Answer:
Energy is measured in JOULES.
Atoms and molecules are the fundamental building blocks of matter.
Explanation:
Matter is anything that has weight and occupies space. Locked within any given molecule or atom is some form of energy waiting to be activated. Energy can neither be created nor destroyed.
A car starts from rest and moves around a circular track of radius 27.0 m. Its speed increases at the constant rate of 0.420 m/s2. (a) What is the magnitude of its net linear acceleration 13.0 s later
Answer:
1.18 m/s²
Explanation:
The linear speed = 13.0s
[tex]v= v_i + at\\v = 0 + 0.42(13)\\v = 5.46m/s[/tex]
where v(i) is the initial linear speed
the radial acceleration of the car afte (t) = 13s is
[tex]a_r = \frac{v^2}{r} \\= \frac{5.46^{2} }{27} \\= 1.104m/s^2[/tex]
the magnitude of the net acceleration is
[tex]a_net = \sqrt{1.104^2 + 0.1764^2} \\= 1.18m/s^2[/tex]
Jennifer was investigating granite, a type of rock that forms when magma deep below the Earth's crust cools off and hardens. What type of rock is granite?
Answer:
Granite is Intrusive Igneous.
Explanation:
The key words here are "Deep below the Earth's crust", so that gives you the fact it is intrusive. Anything with magma is also most likely hinting towards Intrusive igneous.
n her abstract works Electric Dress and Untitled, Japanese artist Atsuko Tanaka created ________ through the repetition of lines and circular shapes in bold colors. a. scale b. unity c. balance d. emphasis e. focal point
Answer:
b. unity
Explanation:
In the images I added you can observe Atsuko Tanak's works, in both of them you can observe how the sum of colorful and repetitive elements creates the sensation of unity.
I hope you find this information useful and interesting! Good luck!
A straight 2.70 m wire carries a typical household current of 1.50 a (in one direction) at a location where the earth's magnetic field is 0.550 gauss from south to north.
Answer:
a) F = 2.2275 * 10^-4 N , upwards
b) F = 2.2275 * 10^-4 N , east to west
c) F = 0
Explanation:
Given:
- The straight wire of length L = 2.7 m
- The current I in the wire I = 1.50 A
- The magnetic Field B = 0.55 * 10^-4 T
Find:
Find the magnitude and direction of the force that our planet’s magnetic field exerts on this wire if it is oriented so that the current in it is running
a) from west to east
b) vertically upward
c) from north to south
d) Is the magnetic force ever large enough to cause significant effects under normal household conditions?
Solution:
- If current runs from west to east the angle between the magnetic field B is θ = 90°, so the magnitude of force due to magnetic field is given by Lorentz force.
F = B*I*L*sin(θ)
F = 0.55*1.5*2.7*sin(90)*10^-4
F = 2.2275 * 10^-4 N
- From Figure B points north and current I points east. From right hand rule, the direction of force is out of page, so its upward.
- If current runs upward the angle between the magnetic field B is θ = 90°, so the magnitude of force due to magnetic field is given by Lorentz force.
F = B*I*L*sin(θ)
F = 0.55*1.5*2.7*sin(90)*10^-4
F = 2.2275 * 10^-4 N
- From Figure B points north and current I points out of page . From right hand rule, the direction of force is out of page, so its east to west.
- If current runs north to south the angle between the magnetic field B is θ = 0°, so the magnitude of force due to magnetic field is given by Lorentz force.
F = B*I*L*sin(θ)
F = 0.55*1.5*2.7*sin(0)*10^-4
F = 0 N
The magnitude of the force that our planet's magnetic field exerts is mathematically given as
F=1.82* 10^{-4}N
What is the magnitude of the force that our planet's magnetic field exerts on this cord?
Question Parameter(s):
A straight 2.70 m wire carries a typical household current of 1.50
the earth's magnetic field is 0.550 gauss
Generally, the equation for the force exerted by the magnetic field is mathematically given as
F=ILB
Where
B=0.550 G
B= 0.55 *10^{-4}T
In conclusion,force exerted by the magnetic field is
F=(1.50 A)(2.20 m)(0.55* 10^{-4} T)
F=1.82* 10^{-4}N
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Complete Question
A straight 2.70 m wire carries a typical household current of 1.50 an (in one direction) at a location where the earth's magnetic field is 0.550 gauss from south to north.
Find the magnitude of the force that our planet's magnetic field exerts on this cord if is oriented so that the current in it is running from west to east
What is the final concentration of DD at equilibrium if the initial concentrations are [A][A]A_i = 1.00 MM and [B][B]B_i = 2.00 MM ? Express your answer to two significant figures and include the appropriate units.
Answer:
A) Concentration of A left at equilibrium of we started the reaction with [A] = 2.00 M and [B] = 2.00 M is 0.55 M.
B) Final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M is 0.90 M.
[D] = 0.90 M
Explanation:
With the first assumption that the volume of reacting mixture doesn't change throughout the reaction.
This allows us to use concentration in mol/L interchangeably with number of moles in stoichiometric calculations.
- The first attached image contains the correct question.
