 2.1 unit.
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