The angular acceleration is 4.44*[tex]10^{-5} rev/s^{2}[/tex] , radial component is 0.016 m/[tex]s^{2}[/tex], tangential component is 0.9347*[tex]10^{-5} m/s^{2}[/tex].
Angular acceleration,
ω=1/60=0.016 rev/s
[tex]\alpha[/tex]=0.016/(6*60)=4.44*[tex]10^{-5} rev/s^{2}[/tex]
The angular acceleration is 4.44*[tex]10^{-5} rev/s^{2}[/tex]
Radial component of the linear acceleration=[tex]\alpha_{r}[/tex]
v=ωr=0.016*4.75=0.076 m/s
[tex]\alpha_{r}[/tex]=0.076 /4.75=0.016 m/[tex]s^{2}[/tex]
The tangential component = [tex]\alpha_{t}[/tex]=4.44*[tex]10^{-5} rev/s^{2}[/tex]/4.75=0.9347*[tex]10^{-5} m/s^{2}[/tex]
Angular accelerationThe temporal rate at which angular velocity changes is known as angular acceleration. Due to the fact that there are two different types of angular velocity—spin angular velocity and orbital angular velocity—there are also two different types of angular acceleration, referred to as spin angular acceleration and orbital angular acceleration, respectively. The terms "orbital angular acceleration" and "spin angular acceleration" describe the angular acceleration of a point particle about a fixed origin and, respectively, the angular acceleration of a rigid body about its center of rotation.
The unit of measurement for angular acceleration is the angle per unit time squared, or, in SI units, radians per second squared. It is typically denoted by the symbol alpha ().
In traveling to the Moon, astronauts aboard the Apollo spacecraft put spacecraft into a slow rotation to distribute the Sun's energy evenly (so one side would not become too hot). At the start of their trip, they accelerated from no rotation to 1.0 revolution every minute during a 12-min time interval. Think of the spacecraft as a cylinder with a diameter of 8.5 m rotating about its cylindrical axis.
a)Determine the angular acceleration of the ship.
Express your answer using two significant figures.
b)Determine the radial component of the linear acceleration of a point on the skin of the ship 9.5 min after it started this acceleration.
Express your answer to two significant figures and include the appropriate units.
c)Determine the tangential component of the linear acceleration of a point on the skin of the ship 9.5 min after it started this acceleration.
Express your answer to two significant figures and include the appropriate units.
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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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The tires of a car make 68 revolutions as the car reduces its speed uniformly from 85.0 km/h to 56.0 km/h. The tires have a diameter of 0.80 m.
a)What was the angular acceleration of the tires?
Express your answer using two significant figures.
b)If the car continues to decelerate at this rate, how much more time is required for it to stop?
Express your answer to two significant figures and include the appropriate units.
c)If the car continues to decelerate at this rate, how far does it go? Find the total distance.
Express your answer to three significant figures and include the appropriate units.
a) The wheel of the car decelerates at an angular acceleration of - 0.577 radians per square second.
b) The wheel of the car requires a time of 51.149 seconds to stop.
c) The wheel of the car travels a distance of 683.225 meters before stopping.
How to analyze a decelerating rotating wheel
a) In this case we have a rotating wheel that decelerates at constant rate. The angular acceleration, in radians per square second, of the tires is determined by the following formula:
α = [ω'² - ω²] / (2 · θ) (1)
Where:
ω - Initial angular velocity, in radians per second.ω' - Final angular velocity, in radians per second.θ - Change in the angular displacement, in radians.The initial and final angular velocities, in radians per second, are now determined:
ω' = v' / R (2)
ω = v / R (3)
Where:
v - Initial linear velocity, in meters per second.v' - Final linear velocity, in meters per second.R - Radius of the tire, in meters.If we know that R = 0.80 m, v = 23.611 m / s, v' = 15.556 m / s and θ ≈ 427.257 radians, then the angular acceleration of the tire is:
ω' = (15.556 m / s) / (0.80 m)
ω' = 19.445 rad / s
ω = (23.611 m / s) / (0.80 m)
ω = 29.513 rad / s
α = [(19.445 rad / s)² - (29.513 rad / s)²] / [2 · (427.257 rad)]
α = - 0.577 rad / s²
b) The time required to stop the car, in seconds, is determined by the following expression:
t = (ω' - ω) / α (4)
t = (0 rad / s - 29.513 rad / s) / (- 0.577 rad / s²)
t = 51.149 s.