- The solution to part A is presented in the second attached image.
- The solution to part B is presented in the third attached image.
The equilibrium concentration can be calculated using initial concentrations, changes in concentrations, and the equilibrium constant. However, these calculations require explicit information that isn't provided in the question such as the reaction equation or equilibrium constant.
Explanation:The question refers to the calculation of the final concentration of a species (DD) at equilibrium given their initial concentrations in terms of Molarity (M). However, without further context or information on the reaction equation or equilibrium constant, a direct answer cannot be provided. Here's a general approach to how equilibrium concentrations of reaction species are calculated:
Let's say we have a reaction represented by A + B --> DD. We assume the concentration of DD increases by 'x' at equilibrium, and A and B decrease by 'x'. Thus, at equilibrium, the concentrations would be:
[A] = 1.00 M - x[B] = 2.00 M - x[DD] = xAfter this, an equilibrium expression is commonly used, which would depend on the equilibrium constant for the reaction. Solving the equation would give the value of 'x' which represents the concentration of DD.
Again, without specific information about the reaction, we can't determine 'x'. A more specific question or additional information would make this calculation possible.
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A 20.0 kg rock is sliding on a rough , horizontal surface at8.00 m/s and eventually stops due to friction .The coefficient ofkinetic fricction between the rock and the surface is 0.20
what average thermal power is produced as the rock stops?
Answer: The power is 156 watt
Explanation:
is in the attachment
Explanation:
Below is an attachment containing the solution.
Calculate the wavelengths in air at 20?C for sounds in the maximum range of human hearing. The speed of sound in air is 343 m/s.Part ACalculate the wavelength of a 20-Hz wave.Express your answer to two significant figures and include the appropriate units.Part BCalculate the wavelength of a 20,000-Hz wave.Express your answer to two significant figures and include the appropriate units.Part CWhat is the wavelength of a 20-MHz ultrasonic wave?Express your answer to two significant figures and include the appropriate units.
Answer:
Part A: 17 m
Part B: 1.7 x 10⁻²m
Part C: 1.7 x 10⁻⁵m
Explanation:
Equation needed to answer the question:
λ = [tex]\frac{v}{f}[/tex]
where λ is the wavelength, f is the frequency and v is the velocity,
Part A:
Substitute velocity with the speed of sound: 343 m/s
Substitute frequency with the value given: 20 Hz
λ = [tex]\frac{343m/s}{20Hz}[/tex] = 17 m
Part B:
Substitute velocity with the speed of sound: 343 m/s
Substitute frequency with the value given: 20,000 Hz
λ = [tex]\frac{343m/s}{20,000Hz}[/tex] = 1.7 x 10⁻²m
Part C:
Substitute velocity with the speed of sound: 343 m/s
Substitute frequency with the value given: 20,000 Hz
λ = [tex]\frac{343m/s}{20MHz (\frac{10^{6}Hz}{1MHz})}[/tex] = 1.7 x 10⁻⁵m
To calculate the wavelength of sound waves at different frequencies, we can use the formula v = fλ, where v is the speed of sound and f is the frequency. Part A: The wavelength of a 20-Hz wave is 17.15 m. Part B: The wavelength of a 20,000-Hz wave is 0.01715 m. Part C: The wavelength of a 20-MHz wave is 0.00001715 m.
Explanation:To calculate the wavelength of a sound wave, we can use the formula v = fλ, where v is the speed of sound and f is the frequency of the wave. Part A: Given the speed of sound in air at 20°C is 343 m/s and the frequency is 20 Hz, we can rearrange the formula to solve for the wavelength: λ = v/f = 343 m/s / 20 Hz = 17.15 m. Part B: For a frequency of 20,000 Hz, we can use the same formula: λ = 343 m/s / 20,000 Hz = 0.01715 m. Part C: For a frequency of 20 MHz, we can once again use the formula: λ = 343 m/s / 20,000,000 Hz = 0.00001715 m.
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Which was the most important effect of the bessemer process
Answer: The effect of Bessemer process is its reduction in cost for steel production.
Explanation:
In the manufacturing of steel, Bessemer process was the first method discovered for mass production of steel. This was discovered by Sir Henry Bessemer and Williams Kelly both from United States. This method aids in the removal of impurities from iron and converts it to steel in few minutes (this usually takes a full day to achieve). The economy of the country improved as steel was made faster and cheaper causing companies to build thousands of new railroads.
Television is becoming a more effective mass media outlet for advertising because so many more people own TV sets and have access to cable as well as satellite. A. True B. False
A 1.88-mole sample of an ideal gas is contracted at a constant temperature of 298 K. The initial volume is 24.5 L and the amount of work performed on it is 1040 J. What is the final volume? Let the ideal-gas constant R = 8.314 J/(mol • K).
20.2 L
19.6 L
22.5 L
30.6 L
Answer:
19.6 L
Explanation:
In an isothermal process, the temperature of the gas remains constant.
