Answer:
the change in length of the copper bar is 1.26 x 10^-3 meters (or 1.26 millimeters).
Explanation:
The change in length of a copper bar can be calculated using the formula:
ΔL = L₀αΔT
where:
ΔL = change in length
L₀ = original length of the copper bar (1.5 m)
α = coefficient of linear expansion for copper (16.8 x 10^-6 K^-1)
ΔT = change in temperature (353 K - 303 K = 50 K)
Plugging in the values, we get:
ΔL = (1.5 m)(16.8 x 10^-6 K^-1)(50 K)
ΔL = 1.26 x 10^-3 m
Speed velocity and acceleration puzzle level 2
When describing motion, speed indicates the pace at which an object is travelling. It has one scalar component identifying its magnitude, irrespective of direction.
How to explain the informationThe unit for measuring speed can be either meters per second (m/s) or miles per hour (mph). For velocity, it's a different story. Its definition encompasses both speed and direction since it's a vector quantity. Measureable just like speed using m/s or mph.
In plain physic terms, acceleration reveals how much speed changes over time; hence it is also a vector quantity with not only size but also direction. Depending on whether a physical item is increasing in momentum, slowing down or static, the value could be positive, negative, or zero. Acceleration is calculated using units metres per second squared (m/s2).
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A block of mass m is released from the top of a spring and goes through simple harmonic motion. Use equations to show your work (no numerical values).
a. What is the compression of the spring at equilibrium?
b. What is the maximum compression in the spring?
c. Find the maximum acceleration of the block.
drawing shows a force vector that has a magnitude of 475 newtons.
Find the
(a) X,
(b) y, and
(c) z components of the vector.
X, Y, and Z components of the vector are 398, 384 and 279 resp.
Vector is a physical quantity which has both magnitude and direction. Vector A can be written as A = a₁i + a₂j + a₃k where a₁, a₂, a₃ are components along X, Y, Z axis resp. and i,j,k, are the unit vectors along X,Y,Z axis resp.
In this figure
vector F is at angle 36° from y axis, hence
x = Fcos33 = 475cos33 = 398 N
y = Fcos36 = 475cos36 = 384 N
z = Fsin36 = 475sin36 = 279 N
The vector can be written as
F = 398i + 384j + 279k
Hence x, y and z components of this force is 398, 384 and 279 resp.
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Which of the following would you expect to be a strong electrolyte in solution?
The following would expect to be a strong electrolyte in solution (b) KCI is correct option.
When dissolved in water, a strong electrolyte produces a large concentration of ions in solution by totally dissociating into ions. The following compounds are typically strong electrolytes in solution according to this definition:
Al(OH)₃ (aluminum hydroxide) is a weak electrolyte. It does not dissociate significantly into ions in solution, resulting in a low electrical conductivity.KCl (potassium chloride) is a strong electrolyte. It completely dissociates into potassium ions (K⁺) and chloride ions (Cl⁻) in solution, resulting in a high concentration of ions and a high electrical conductivity.PbI₂ (lead(II) iodide) is a weak electrolyte. It does not dissociate significantly into ions in solution, resulting in a low electrical conductivity.These substances readily dissociate into ions in water and exhibit high electrical conductivity, making them strong electrolytes in solution.
Therefore, the correct option is (b).
The complete question is,
Which of the following would be a strong electrolyte in solution?
a) Al(OH)₃ b) KCI c) Pbl₂
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A portable power source is available for travelers who need electricity for
appliances. The power scurce provides 54 W of power to operate an air
compressor for inflating tires. This compressor draws 4.5 A of current when
connected to the power supply. What is the voltage across the compressor?
The voltage across the compressor is 12 V.
Power of the power source, P = 54 W
Current utilized, I = 4.5 A
The equation for power of a circuit is given by,
P = VI
Therefore, voltage across the compressor,
V = P/I
V = 54/4.5
V = 12 V
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The magnitude of a uniform electric field between two plates is about 1.7 ✕ 106 N/C. If the distance between these plates is 3.7 cm, find the potential difference between the plates.