c) First, we find the change in angular displacement of the tire:
θ = [ω'² - ω²] / (2 · α) (5)
θ = [(0 rad / s)² - (29.513 rad / s)²] / [2 · (- 0.577 rad / s²)]
θ = 754.781 rad
Lastly, the distance traveled by the vehicle is:
s = R · θ (6)
s = (0.80 m) · (754.781 rad)
s = 683.825 m
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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 uniform plank of length 2.00 m and mass 30.23 kg is supported by three ropes. A 700 N person is a distance, of 0.55 m from the left end. A) find the magnitude of the tension in the vertical rope on the left end. Give your answers in newtons. B) find the magnitude of the tension in the rope in the right end. Give your answers in newtons C) find the magnitude of the tension in the horizontal rope on the left end. Give your answers in newtons
a. The magnitude of the tension, T₂ in the vertical rope on the left end is T₂ = 655.62 N
b. The magnitude of the tension in the rope in the right end is T₁ = 530.4 N
c. The magnitude of the tension in the horizontal rope on the left end, T₃ is T₃ = 406.3 N
What is tension force?Tension force refers to a pulling force that is exerted by a string or cable about an axis.
a. The magnitude of the tension, T₂ in the vertical rope on the left end is given as follows:
Taking moment about the vertical axis
T₂ = 30.23 * 9.81 + 700 - T₁ * Sin40°
Solving for T₁ by taking the left end as the pivot;
T₁ Sin 40° * 2.00 = 700 * 0.55 + (30.23 * 9.81) * 1.0
T₁ * 1.285 = 681.5563
T₁ = 530.4 N
Therefore;
T₂ = 30.23 * 9.81 + 700 - 530.4 * Sin 40°
T₂ = 655.62 N
b. From calculation, the magnitude of the tension in the rope in the right end is T₁.
T₁ = 530.4 N
c. The magnitude of the tension in the horizontal rope on the left end, T₃ is determined thus:
Taking moments about the left end in the horizontal direction;
T₃ = T₁ * cos 40°
T₃ = 530.4 N * cos 40°
T₃ = 406.3 N
In conclusion, the tension at the rope in the various ends is determined by taking moments about the left end.
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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:
the meaninn of dynamics
Answer:
Dynamics is just a nice word used in physics denoting a branch of physics, related to the study of forces. Usually these forces are not in mechanical equilibrium, else the branch would be statics.
A phone with a mass of 0.2 kg is dropped from a height of 30 m what is it’s speed when it hits the ground?the acceleration of gravity is 9.8ms
Answer:
24.2 m/s
Explanation:
Mass is irrelevant in this situation....
Displacement: ( to find time)
x = xo + vo t - 1/2 at^2
30= 0 + 0 - 1/2 (9.8)t^2
t = 2.47 seconds
Velocity:
vf = a t = 9.8 (2.473) = 24.2 m/s
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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(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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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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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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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.
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.
Liquids are used in thermometer but not the solid.
Solids have very small intermolecular space and when heated, do not expand as per the container volume. Basically, their expansion is not good.
Liquids, on the other hand, have a definite volume and their expansion is ideal. Also, they fill the container only to the desired level, unlike gases.
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.
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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A ray of light incident in air strikes a rectangular glass block of refractive index 1.50, at an angle of incidence of 45°. Calculate the angle of refraction in the glass.
Answer:
Approximately [tex]28^{\circ}[/tex].
Explanation:
The refractive index of the air [tex]n_{\text{air}}[/tex] is approximately [tex]1.00[/tex].
Let [tex]n_\text{glass}[/tex] denote the refractive index of the glass block, and let [tex]\theta _{\text{glass}}[/tex] denote the angle of refraction in the glass. Let [tex]\theta_\text{air}[/tex] denote the angle at which the light enters the glass block from the air.
By Snell's Law:
[tex]n_{\text{glass}} \, \sin(\theta_{\text{glass}}) = n_{\text{air}} \, \sin(\theta_{\text{air}})[/tex].
Rearrange the Snell's Law equation to obtain:
[tex]\begin{aligned} \sin(\theta_{\text{glass}}) &= \frac{n_{\text{air}} \, \sin(\theta_{\text{air}})}{n_{\text{glass}}} \\ &= \frac{(1.00)\, (\sin(45^{\circ}))}{1.50} \\ &\approx 0.471\end{aligned}[/tex].
Hence:
[tex]\begin{aligned} \theta_{\text{glass}} &= \arcsin (0.471) \approx 28^{\circ}\end{aligned}[/tex].
In other words, the angle of refraction in the glass would be approximately [tex]28^{\circ}[/tex].
[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).
*photo attached* The diameters of the main rotor and tail rotor of a single-engine helicopter are 7.67 m and 1.01 m, respectively. The respective rotational speeds are 444 rev/min and 4,130 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 178.3 m/s and tail rotor 218.4 m/s.
(b) The speed of the main rotor is 0.52 speed of sound, and the speed of the tail rotor is 0.64 speed of sound.