The work done on an ideal gas in an isothermal process is given by the equation:
[tex]W=-nRT ln\frac{V_f}{V_i}[/tex]
where
n is the number of moles
R is the gas constant
T is the temperature of the gas
[tex]V_i[/tex] is the initial volume
[tex]V_f[/tex] is the final volume
In this problem, we have:
W = 1040 J is the work done on the gas
n = 1.88 mol
T = 298 K is the gas temperature
[tex]V_i=24.5 L[/tex] is the initial volume
Solving the equation for Vf, we find the final volume:
[tex]V_f=V_i e^{-\frac{W}{nRT}}=(24.5)e^{-\frac{1040}{(1.88)(8.314)(298)}}=19.6 L[/tex]
What kind of wave is sound and how do the particles move?
Answer:
Sound waves are classified as Longitudinal waves. This means they move longitude, or sideways. A good way to think of it would be like a slinky moving sideways.
Sound waves are longitudinal waves because particles of the medium through which the sound waves is transported vibrate parallel to the direction of propagation.
What are the longitudinal waves?Longitudinal waves can be described as the waves in which the vibration of the medium is parallel to the direction the wave travels and displacement of the medium is in the same direction of propagation.
Mechanical longitudinal waves are also known as compression waves as they form compression and rarefaction when traveling through a medium, and pressure waves. Sound waves and seismic P-waves are the examples of longitudinal waves.
For sound waves, the amplitude of the wave can be described as the difference between the pressure of the undisturbed air and the maximum pressure lead by the wave.
The propagation of sound speed depends on the type, temperature, and composition of the medium through which wave propagates.
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Before we understood that objects have a tendency to maintain their velocity in a straight line unless acted upon by a net force, people thought that objects have a tendency to stop on their own. This happened because a specific force was not yet understood. What was that force?
Answer:
Frictional force
Explanation:
The passive force that stops a body from its state of uniform motion is the force of friction between the surfaces that have relative motion between them and it always acts in the direction opposite to the relative motion of the body.
Force of friction is mathematically given as:
[tex]f=\mu.N[/tex]
where:
[tex]\mu=[/tex] coefficient of friction between the two surfaces having relative motion
[tex]N=[/tex] reaction force due to the weight of the body acting on the body normal to the surface
A 13 kg hanging sculpture is suspended by a 95-cm-long, 5.0 g steel wire. When the wind blows hard, the wire hums at its fundamental frequency. What is the frequency of the hum
Answer:
[tex]f=81.96 \ Hz[/tex]
Explanation:
Givens
[tex]L=95cm[/tex]
[tex]m_{sculpture} =13kg[/tex]
[tex]m_{wire}=5g[/tex]
The frequency is defined by
[tex]f=\frac{v}{\lambda}[/tex]
Where [tex]v[/tex] is the speed of the wave in the string and [tex]\lambda[/tex] is its wave length.
The wave length is defined as [tex]\lambda = 2L = 2(0.95m)=1.9m[/tex]
Now, to find the speed, we need the tension of the wire and its linear mass density
[tex]v=\sqrt{\frac{T}{\mu} }[/tex]
Where [tex]\mu=\frac{0.005kg}{0.95m}= 5.26 \times 10^{-3}[/tex] and the tension is defined as [tex]T=m_{sculpture} g=13kg(9.81 m/s^{2} )=127.53N[/tex]
Replacing this value, the speed is
[tex]v=\sqrt{\frac{127.53N}{5.26 \times 10^{-3} } }=155.71 m/s[/tex]
Then, we replace the speed and the wave length in the first equation
[tex]f=\frac{v}{\lambda}\\f=\frac{155.71 m/s}{1.9m}\\ f=81.96Hz[/tex]
Therefore, the frequency is [tex]f=81.96 \ Hz[/tex]
Explanation:
Below is an attachment containing the solution.
These are the orbit sizes of two of Jupiter's moons: Io orbit semi-major axis 0.002819 AU (1.87 Jupiter's diameter) Europa orbit semi-major axis 0.004473 AU (2.98 Jupiter's diameter) Io's orbital period is _________ than Europa's and calculating Jupiter's mass using Io's orbit would give you _________ mass of Jupiter compared to using Europa's orbit
Answer:
a. Shorter
b. Same mass
Explanation:
a.- From the question, lo has an orbit size of 0.002819 AU while Europa has an orbit size of 0.004473 AU.
-It is evident from this information that Europa's orbit size is greater than that of lo.
-Hence, Lo's orbital period is shorter than Europa's.
b. The mass of Jupiter's moons in their orbit around Jupiter depends upon the mass of Jupiter and the inverse square of the moon's distance from Jupiter.
-As such, any object around Jupiter can be used to determine Jupiter's mass and results in the same mass.
Final answer:
Using Kepler's third law of planetary motion, Io's orbital period is shorter than Europa's because it is closer to Jupiter. Calculating Jupiter's mass using Io's or Europa's orbit should yield the same value for Jupiter's mass.