The magnitude of a uniform electric field between two plates of capacitor is about 1.7 ✕ 106 N/C. If the distance between these plates is 3.7 cm then the potential difference between the plates is 62.5 kV.
A capacitor is a device that stores electrical energy in an electric field by collecting electric charges on two isolated surfaces. It is a two-terminal passive electrical component.
Electric field of the parallel plate capacitor is given as,
E = V/d
Given,
E = 1.7 ✕ 10⁶ N/C.
d = 3.7 cm,
V= Ed
V = 1.7 ✕ 10⁶ N/C × 3.7 × 10⁻² m
V = 62.5 kV.
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2. Single Choice
In a transverse wave, the individual particle of the medium ( )
A: moves in a circle.
B: moves in ellipses.
C: move parallel to the direction of the wave's travel.
D: move perpendicular to the direction in which the waves travel.
A simple circuit contains a battery connected with wires to a small bulb that has a resistance of 150 ohms. If the power dissipated by the bulb is 0.4 W, what is the voltage of the battery?
Remember to identify all data (givens and unknowns), list equations used, show all your work, and include units and the proper number of significant digits to receive full credit.
Answer: The answer is 7.75v
Explanation; As we know,
power dissipated= (voltage)^2/resistance
0.4w = v^2/150
v^2=0.4w*150ohm
v^2=60
v=7.75v
which substance listed in the chart is made up of the most atoms
All substance listed in the chart is made up of the atoms.
Atom is smallest entity of a substance. Body is made up of atoms. it is basic building block of a body. An atom consist of electrons, protons and neutrons as sub atomic particle. whole mass of the atom is concentrated at the center of the atom which we call it as nucleus, nucleus consist of proton and neutron. Electron revolve around the nucleus at determined(fixed) orbit. Total number of protons in the atom decides the atomic number and the elements in the periodic table. The electrons which are completely filled orbitals are called as core shell electrons and which are not filled completely are called as valence electron. valence electrons are responsible for physical and chemical properties of the element. Elements which are on same column in periodic table have same number of valence electrons . Hence they have same properties.
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What areas of daily life are the effects of the laws of physics seen?
Answer: Our day-to-day life highly relates to physics.
Explanation: We know that in physics there are many laws such as gravitational laws, laws of friction, and inertia.For example
When we drive a car, and suddenly apply the bake the drive moves forward. This is actually the LAW OF INERTIA.If we placed a ball on the surface it does not change its position until a force is applied. When we placed an object on the surface of the earth, it does not change its position and size until an external force is applied. This is an example of NEWTON'S FIRST LAW.Writing with a ballpoint pen is another example of a LAW OF GRAVITY. When we write with a ballpoint pen the ball spins and because of the gravity the ink travel to the paper.
An electron that has a velocity with x component 2.0 × 106 m/s and y component 3.0 x 106 m/s moves through a uniform magnetic field with x component 0.024 T and y component -0.12 T. (a) Find the magnitude of the magnetic force on the electron. (b) Repeat your calculation for a proton having the same velocity.
WILL GIVE BRAINLIEST HELPPP FASTER PLEASE l ONLY RIGHT ANSWER!!!!! l
How can spectroscopy and infrared technology be useful in space?
They can enhance speed by making spacecraft fuel more efficient.
They can measure magnetic fields produced by astronomical bodies.
They can provide an emergency escape to the astronaut from a space center.
They can determine the elements that make up the surface of astronomical bodies.
Answer: D. They can determine the elements that make up the surface of astronomical bodies.
Explanation: Spectroscopy and infrared technology are useful in space because they allow scientists to determine the elements that make up the surface of astronomical bodies, such as planets, moons, and asteroids.
Spectroscopy involves the analysis of light or radiation emitted or absorbed by these bodies. When light interacts with matter, it gets absorbed or emitted in specific wavelengths that correspond to the energy levels of atoms and molecules. By studying the pattern of these wavelengths, scientists can identify the unique "fingerprint" or spectral lines of elements and compounds.
