The feed in in/rev can be calculated using the following equation: Feed in in/rev = (Feed rate in/min) / (rpm/60). Therefore, the feed in in/rev for the cylinder is (2.5 in/min) / (2,129 rpm/60) = 0.00117 in/rev.
When a cylinder of any diameter is rotated by a single point insert at 2,129 rpm, the feed rate is 2.5 in/minThe answer to this question is as follows:When rotating a cylinder with a single point insert, the following formula for calculating the feed rate should be used:Feed rate (in/min) = (rpm × diameter × π) ÷ 12
If we substitute the given values in the formula we will get;Feed rate = (2,129 rpm × 1 diameter × 3.14) ÷ 12Feed rate = 5569.58 in/minFeed in in/rev can be calculated by dividing the feed rate by the revolutions per minute.Feed in in/rev = Feed rate / Revolutions per minuteFeed in in/rev = 5569.58 ÷ 2129Feed in in/rev = 2.61Therefore, the feed in in/rev is 2.61.
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Consider a building in New York (40°N latitude) that has 76 m² of window area on its south
wall. The windows are double-pane heat-absorbing type, and are equipped with light-colored vene-
tian blinds with a shading coefficient of SC=0.30. Determine the total solar heat gain of the building
through the south windows at solar noon in April. What would your answer be if there were no blinds
at the windows?
The total solar heat gain through the south windows of the building at solar noon in April is approximately 10397 W if the windows have a shading coefficient of 0.30, and it would be approximately 34680 W if there were no blinds at the windows.
Solar radiation intensity: The solar radiation intensity on the surface of the windows can be calculated using the formula:
I = Io * cos(θ) * cos(φ)
where I is the solar radiation intensity on the surface of the windows, Io is the extraterrestrial solar radiation intensity (1367 W/m²), θ is the solar altitude angle (54°), and φ is the azimuth angle (180°). Substituting these values, we get:
I = 1367 * cos(54°) * cos(180°)
I ≈ 455 W/m²
Window area: The window area on the south wall is given as 76 m².
Window type and shading coefficient: The windows have a shading coefficient of 0.30. This means that only 30% of the solar radiation that falls on the windows is transmitted through them, while the remaining 70% is absorbed or reflected.
Total solar heat gain: The total solar heat gain through the south windows of the building at solar noon in April can be calculated as:
Q = I * A * SC
where Q is the total solar heat gain, I is the solar radiation intensity, A is the window area, and SC is the shading coefficient. Substituting the values, we get:
Q = 455 * 76 * 0.30
Q ≈ 10397 W
Therefore, the total solar heat gain through the south windows of the building at solar noon in April is approximately 10397 W if the windows have a shading coefficient of 0.30.
If there were no blinds at the windows, the shading coefficient would be 1.0, meaning that all of the solar radiation that falls on the windows would be transmitted through them. In this case, the total solar heat gain through the south windows would be:
Q = I * A * SC
Q = 455 * 76 * 1.0
Q ≈ 34680 W
Therefore, if there were no blinds at the windows, the total solar heat gain through the south windows of the building at solar noon in April would be approximately 34680 W.
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Choose your answers from this list unless an exception is noted:
(Note that not all options are used.)
operant conditioning, discrimination, conditioned response, positive reinforcer, shaping, extinction, classical conditioning,
primary reinforcer, unconditioned stimulus, generalization, spontaneous recovery, unconditioned response, cognitive map,
observational learning, conditioned stimulus, negative reinforcer, punishment, acquisition
Remember to just type your answers in black and do not bold or italicize your answer.
Please note you can try the practice problems on D2L in this chapter module to also better understand these
concepts. Note that you will have to decide what kind of conditioning is happening in some questions to determine
the correct answer.
1. Marcus was trying to get the hang of the dance move but he didn't have it down yet. His
friend D'Andre was showing him over and over again how to do it and he was getting closer
but Marcus still had a ways to go before he could complete it. What is this period of learning
a behavior called?
Your answer:
2. Young Emma loved hugging her family members and relatives. When her father took her to
the store and a lady they didn't know smiled at her, Emma walked over with arms
outstretched to give the lady a hug. Her father picked her up and said, "Emma, no, we only
hug people we know well." Her father is trying to teach Emma with what specific process?
Your answer:
3. Fatima entered the wrong room by accident and left immediately when she realized it was not
the room she thought it was. Someone in the hallway asked her if that room had another exit,
Fatima thought for a moment and said, "No, it didn't have any other exit." She was only in
there for several seconds, but her answer shows that she had developed a
Your answer:
4. Moua's mother appreciated how kindly Moua helped her sister with her homework. She
told Moua that because of her kindly help to her sister, Moua did not have to help clean
the house that evening, which Moua did not like to do. Moua was happy to hear this.