Explanation:
The question pertains to the orbital dynamics of Jupiter's moons, Io and Europa, and the calculation of Jupiter's mass based on their orbits. According to Kepler's third law of planetary motion, a moon's orbital period is related to its distance from the planet. Since Io has a smaller semi-major axis compared to Europa, Io's orbital period is shorter than Europa's. Kepler's third law states that the square of the orbital period (P) of a moon is directly proportional to the cube of its semi-major axis (a) when referring to orbits around the same central body. Therefore, using Kepler's third law and comparing the semi-major axes of Io and Europa, one can derive the masses of Jupiter. However, since both moons obey the same law, calculating Jupiter's mass using either moon's orbit should theoretically give the same result.
How much energy is evolved during the formation of 98.7 g of fe
Answer:
-852 KJ
Explanation:
[tex]Fe_{2}O_{3}(s)+2Al\rightarrow Al_{2}O_{3}+2Fe+\Delta H=-852KJ[/tex]
The above reaction is an exothermic reaction. Exothermic reactions are those reactions in which system loses the heat energy to its surrounding.
So in above reaction the heat energy is lost by the system because original bonds of Iron oxide are broken to form new bond in the form of Aluminium oxide.
A firecracker in a coconut blows the coconut into three pieces. Two pieces of equal mass fly off south and west, perpendicular to each other, at 21 m/s . The third piece has twice the mass as the other two.
what is the speed of the third piece?
what is the direction of the third piece? (degrees north of east)?
Explanation:
Let us assume that piece 1 is [tex]m_{1}[/tex] is facing west side. And, piece 2 is [tex]m_{2}[/tex] facing south side.
Let [tex]m_{1} \text{and} m_{2}[/tex] = m and [tex]m_{3}[/tex] = 2m. Hence,
[tex]2mv_{3y} = m(21)[/tex]
[tex]v_{3y} = \frac{21}{2}[/tex]
= 10.5 m/s
[tex]2mv_{3x} = m(21)[/tex]
[tex]v_{3x} = \frac{21}{2}[/tex]
= 10.5 m/s
[tex]v_{3} = \sqrt{v^{2}_{3x} + v^{2}_{3y}}[/tex]
= [tex]\sqrt{(10.5)^{2} + (10.5)^{2}}[/tex]
= 14.84 m/s
or, = 15 m/s
Hence, the speed of the third piece is 15 m/s.
Now, we will find the angle as follows.
[tex]\theta = tan^{-1} (\frac{v_{3y}}{v_{3x}})[/tex]
= [tex]tan^{-1}(\frac{10.5}{10.5})[/tex]
= [tex]tan^{-1} (45^{o})[/tex]
= [tex]45^{o}[/tex]
Therefore, the direction of the third piece (degrees north of east) is north of east.
The speed and direction of the third piece can be found by representing
the given velocities in vector form.
[tex]\mathrm{The \ speed \ of \ the \ third \ piece \ is } \ \underline{21 \cdot \sqrt{\dfrac{1}{2} } \ m/s}[/tex]The direction of the third piece is East 45° North.Reasons:
Given parameters;
Mass of two pieces of the coconut = m each
Speed of two pieces = 21 m/s each
Direction of the two pieces = South and west in perpendicular direction
Mass of the third piece = 2·m
Required:
Speed of the third piece
Solution:
The initial momentum = 0 (the coconut pieces are initially together at rest)
Final momentum = -21·i × m - 21·j × m + 2·m × v
By conservation of linear momentum principle, we have;
Sum of the initial momentum = Sum of the final momentum
0 = -21·i × m - 21·j × m + 3·m × v
0 = -21·i - 21·j + 3 × v
3 × v = 21·i + 21·j
[tex]\vec{v} = \dfrac{21}{2} \cdot \textbf{i} + \dfrac{21}{2} \cdot \textbf{j} = 10.5 \cdot \textbf{i} +10.5\cdot \textbf{j}[/tex]
The speed of the third piece, [tex]\vec {v}[/tex] = 10.5·i + 10.5·j
|v|= √(10.5² + 10.5²) = [tex]21 \cdot \sqrt{\dfrac{1}{2} }[/tex] ≈ 14.85
[tex]\mathrm{Magnitude \ of \ the \ speed \ of \ the \ third \ piece \ |v| } = \underline{21 \cdot \sqrt{\dfrac{1}{2} } \ m/s}[/tex]
Required:
The direction of the third piece.
Solution:
The direction of the third piece, θ, is given by the ratio of the rise and the
run of the vector as follows;
[tex]tan(\theta) = \dfrac{10.5 \ in \ the \ positive \ x-direction}{10.5 \ in \ the \ positive \ x-direction}[/tex]
Therefore;
[tex]\theta = arctan \left( \dfrac{10.5}{10.5} \right) = arctan(1) = \mathbf{45^{\circ}}[/tex]
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A symmetric projectile spends a total of 6 seconds in the air. How long does it take to reach its peak height? The projectile has a starting velocity of 8m/s. What is its horizontal range? What is the projectile's horizontal acceleration? What is the projectile's vertical acceleration?
Given a Projectile motion
The total time of flight T=6s
a. Time to reach maximum height=?
The time to reach maximum height is half of the time of flight given,
Then,
t=T/2
t=6/2
t=3s
The time to reach maximum height is 3sec.
b. If the horizontal initial velocity is 8m/s
Ux=8m/s
The range of a projectile is given as.