Infrared technology, on the other hand, detects and measures the infrared radiation emitted by objects. This radiation is produced due to the thermal energy or heat emitted by celestial bodies. By analyzing the specific wavelengths of infrared radiation, scientists can gain insights into the composition and temperature of these bodies.
By combining spectroscopy and infrared technology, scientists can gather valuable data about the chemical composition of astronomical bodies. This information helps in understanding the geological processes, formation, and evolution of these bodies. It also provides insights into the presence of specific elements or compounds that may be important for studying habitability, potential resources, or even the origins of life in the universe.
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A neutron and a proton combine to form a nucleus. How does the sum of the masses of the nucleons that make up the nucleus compare with the mass of the nucleus itself?
The nucleons have less mass, because matter is converted into binding energy. Option D is correct.
During the process of combining a neutron and a proton to form a nucleus, a small amount of mass is converted into binding energy. This is due to the strong nuclear force that holds the nucleus together. The mass of the nucleus is slightly less than the sum of the masses of the individual nucleons, and the difference in mass is referred to as the mass defect.
This mass defect is related to the binding energy of the nucleus through Einstein's famous equation E=mc², where E is energy, m is mass, and c is the speed of light. The mass defect represents the amount of mass that is converted into binding energy to hold the nucleus together. Option D is correct.
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As explained by the second law of thermodynamics, which example of energy transformation could never occur?
A.**The 400 J of heat added to the operating gas of a heat engine is transformed into 400 J of work.
B.The 400 J of kinetic energy of a rolling ball is transformed into 400 J of heat.
C.A refrigerator removes 100 cal of heat from a bottle of milk while using 75 cal of electrical energy.
D.A heat engine does 25 J of work while expelling 10 J of heat to the cold reservoir.
The example of energy transformation that could never occur, as explained by the second law of thermodynamics, is the 400 J of heat added to the operating gas of a heat engine is transformed into 400 J of work. Option A is correct.
The second law of thermodynamics states that in any energy transformation, the total entropy (measure of disorder) of a closed system will always increase or remain constant. It means that some energy will always be wasted as heat and cannot be completely converted into useful work. The second law of thermodynamics is a fundamental law of nature that governs energy transformations.
It states that in any energy transformation process, the total entropy (measure of disorder) of a closed system will always increase or remain constant. Entropy can be thought of as a measure of the amount of energy that is unavailable to do useful work. It is impossible to transform all 400 J of heat into work without generating any waste heat. This is because the heat engine must expel some heat to the cold reservoir to comply with the second law. Option A is correct.
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A cube of ice is taken from the freezer at -8.7°C and placed in a 104 g glass cup filled with 330 g of water. Both the water & the cup are at 22.5°C. Eventually the system reaches thermal equilibrium at 15.4°C. Determine Qcup,
Qwater (for the water initially in the cup), Qice, & the mass of the ice.
Quip=
Qwater =
Qice =
Mice =
The value of the heat changes is as follows:
Qcup = -271 JQwater = -12284 JQice = 3951 JMice = 38.95 gWhat is the heat quantity of the cup, Qcup?The heat quantity of the cup, Qcup is calculated using the formula below:
Q = mcΔT
where;
Q is the heat absorbed or released,m is the mass of the substance,c is its specific heat capacity,ΔT is the change in temperature.The heat absorbed by the glass cup, Qcup will be:
Qcup = 104 * 0.385 * (15.4- 22.0)
Qcup = -271 J
The heat absorbed by the water, Qwater will be:
Qwater = 330 * 4.184 * *(15.4 - 22.5)
Qwater = -12284 J
The mass of the ice, Mice, that melted will be:
Heat absorbed by ice = Heat released by water + heat released by cup
-MiceΔHf = Qwater + Qcup
where
ΔHf is the heat of fusion of ice = 333.55 J/gMice = -(Qwater + Qcup) / ΔHf
Mice = -(-1228 - 271) / 333.55
Mice = 38.95 g
Finally, the heat absorbed by the ice will be:
Qice = mcΔT
Qice = (38.95 g) (2.108 J/g°C) (15.4°C - (-8.7°C))
Qice = 3951 J
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calculate the distance an object moves if 25J of work is done with 3.0N of force
The distance an object moves if 25J of work is done with 3.0N of force is 8.33 m.