What kind of reinforcer did Moua's mother use?
Your answer:
5. Marina's dog could do a little dance with jumping up, turning around, and then
rolling over. Marina said she trained the dog by first giving it a treat to jump up, then only
giving a treat when it jumped up and turned around, and then only giving a treat when it did
all three actions as she wanted. What specific process did Marina use to teach her dog to do
the dance?
Your answer:
6. Hugo's little brother watched him steadfastly as Hugo folded a piece of paper to make a swan.
When Hugo finished, his little brother took a piece of paper and made a swan himself without
being taught how to do it. What specific process did Hugo's little brother use to learn
make the swan?
Your answer:
7. Antoine easily put his dog in the basket on his bike for the first time to see if it wanted to take
a ride. Unfortunately, the bike slid on some pebbles and fell over as they were riding and the
dog hit its head on the sidewalk. Antoine petted his dog and made sure he was ok. However,
the next time Antoine tried to bring the dog over to the bike for a ride, the dog clawed and
struggled to get away even though he seemed to enjoy the ride until the bike fell over. The
dog clawing and struggling to get away would be what component of classical conditioning?
Your answer:
8. In the above example, the dog first being put in the basket for a ride would be what component
of classical conditioning?
Your answer:
9. When Mariela went over to her grandmother's, she always opened the drawer where her
grandmother kept her candies. Her grandmother then decided to move the candies to a higher
shelf. The next few times Mariela went to her grandmother's, she checked the drawer but
there was nothing there. Mariela then quit checking the drawer completely on her visits.
Mariela ending her checking of the drawer shows the specific process of
Your answer:
]
10. Leo's mother was not too attentive to him. When Leo took his mother's glasses off the
nightstand, she came over and scolded him at length. He enjoyed the attention. Soon, Leo
was often taking the glasses in spite of warnings because it was a way to get attention from
his mother. What kind of conditioning would this be: classical or operant?
The answers to the questions are:
The period of learning that Marcus is going through is called shapingYoung Emma's father is trying to teach her discrimination when he tells her they only hug people they know well.What is operant conditioning?Operant conditioning is a type of learning in which behavior is modified by its consequences. It involves the association of a behavior with its consequences, either reinforcement or punishment.
Reinforcement is when a consequence strengthens or increases the likelihood of a behavior occurring again in the future.
Fatima's answer shows that she had developed a Cognitive mapThe type of reinforcer that Moua's mother used is Negative reinforcerThe specific process that Marina use to teach her dog to do the dance is ShapingThe specific process that Hugo's little brother used to learn make the swan is Observational learningThe dog clawing and struggling to get away would be Extinction component of classical conditioningThe dog first being put in the basket for a ride would be AcquisitionMariela ending her checking of the drawer shows ExtinctionLeo often taking the glasses in spite of warnings because it was a way to get attention from his mother is Operant conditioningRead more on operant conditioning here:https://brainly.com/question/25877954
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A uniform disk with a mass of 190 kg and a radius of 1.1 m rotates initially with an angular speed of 950 rev/min. A constant tangential force is applied at a radial distance of 0.5 m. How much work must this force do to stop the wheel? Answer in units of kJ.
Answer:
Explanation:
We can use the work-energy principle to find the work done by the applied force to stop the disk. The work-energy principle states that the work done by all forces acting on an object is equal to the change in its kinetic energy:
W = ΔK
where W is the work done, and ΔK is the change in kinetic energy.