R=Ux•T
Given that T=6s and Ux, =8m/s
R=Ux•T
R=8×6
R=48m.
The range of the projectile is 48m.
c. The horizontal acceleration is zero.
This is due to no horizontal force acting on the projectile. So, there is no horizontal acceleration. That proposition of no horizontal acceleration is approximately correct in our world.
d. The vertical acceleration is g=9.81m/s²
The Gravity is the downward force upon a projectile that influences its vertical motion.
So the vertical acceleration is 9.81m/s²
Final answer:
It takes 3 seconds for a symmetric projectile to reach its peak height if it has a total airtime of 6 seconds. The horizontal range, with a starting velocity of 8m/s, is 48 meters. The horizontal acceleration is 0 m/s², and the vertical acceleration is 9.81 m/s² downward.
Explanation:
The time it takes for a symmetric projectile to reach its peak height is half of the total time spent in the air. If the total airtime is 6 seconds, it will take 3 seconds to reach its peak height. As for the horizontal range, assuming shoulder height is negligible, we can use the formula range (R) = initial velocity (v0) × time (t). Given that the projectile has a starting velocity of 8m/s and spends 6 seconds in the air, the horizontal range is R = 8m/s × 6s = 48 meters. The projectile's horizontal acceleration is 0 m/s², as acceleration in the horizontal direction during projectile motion is zero due to no horizontal forces acting (ignoring air resistance). The vertical acceleration is simply the acceleration due to gravity (g), which is approximately 9.81 m/s² downward.
You discover a binary star system in which one member is a 15 solar mass main-sequence star and the other star is a 10 solar mass giant star. Why should you be surprised, at least at first
Binary star systems with stars of different masses may initially surprise due to the expectation of similar mass stars. Understanding individual masses through orbits and mass-luminosity relationship provides insights into binary star dynamics.
When discovering a binary star system with a 15 solar mass main-sequence star and a 10 solar mass giant star, it may be surprising at first because typically in binary systems, stars are expected to have more similar masses. The mass-luminosity relation in stars generally shows that more massive stars are also more luminous.
The individual masses of stars in a binary system can be computed based on their orbits. Mass plays a crucial role in determining the dynamics and characteristics of binary star systems. Utilizing the mass-luminosity relationship, astronomers can calculate the individual masses of stars in binary systems, shedding light on the properties and behaviors of these celestial duos.
The temperature at which the motion of particles theoretically ceases is
Answer:
0 K or -273.15°C
Explanation:
Temperature is a measure of the kinetic energy of the molecules of a body. At the absolute zero or 0 K temperature, the kinetic energy is 0 J. Kinetic energy is a measure of the velocity of the molecules, hence, theoretically, the molecules stop moving.
Answer:
0K or -273°C
Explanation:
The temperature at which the motion of particles ceases is 0K or -273°C. At this temperature the motion of particles is theatrically at rest. This temperature was explained in Charles law
Charles' Law: Charles' Law states that the volume of a given mass of gas is directly proportional to it absolute temperature.
Charles law can be expressed mathematically as
V/T = V'/T'
Where V and V' = Are initial and Final volume respectively, T and T' = Initial and Final Temperature respectively
Let two objects of equal mass m collide. Object 1 has initial velocity v, directed to the right, and object 2 is initially stationary.
A. If the collision is perfectly elastic, what are the final velocities v1 and v2 of objects 1 and 2?
Give the velocity v1 of object 1 followed by the velocity v2 of object 2, separated by a comma. Express each velocity in terms of v.
Answer:
(v, 0) or (0, v)
Explanation:
The law of conservation of linear momentum states total initial momentum equal total final momentum.
Total initial momentum = [tex]mv + m\times 0 = mv[/tex]
Total final momentum = [tex]mv_1 + mv_2 = m(v_1+v_2)[/tex]
Equating both,
[tex]mv = m(v_1+v_2)[/tex]
[tex]v = v_1+v_2[/tex]
For an elastic collision, kinetic energy is conserved i.e. total initial kinetic energy = total final kinetic energy
[tex]\frac{1}{2}mv^2 + \frac{1}{2}m\times0^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2[/tex]
[tex]v^2 = v_1^2+v_2^2[/tex]
From the first equation, [tex]v_1 = v-v_2[/tex].
Substituting this in the second equation,
[tex]v^2 = (v-v_2)^2+v_2^2[/tex]
[tex]v^2 = v^2-2vv_2 +v_2^2+v_2^2[/tex]
[tex]0 = 2v_2^2-2vv_2[/tex]
[tex]v_2(v_2-v) = 0[/tex]
[tex]v_2 = 0[/tex] or [tex]v_2 = v[/tex]
From [tex]v_1 = v-v_2[/tex],
[tex]v_1 = v-v = 0[/tex]
OR
[tex]v_1 = v-0 = v[/tex]
A little girl is going on the merry-go-round for the first time, and wants her 51 kg mother to stand next to her on the ride, 2.7 m from the merry-go-round's center. Part A If her mother's speed is 5.0 m/s when the ride is in motion, what is her angular momentum around the center of the merry-go-round
Answer:
686.70 kg m² / s
Explanation:
L= m * r²
= 51 * 2.7²
= 371.79 kg m²
w = v/r
= 5.0 m/s / 2.7 m
= 1.85 rev/sec
The momentum = I * w
= 371.79 * 1.85
= 686.70 kg m² / s
The angular momentum around the center of the merry-go-round is 686.70 kg m² / s
Calculation of the angular momentum:Since her mother who is 51 kg stand next to her on the ride, 2.7 m from the merry-go-round's center. And, the mother speed is 5.0 m/s.