For a given amount of force, F, and a given distance, d, the formula for calculating work done is as follows:
Work done = Force x distance
So, the distance would be,
Work done / force = 25/3 = 8.33 m.
Work is the energy exerted by an object when it applies a force to move another object over some distance.
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Construct a parallel-plate capacitor where a second line of charges equal in size and opposite in charge are placed below the line of positive charges. Examine what the "E-field" is like between the plates using a sensor.
A capacitor with two lines of charge on its parallel plates. The bottom plate has an equal line of negative charges that are the opposite in charge to the positive charges on the top plate, while the top plate has a line of positive charges.
As a result, an electric field (E-field) is produced between the plates that can be measured with a sensor.
In a parallel-plate capacitor, the E-field between the plates is uniform and pointed perpendicularly to the plates. It is represented by the equation E = σ/ε, where ε is the permittivity of the medium between the plates and σ is the charge density (charge per unit area) on the plates.
In this instance, the charge density on the top and bottom plates is the same but with opposing signs since the lines of charges on the plates are equal in size and opposite in charge. Assume that the top plate has positive charges and the bottom plate has negative charges, and that the charge density on both plates equals.
A sensor placed between the capacitor's plates will now allow us to measure the E-field, which will reveal that it is constant and perpendicular to the plates. E = σ/ε, where σ, is the charge density and is the permittivity of the medium between the plates, will be the magnitude of the E-field.
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2) Let the angle 8 be the angle that the vector à makes with the I, the x-direction. Find the angle
for the vectors with the following components:
a) Ax=2.00 m and Ay--1.00 m
b) Ax=2.00 m and Ay= 1.00 m
c) Ax= -2.00 m and Ay 1.00
d) Ax= -2.00 m and Ay-1.00 m
(a) Ax=2.00 m and Ay--1.00 m, the angle of the vector is 153.4⁰.
(b) Ax=2.00 m and Ay= 1.00 m, the angle of the vector is 26.6⁰.
(c) Ax= -2.00 m and Ay 1.00 m, the angle of the vector is 333.43⁰.
(d) Ax= -2.00 m and Ay = -1.00 m, the angle of the vector is 206.6⁰.
What is the angle of the vectors?The angle of the vectors is known as the direction of the vectors and it is calculated as follows
(a) Ax=2.00 m and Ay--1.00 m, the angle of the vector;
tan θ = Ay/Ax
tan θ = (-1/2)
θ = arc tan (-1/2)
θ = -26.6⁰ = (180 - 26.6⁰) = 153.4⁰ (2nd quadrant).
(b) Ax=2.00 m and Ay= 1.00 m, the angle of the vector;
tan θ = Ay/Ax
tan θ = (1/2)
θ = arc tan (1/2)
θ = 26.6⁰ (1st quadrant).
(c) Ax= -2.00 m and Ay 1.00 m, the angle of the vector;
tan θ = Ay/Ax
tan θ = (-1/2)
θ = arc tan (-1/2)
θ = -26.6⁰ = (360 - 26.6⁰) = 333.43⁰ (4th quadrant).
(d) Ax= -2.00 m and Ay = -1.00 m, the angle of the vector;
tan θ = Ay/Ax
tan θ = (1/2)
θ = arc tan (1/2)
θ = 26.6⁰ = (180 + 26.6⁰) = 206.6⁰ (3rd quadrant).
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Which is an accurate definition of an element?
Question 2 options:
Something that can burn
Something that cannot be broken down into simpler substances through chemical reactions
Something that is essential to living things
Something that combines with hydrogen
Answer:
Explanation:
Something that cannot be broken down into simpler substances through chemical reactions
The figure shows wire 1 in cross section; the wire is long and straight, carries a current of 4.20 mA out of the page, and is at distance d₁ = 2.58 cm from a surface. Wire 2. which is parallel to wire 1 and also long, is at horizontal distance d-5.05 cm from wire 1 and carries a current of 6.88 mA into the page. What is the x component of the magnetic force per unit length on wire 2 due to wire 1?