Linear speed of main motor and tail rotorv = ωr
where;
ω is the angular speed (rad/s)r is radius (m)v(main rotor) = (444 rev/min x 2π rad x 1 min/60s) x (0.5 x 7.67 m)
v(main rotor) = 178.3 m/s
v(tail rotor) = (4,130 rev/min x 2π rad x 1 min/60s) x (0.5 x 1.01 m)
v(tail rotor) = 218.4 m/s
Speed of the rotors with respect to speed of sound% speed (main motor) = 178.3/343 = 0.52 = 52 %
% speed (tail motor) = 218.4/343 = 0.64 = 64 %
Thus, the speed of the main rotor is 0.52 speed of sound, and the speed of the tail rotor is 0.64 speed of sound.
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A mechanic changing the spark plugs in a car notes that the instruction manual calls for a torque with a magnitude of
39 N · m.
If the mechanic grasps the wrench as shown in the figure below, determine the magnitude (in N) of the force she must exert on the wrench.
______N
The magnitude (in N) of the force she must exert on the wrench is 150.1 N.
Force exerted by the wrench
The force exerted by the wrench is calculated using torque formula as follows;
torque, τ = F x r x sinθ
where;
F is the applied forcer is the perpendicular distance if force appliedF = τ /(r sinθ)
F = (39) / (0.3 sin 60)
F = 150.1 N
Thus, the magnitude (in N) of the force she must exert on the wrench is 150.1 N.
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A piston–cylinder device with a set of stops initially contains 0.6 kg of steam at 1.0 MPa and
400°C. The location of the stops corresponds to 40 percent of the initial volume. Now the
steam is cooled. Determine the compression work if the final state is
a) 1.0 MPa and 250°C and
b) 500 kPa.
c) Also determine the temperature at the final state in part b.
A piston-work cylinder device with a set of stops initially contains 0.6 kg of steam at 1.0 MPa and 400°C. The location of the stops corresponds to 40 percent of the initial volume. Now the steam is cooled. compression work is 44.32 KJ,
The amount of labor put into the piston to cause its reciprocating motion is known as the piston work. It is calculated by multiplying the piston's displacement by the net force.
An expanding gas cylinder's force output is transferred by pistons to the crankshaft, which then drives the flywheel's rotation. A reciprocating engine is a device like this.
Piston work is the effort made by the piston to make its reciprocating motion. The piston's displacement is calculated by multiplying it by the net force.
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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.
Peter has started a workout where he jumps rope, does push-ups, kicks a ball through a pattern of cones, does sit-ups, and then repeats the cycle. This is an example of:
Specificity training
Overload training
Circuit training
Interval training
The workout of Peter that involves jumping rope, doing push-ups, kicking a ball through a pattern of cones, doing sit-ups, and then repeats the cycle is an example of circuit training.
What is circuit training?A workout is any activity that requires much physical or mental effort, or produces strain.
There are several approaches that can be employed to achieve workout procedures and they are as follows:
Circuit trainingInterval trainingSpecificity trainingCircuit training is a combination of six or more exercises performed with short recovery periods between them for either a set number of repetitions or a prescribed amount of time.
On the other hand, interval training is similar to circuit training but differs in the sense that a period of rest is allowed in between the exercises.
Therefore, the workout of Peter that involves jumping rope, doing push-ups, kicking a ball through a pattern of cones, doing sit-ups, and then repeats the cycle is an example of circuit training.
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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.
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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A 60 year old person has a threshold of hearing of 79.0 dB for a sound with frequency f=10,000 Hz. By what factor must the intensity of a sound wave of that frequency, audible to a typical young adult, (sound level=43.0 dB) be increased so that it is heard by the older person.
43.DB(decibels)
⇉ Unit conversion
4.3(bels)
4.951NP
4.3(decades)
14.28(octaves)
Interpretations
⇉ Power quantity level
Basic unit dimensions
⇉ ( logarithmic quantity ).
They provided the intensity in decibels for the problem, but they are unsure by what factor to increase it (I) to make the sound loud enough for the elderly person to hear.
Neglect f entirely.
The following equation must be used to convert decibels (dB) to I:
I=(10^(dB/10))*10^-12
Divide the elder person's dB by the younger person's dB after doing this for each dB.
1.8372093023.
Decibels are used to measure sound (dB). A motorcycle engine operating is roughly 95 dB louder than regular conversation, which is around 60 dB louder than a whisper. Your hearing may begin to be harmed if exposed to noise over 70 dB for an extended period of time. Your ears can suffer instant damage from loud noise above 120 dB.
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light has a frequency of about 6 x1014 Hz.
Answer:
The visible light frequency is 400 THz to 700 THz, approximately. A THz is a Terahertz, which is a unit of frequency equal to one trillion Hertz.
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.
Hope this helps!
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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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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