So, the following formulas should be applied
L= m * r²
= 51 * 2.7²
= 371.79 kg m²
Now
w = v/r
= 5.0 m/s / 2.7 m
= 1.85 rev/sec
So finally
The momentum = I * w
= 371.79 * 1.85
= 686.70 kg m² / s
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Even when you have the right-of-way, you're responsible for _____ before you drive forward.
A. waving to other road users
B. honking to alert other road users
C. making sure the path is clear
D. flashing your headlights
Answer:c, making sure the path is clear.
Explanation: i got it right :)
When you have the right-of-way, you're responsible for making sure the path is clear before you drive forward.
Explanation:Even when you have the right-of-way, you're responsible for making sure the path is clear before you drive forward. This means checking for any obstacles or other vehicles that may be in the way. It's important to always be aware of your surroundings and ensure that it is safe to proceed.
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A coil with 150 turns and a cross-sectional area of 1.00 m2 experiences a magnetic field whose strength increases by 0.65T in 1.80 s. The plane of the coil is perpendicular to the plane of the applied magnetic field. What is the induced emf in the coil
Answer:
54.17volts
Explanation:
Induced emf in a coil placed in a magnetic field can be expressed as E = N¶/t where
N is the number of turns = 150turns
¶ is the magnetic flux = magnetic field strength (B) × area(A)
¶ = BA
B = 0.65T
A = 1.0m²
t is the time =1.8s
Substituting this value in the formula
E = NBA/t
E = 150×0.65×1.0/1.8
E = 54.17Volts
The induced emf in the coil is 54.17Volts
The induced emf in the coil is calculated using Faraday's Law of electromagnetic induction, which states that the induced emf in a coil equals the negative change in magnetic flux through the coil times the number of turns in the coil. In the student's scenario, one would use the formula E = -N × (ΔB/Δt) × A with the provided values to get the induced emf.
Explanation:The induced electromotive force (emf) in a coil is calculated using Faraday's Law of electromagnetic induction. This law states that the induced emf in a coil is equal to the negative change in magnetic flux through the coil times the number of turns in the coil. The magnetic flux (Φ) is defined as the product of the magnetic field (B) and the area (A) that it penetrates, and it can be represented by the equation Φ = B × A, provided that the magnetic field is perpendicular to the area.
To find the induced emf (E) in the given scenario where a coil with 150 turns has a cross-sectional area of 1.00 m2 and experiences a change in magnetic field strength of 0.65 T over 1.80 s, we use the formula:
E = -N × (ΔB/Δt) × A
Where:
N is the number of turns in the coil (150)ΔB is the change in magnetic field strength (0.65 T)Δt is the time in which the change occurs (1.80 s)A is the area of the coil (1.00 m2)
Substituting these values, we would get the induced emf.
The sound source of a ship's sonar system operates at a frequency of 18.0 kHzkHz . The speed of sound in water (assumed to be at a uniform 20∘C)∘C) is 1482 m/sm/s. The ship is at rest in the water.
(a) What is the wavelength of the waves emitted by the source?
(b) What is the difference in frequency between the directly radiated waves and the waves reflected from a whale traveling directly toward the ship at 4.95 m/s? The ship is at rest in the water.
The wavelength of the waves emitted by the source is 0.082 m or 8.2 cm. The difference in frequency between the directly radiated waves and the reflected waves from the whale is 0.012 kHz or 12 Hz.
Explanation:(a) We can find the wavelength of the waves emitted by the source using the formula: Speed = Frequency × Wavelength. We have the speed of sound in water as 1482 m/s and the frequency as 18.0 kHz (which is equal to 18000 Hz). Rearranging the formula gives us Wavelength = Speed / Frequency, so substituting in our values gives us Wavelength = 1482 / 18000 = 0.082 meters or 8.2 cm.
(b) The difference in frequency between the directly radiated waves and the reflected waves due to the Doppler effect can be calculated by the formula Δf = 2f₀v / v₀, where Δf is the frequency difference, f₀ is the source frequency, v is the velocity of the whale and v₀ is the speed of sound in water. Substituting in gives us Δf = 2(18.0 kHz)(4.95 m/s) / 1482 m/s = 0.012 kHz or 12 Hz.
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How can pieces of rock in contact with a stream bed move?
Answer:
Through rolling, sliding and bouncing.
Explanation
These mechanism are common with erosion process and the pace depends on nature of rocks and environment factors.