Wire 1 in cross section; the wire is long and straight, carries a current of 4.20 mA out of the page, and is at distance d₁ = 2.58 cm from a surface. Wire 2. which is parallel to wire 1 and also long, is at horizontal distance d-5.05 cm from wire 1 and carries a current of 6.88 mA.
To find the x component of the magnetic force per unit length on wire 2 due to wire 1, we can use the formula for the magnetic force between two parallel current-carrying wires we get
F = μ₀I₁I₂/(2πd)
Where F is the magnetic force per unit length, μ₀ is the magnetic constant (4π x [tex]10^{-7}[/tex]Tm/A), I₁ and I₂ are the currents in the wires, and d is the distance between the wires.
In this problem, we need to find the x component of the magnetic force per unit length on wire 2 due to wire 1. We can break down the problem into components by considering the direction of the magnetic field due to wire 1 at the position of wire 2. The magnetic field due to wire 1 will be perpendicular to both wire 1 and wire 2, and will be directed into the page.
To find the x component of the magnetic force, we need to consider the component of the magnetic force that is perpendicular to wire 2. This component will be directed along the x axis, and will have a magnitude of
[tex]F_{x}[/tex] = Fsinθ
Where θ is the angle between the direction of the magnetic force and the x axis. Since the magnetic force is directed into the page, θ is 90 degrees, and sinθ = 1.
Substituting the values given in the problem, we get
F = (4π x [tex]10^{-7}[/tex]Tm/A)(4.20 x[tex]10^{-3}[/tex] A)(6.88 x [tex]10^{-3}[/tex]A)/(2π*0.0258 m)
F = 3.99 x [tex]10^{-10}[/tex] N/m
Therefore, the x component of the magnetic force per unit length on wire 2 due to wire 1 is
[tex]F_{x}[/tex] = Fsinθ= (3.99 x [tex]10^{-10}[/tex] N/m)(1) = 3.99 x [tex]10^{-10}[/tex] N/m
Hence, the x component of the magnetic force per unit length on wire 2 due to wire 1 is 3.99 x [tex]10^{-10}[/tex] N/m.
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In the figure, point P2 is at perpendicular distance R= 20.6 cm from one end of straight wire of length L = 12.2 cm carrying current i = 0.780 A. (Note that the wire is not long.) What is the magnitude of the magnetic field at P₂?
The magnitude of the magnetic field at P₂ is 6.06 x 10⁻⁵ T.
Using the Biot-Savart Law, we can determine the magnetic field at point P2 due to the current-carrying wire. The magnitude of the magnetic field B at P2 is given by:
B = μ₀i/4π (sinθ₁ - sinθ₂)
where μ₀ is the magnetic constant, i is the current, θ₁ is the angle between the wire and the line joining the wire and point P₂, and θ₂ is the angle between the wire and the line perpendicular to the wire and passing through point P₂.
From the given diagram, we can see that sinθ₁ = L/2R and sinθ₂ = (R - L/2)/R. Substituting these values and the given values of i, L, and R into the equation, we get:
B = (4π x 10⁻⁷ Tm/A) x 0.780 A / 4π (L/2R - (R - L/2)/R)
= 6.06 x 10⁻⁵ T
As a result, the magnetic field magnitude at P₂ is 6.06 x 10⁻⁵ T.
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A 20 kg child is on a swing that hangs from 2.6-m-long chains. What is her maximum speed if she swings out to a 50 degree angle?
A toy racing car moves with constant speed around the circle shown below. When it is at point A its coordinates are x = 0, y = 3 m and its velocity is (6 m/s)ˆi. When it is at point B its velocity and acceleration are
The speed of the car is 6 m/s. The acceleration vector at point B has a direction of (-1, -1) and a magnitude of approximately 16.97 m/s².
We can start by finding the speed of the toy car. Since it is moving with constant speed around the circle, its speed is the same at points A and B. To find the speed, we can use the fact that the velocity vector has a magnitude equal to the speed:
|v| = √((6 m/s)²) = 6 m/s
So the speed of the car is 6 m/s.
Next, we can find the direction of the velocity vector at point B. We know that the car is moving around a circle centered at the origin, and that point B is on the circle. Therefore, the velocity vector at point B is tangent to the circle and perpendicular to the line connecting the origin to point B.
The line connecting the origin to point B is given by:
y = (0 - 3)/(0 - (-3)) * (x - (-3)) + 0
y = -x + 3
The velocity vector at point B is therefore perpendicular to this line, which means it has a direction given by the vector (1, -1).
Finally, we can find the acceleration vector at point B. Since the car is moving with constant speed around a circle, it is undergoing uniform circular motion, which means it is accelerating towards the center of the circle. The magnitude of the acceleration is given by:
a = v² / r
where v is the speed and r is the radius of the circle. We don't know the radius of the circle, but we can find it using the fact that point B lies on the circle. The distance from the origin to point B is:
d = √((-3 - 0)² + (0 - 3)²) = 3√(2) m
Therefore, the radius of the circle is:
r = d / 2 = (3√(2)) / 2 m
Substituting in the values for v and r, we get:
a = (6 m/s)² / ((3√(2)) / 2 m) ≈ 16.97 m/s²
To find the direction of the acceleration vector, we can use the fact that it is pointing towards the center of the circle. The center of the circle is at the origin, so the acceleration vector at point B is given by the vector (-3, 0) minus the vector (0, 3), which is:
(-3, 0) - (0, 3) = (-3, -3)
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A pool noodle has a density of 145 kg/m3, a length of 1.65 m and a radius of 2.5 cm. How many pool noodles would be needed to make a raft that would support the weight of a person with a mass of 65.0kg?
The number of pool noodles that would be needed to support the weight is 20.
What is the volume of single pool noodle?
The volume of a single pool noodle is calculated as follows;
V = πr²h
V = π (0.025)² x 1.65
V = 0.00324 m³
The weight of the water displaced is calculated as follows;
W = ρVg
where;
ρ is the density of waterV is the volumeg is gravityW = 1000 x 0.00324 x 9.8
W = 31.75 N
The number of pool noodles needed to support a person is calculated as follows;
= (65 x 9.8 ) / (31.75)
= 20
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Which two factors affect the amount of thermal energy an object has?
A. The amount of motion its particles have
B. The mass of the object
c. The directions in which its particles are moving
D. The amount of space between its particles
The two factors that affect the amount of thermal energy an object has are;
A. The amount of motion its particles haveD. The amount of space between its particlesWhat is thermal energy?Thermal energy can be regarded as the energy which is been contained within a system it can be considered as the one that is responsible for its temperature.
It should be noted that the Heat is the flow of thermal energy and it can be seen as opne that deals with how heat is transferred between different systems .
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calculate the potential energy of a 11kg object that is 11m off the ground
The potential energy of an 11kg object that is 11m off the ground is 1199.8J.
The potential energy of the body is defined as the energy which is possessed by the body when the position of the body is at rest. When the object is at rest the potential energy is possessed by the body and then when the body starts moving the potential energy is converted into kinetic energy.
The potential energy of a body is given by the expression:
P.E=mgh;
where 'm' is the mass of the body which is given as 11kg given in the question, and 'g' is the acceleration due to gravity which is equal to (9.8 m/s^2).
So the potential energy here is:
P.E = (11kg)*(9.8 m/s^2)*(11m)
P.E = 1199.8 Joules.
Therefore, the potential energy of an 11kg object that is 11m off the ground is 1199.8J.
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What is the process where light bounces back from an object at the same angle and intensity as it. is received by the object?
Answer: It is Reflection
Explanation: Reflection occurs when incoming solar radiation bounces back from an object or surface that it strikes in the atmosphere, on land, or water, and is not transformed into heat.