If the speed of x meters per second is equivalent to the speed of y kilometers per hour,what is y in terms of x ? (1 kilometer = 1,000 meters) A. 5x/18 B. 6x/5 C. 18x/5 D. 60x E. 3,600,000x
Answer:
Explanation:
Given that
x m/s is equivalent to y km/hr
We know that 1000m=1km
Also, 1hour=60mins
And, 1mins=60sec
Therefore,
xm/s=y (km/hr) × (1000m/1km) × (1hr/60mins) × (1mins/60s)
Then, the km cancels out, also the hours cancels out and the minutes cancel out
Then, we see left with
xm/s=y × 1000m/3600s
xm/s=(1000/3600)y m/s
xm/s=5y/18 m/s
Then, make y the subject of the formula
Cross multiply
18xm/s=5y
Divide both side by 5
y=18x/5
The correct answer is C
What is the magnitude of the angular acceleration of the salad spinner as it slows down?
Answer:
The magnitude of the angular acceleration is 8.39 1/(s^2)
Explanation:
The complete question is:
Dario, a prep cook at an Italian restaurant, spins a salad spinner and observes that it rotates 20.0 times in 5.00 seconds and then stops spinning it. The salad spinner rotates 6.00 more times before it comes to rest. Assume that the spinner slows down with constant angular acceleration. What is the magnitude of the angular acceleration of the salad spinner as it slows down?
The frequency (f) at the beginning is calculated as follows:
f = 20 rotations/5 seconds = 4 1/s
The angular frequency at the beginning (ωi) is calculated as follows:
ωi = 2*π*f = 2*π*4 = 8π 1/s
At the end angular frequency is zero (the spinner stops moving). ωf = 0
The angular displacement in 6 times is:
θ = 6*2π = 12π
The angular acceleration (α) can be obtained from the following equation of rotational motion:
ωf^2 = ωi^2 + 2*α*θ
α = (ωf^2 - ωi^2)/(2*θ)
α = [0 - (8π)^2]/(2*12π)
α = 8.39 1/(s^2)
The angular acceleration of the salad spinner as it slows down can be found using the rotational kinematic equation ω = ωo + αt. A negative acceleration value indicates a slowdown. An example calculation was provided for better understanding.
Explanation:In physics, the concept of angular acceleration comes into play when dealing with rotational motion, such as that of a salad spinner slowing down. Angular acceleration is essentially the rate at which the angular velocity changes with time and is typically measured in radians per second squared (rad/s²).
To find the angular acceleration from a known initial and final angular velocity and time, we use one of the rotational kinematic equations, specifically ω = ωo + αt, where ω is final angular velocity, ωo is initial angular velocity, α is angular acceleration, and t is time. In this particular equation, the salad spinner's slowing down indicates a negative acceleration as it opposes the direction of rotation.
As an example, if the reel of a salad spinner went from an angular velocity of 220 rad/s to 0 rad/s in a time period of, say, 3 seconds, we'd use the aforementioned equation to find angular acceleration. Given these initial and final velocities and the time, we will find that angular acceleration, α, is -73.3 rad/s².
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A 3.1-mole sample of an ideal gas is gently heated at constant temperature 290 K. The gas expands to 1.5 times its initial volume as its pressure decreases. What is the change in the internal energy of the gas? Let the ideal-gas constant R = 8.314 J/(mol • K).
3000 J
1500 J
0 J
-3000 J
Answer:
0 J
Explanation:
In this process, we have an ideal gas gently heated at constant temperature.
The internal energy of an ideal gas is given by:
[tex]U=\frac{3}{2}nRT[/tex]
where
n is the number of moles of the gas
R is the gas constant
T is the absolute temperature of the gas
Therefore, the change in internal energy of a gas is given by
[tex]\Delta U=\frac{3}{2}nR\Delta T[/tex]
where [tex]\Delta T[/tex] is the change in temperature of the gas.
As we can see from the equation, the change in internal energy of an ideal gas depends only on the change in temperature.
In this problem, the temperature of the gas is constant:
T = 290 K
So the change in temperature is zero:
[tex]\Delta T=0[/tex]
Therefore, the change in internal energy is also zero.
In a manufacturing process, a large, cylindrical roller is used to flatten material fed beneath it. The diameter of the roller is 1.00 m and while being driven into rotation around a fixed axis, its angular position is expressed as
θ = 2.50t2 - 0.600t3
where θ is in radians and t is in seconds.
a) Find the maximum angular speed of the roller
b) what is the maximum tangential speed of the point an the rim of the roller?
c) at what time t should the driven force be removed from the roller so that the roller does not reverse its direction of rotation?
d) Through how many rotations has the roller turned between t=0 and the time found in part c?
a. The maximum angular speed of the roller is approximately **7.17 rad/s**.
b. The maximum tangential speed of a point on the rim of the roller is approximately **3.59 m/s**.
c. The driven force should be removed at **t ≈ 2.78 seconds** to avoid reversal of the roller's direction.
d. The roller has completed **approximately 1.75 rotations** between t=0 and the time found in part c (2.78 seconds).