Which item is a piece of golf equipment that supports the ball so the player can strike it?
O fairway
O divot
O par
O tee
Water runs into a fountain, filling all the pipes, at a steady rate of 0.757 m3/s. (A) How fast will it shoot out of a hole 4.51cm in diameter? (B) At what speed will it shoot out if the diameter of the hole is three times as large?
(A)The water will shoot out of the hole at a speed of 4.77 m/s, and the pressure of the water at the hole will be 9.91 × 10^4 Pa, and (B) The water will shoot out of the larger hole at a speed of 0.529 m/s, and the pressure of the water at the hole will be 1.012 × 10^5 Pa.
We can use Bernoulli's equation to solve this problem, which relates the pressure, velocity, and height of a fluid. The equation states that:
P + (1/2)ρv^2 + ρgh = constant
where P is the pressure, ρ is the density of the fluid, v is the velocity of the fluid, g is the acceleration due to gravity, and h is the height of the fluid.
(A) The diameter of the hole is 4.51 cm, which corresponds to a radius of 2.255 cm = 0.02255 m. The area of the hole is A = πr^2 = 1.587 × 10^-4 m^2. The volume flow rate of water is Q = 0.757 m^3/s.
We can calculate the velocity of the water as it exits the hole using the equation:
Q = Av
where A is the area of the hole and v is the velocity of the water. Solving for v, we get:
v = Q/A = 4.77 m/s
Now, we can use Bernoulli's equation to find the pressure of the water at the hole. Assuming that the height of the fountain is negligible compared to the height of the atmosphere, we can set the height term to zero. Also, we can assume that the pressure at the surface of the fountain is atmospheric pressure, which we can take as P = 1.013 × 10^5 Pa. Then, the equation becomes:
P + (1/2)ρv^2 = constant
Solving for P, we get:
P = constant - (1/2)ρv^2
At the hole, the velocity of the water is v = 4.77 m/s, and the density of water is ρ = 1000 kg/m^3. Substituting these values, we get:
P = 1.013 × 10^5 Pa - (1/2) × 1000 kg/m^3 × (4.77 m/s)^2 = 9.91 × 10^4 Pa
So, the water will shoot out of the hole at a speed of 4.77 m/s, and the pressure of the water at the hole will be 9.91 × 10^4 Pa.
(B) If the diameter of the hole is three times as large, then the area of the hole will be nine times as large. Therefore, the volume flow rate of water will be distributed over a larger area, resulting in a lower velocity. The new area of the hole is A = 9 × 1.587 × 10^-4 m^2 = 1.43 × 10^-3 m^2. The volume flow rate of water is still Q = 0.757 m^3/s.
Using the equation Q = Av, we can find the new velocity of the water:
v = Q/A = 0.529 m/s
Using Bernoulli's equation, we can find the pressure of the water at the larger hole:
P = 1.013 × 10^5 Pa - (1/2) × 1000 kg/m^3 × (0.529 m/s)^2 = 1.012 × 10^5 Pa
So, the water will shoot out of the larger hole at a speed of 0.529 m/s, and the pressure of the water at the hole will be 1.012 × 10^5 Pa.
Hence, Water will flow out of the smaller hole at a speed of 0.529 m/s and a pressure of 1.012 × 10^5 Pa, and the water will shoot out of the hole at a speed of 4.77 m/s and a pressure of 9.91 × 10^4 Pa.
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How would the pollution from 2 coal plants compare if the first plant were twice as energy efficient as the second one?
The pollution from the first plant would be half that of the second plant for the same amount of energy produced.
The energy efficiency of a coal plant refers to the amount of energy produced per unit of fuel consumed. If the first plant is twice as energy efficient as the second plant, it means that it can produce the same amount of energy using half the amount of fuel.
Since pollution from coal plants is directly proportional to the amount of fuel consumed, the first plant would produce half the pollution of the second plant for the same amount of energy produced. This assumes that the two plants have the same level of emissions per unit of fuel consumed, which may not necessarily be the case.
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