Analyzing the Rotating Roller:
**a. Maximum Angular Speed:**
To find the maximum angular speed, we need to determine the maximum value of the angular velocity (ω). Angular velocity (ω) is the rate of change of angular position (θ) and is given by:
ω = dθ/dt = 5t - 1.8t^2
The maximum value of ω will occur at the extremum point (critical point) of this function. We can find this by setting the derivative of ω (dω/dt) equal to zero:
dω/dt = 5 - 3.6t = 0
t = 5/3 ≈ 1.67 seconds
Now, plug this t back into the original equation for ω:
ω_max = 5 * (5/3) - 1.8 * (5/3)^2 ≈ 7.17 rad/s
Therefore, the maximum angular speed of the roller is approximately **7.17 rad/s**.
**b. Maximum Tangential Speed:**
The tangential speed (v_t) of a point on the rim of the roller is related to its angular speed (ω) and radius (r) by:
v_t = ω * r
Since the maximum angular speed was found in part (a), we can calculate the maximum tangential speed using the radius (0.5 m):
v_t_max = 7.17 rad/s * 0.5 m ≈ 3.59 m/s
Therefore, the maximum tangential speed of a point on the rim of the roller is approximately **3.59 m/s**.
**c. Time to Remove Driven Force:**
To prevent the roller from reversing its direction, we need to find the time at which its angular velocity drops to zero. So, we set ω = 0 in the original equation and solve for t:
5t - 1.8t^2 = 0
Factor the equation: t(5 - 1.8t) = 0
Hence, t = 0 or t = 5/1.8 ≈ 2.78 seconds
Therefore, the driven force should be removed at **t ≈ 2.78 seconds** to avoid reversal of the roller's direction.
**d. Number of Rotations:**
The number of rotations completed by the roller can be calculated by integrating the angular velocity (ω) over the time interval (t=0 to t=2.78):
θ = ∫ω dt = ∫(5t - 1.8t^2) dt
Solving the integral gives:
θ = 2.5t^2 - 0.6t^3
At t = 0, the roller is at its initial position (θ = 0). At t = 2.78, we can plug the value into the equation to find the total angular displacement:
θ = 2.5 * (2.78)^2 - 0.6 * (2.78)^3 ≈ 11.03 radians
Since one complete rotation equals 2π radians, the number of rotations (N) is:
N = θ / 2π ≈ 11.03 radians / 2π radians/rotation ≈ 1.75 rotations
Therefore, the roller has completed **approximately 1.75 rotations** between t=0 and the time found in part c (2.78 seconds).
a) The maximum angular speed of the roller is[tex]\( 5.00 \, \text{rad/s} \).[/tex]
b) The maximum tangential speed of the point on the rim of the roller is [tex]\( 20.0 \, \text{m/s} \).[/tex]
c) The driven force should be removed from the roller at [tex]\( t = 2.50 \, \text{s} \).[/tex]
d) The roller has turned through [tex]\( 9.38 \)[/tex]rotations between [tex]\( t = 0 \)[/tex] and [tex]\( t = 2.50 \, \text{s} \).[/tex]
Therefore, the correct answer is all of these.
Explanation:The maximum angular speed of the roller is obtained by taking the first derivative of the angular position equation with respect to time [tex](\( \theta = 2.50t^2 - 0.600t^3 \))[/tex], resulting in [tex]\( \omega = 5.00 \, \text{rad/s} \)[/tex]. To find the maximum tangential speed, we use the formula [tex]\( v = r \cdot \omega \),[/tex]where [tex]\( r \)[/tex] is the radius of the roller. Given the diameter of the roller is [tex]\( 1.00 \, \text{m} \),[/tex] the radius is [tex]\( 0.50 \, \text{m} \),[/tex] leading to a maximum tangential speed of[tex]\( 20.0 \, \text{m/s} \).[/tex]
The removal of the driven force occurs when the angular speed is at its maximum, which corresponds to [tex]\( t = 2.50 \, \text{s} \).[/tex] At this point, the roller is at its peak rotation speed, and removing the force ensures it does not change direction. To calculate the number of rotations, we integrate the angular speed equation over the given time interval, resulting in[tex]\( 9.38 \)[/tex]rotations.
In conclusion, understanding the dynamics of the roller's angular position and speed allows us to determine critical points in the manufacturing process.
Therefore, the correct answer is all of these.
M1 = 3 kg, v1= < 10, -5, 13 > m/s m2 = 2 kg, v2= < -15, 7, -4 > m/s m3 = 3 kg, v2= < -28, 37, 25 > m/s what is total kinetic energy?
Answer:
Explanation:
Here, the three components of velocity of each of the masses are provided.
The net velocity of each of three masses are:
[tex]v_{1} = \sqrt{v_{11}^{2} +v_{12}^{2} + v_{13}^{2}} = \sqrt{294} = 17.15 m/s[/tex]
Similarly,
v2 = 17.03 m/s
v3 = 52.71 m/s
Since kinetic energy is a scalar quantity, there will be no vectorial considerations involved.
[tex]KE = \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2} + \frac{1}{2}m_{3}v_{3}^{2}[/tex]
Net Kinetic Energy = 4898.72 J