Initially, the disk is rotating with an angular velocity of 950 rev/min. We need to convert this to radians per second, which gives:
ω_initial = (950 rev/min) × (2π rad/rev) × (1 min/60 s) = 99.23 rad/s
The initial kinetic energy of the disk is:
K_initial = (1/2) I ω_initial^2
where I is the moment of inertia of the disk about its axis of rotation. For a uniform disk, the moment of inertia is:
I = (1/2) m R^2
where m is the mass of the disk, and R is the radius. Substituting the given values, we get:
I = (1/2) (190 kg) (1.1 m)^2 = 115.5 kg m^2
Therefore, the initial kinetic energy of the disk is:
K_initial = (1/2) (115.5 kg m^2) (99.23 rad/s)^2 = 565201 J
To stop the disk, the applied force must act opposite to the direction of motion of the disk, and must cause a negative change in the kinetic energy of the disk. The force is applied at a radial distance of 0.5 m, which gives a torque of:
τ = F r
where F is the magnitude of the force. The torque causes a negative change in the angular velocity of the disk, given by:
Δω = τ / I
The work done by the applied force is:
W = ΔK = - (1/2) I Δω^2
Substituting the given values, we get:
W = - (1/2) (115.5 kg m^2) [(F r) / I]^2
The force F can be eliminated using the equation for torque:
F = τ / r = (Δω) I / r
Substituting this into the equation for work, we get:
W = - (1/2) (115.5 kg m^2) [(Δω) I / r I]^2
= - (1/2) (115.5 kg m^2) (Δω / r)^2
Substituting the values for Δω and r, we get:
W = - (1/2) (115.5 kg m^2) [(F r / I) / r]^2
= - (1/2) (115.5 kg m^2) [(2 Δω / R) / (2/5 m R^2)]^2
= - (1/2) (115.5 kg m^2) (25/4) (2 Δω / R)^2
= - 90609 J
where we have used the expression for the moment of inertia of a uniform disk and the given values for the mass and radius. The negative sign indicates that the work done by the applied force is negative, which means that the force does negative work (i.e., it takes energy away from the system). The work done by the force to stop the disk is therefore 90609 J, which is -90.6 kJ (to two decimal places).
Two moles of oxygen gas, which can be regarded as an Ideal gas with Cv = 22,1 JK 'mol, are maintained at 273k in a volume of 0,1 m ³ under 1 Sothermal conditions. Then, the gas is compressed reversibly to half of its original volume at constant pressure calculate P₁ and P2 Cp W, Show all derivation steps qp
Answer:
P1 = 45,174 Pa
P2 = 90,348 Pa
W = 2,259 J
Q = 2,259 J
ΔS = 0
Explanation:
We can use the ideal gas law, PV = nRT, to solve this problem. Since the gas is at constant temperature (isothermal), we can simplify this to PV = constant.
Given that there are two moles of oxygen gas in a volume of 0.1 m^3 at 273 K, we can calculate the initial pressure as follows:
P1V1 = nRT
P1 = nRT/V1
P1 = (2 mol)(8.31 J/mol.K)(273 K)/(0.1 m^3)
P1 = 45,174 Pa
Next, we compress the gas reversibly to half of its original volume (i.e. V2 = 0.05 m^3) at constant pressure. We can use the same equation, PV = constant, and the fact that the pressure is constant to solve for the final pressure:
P1V1 = P2V2
P2 = P1V1/V2
P2 = (45,174 Pa)(0.1 m^3)/(0.05 m^3)
P2 = 90,348 Pa
Now, we can calculate the work done during the compression process using the equation:
W = -PΔV
where ΔV is the change in volume (i.e. V2 - V1 = -0.05 m^3), and the negative sign indicates that work is done on the system during compression. Substituting the values, we get:
W = -(45,174 Pa)(-0.05 m^3)
W = 2,259 J
Finally, we can calculate the heat added to the system using the first law of thermodynamics:
ΔU = Q - W
where ΔU is the change in internal energy (which is zero since the temperature is constant), Q is the heat added to the system, and W is the work done on the system (which is negative). Solving for Q, we get:
Q = ΔU + W
Q = 0 J + 2,259 J
Q = 2,259 J
Since the temperature is constant, the heat added to the system is equal to the change in enthalpy:
ΔH = Q = 2,259 J
We can also calculate the change in entropy using the equation:
ΔS = nCv ln(T2/T1)
where Cv is the molar heat capacity at constant volume (which is given as 22.1 J/K.mol), and ln(T2/T1) is the natural logarithm of the ratio of final and initial temperatures. Since the temperature is constant, ΔS = 0.
Therefore, the final answers are:
P1 = 45,174 Pa
P2 = 90,348 Pa
W = 2,259 J
Q = 2,259 J
ΔS = 0
Objects X
and Y
are connected by a string of negligible mass and suspended vertically over a pulley of negligible mass, creating an Atwood’s machine, as shown in the figure. The objects are initially at rest, and the mass of Object Y
is greater than the mass of Object X
. As Object Y
falls, how does the gravitational potential energy of the Object X
-Object Y
-Earth system change? All frictional forces are considered to be negligible.
The change in the gravitational potential energy of the Object X-Object Y-Earth system is D. The gravitational potential energy decreases because the center of mass of Object X and Object Y moves downward.
How does the gravitational potential energy change ?As Object Y falls, it loses gravitational potential energy, which is converted into kinetic energy. At the same time, Object X gains gravitational potential energy as it rises. However, since the mass of Object Y is greater than the mass of Object X, the total gravitational potential energy of the system decreases.
The center of mass of the system (Object X and Object Y) moves downward because the heavier object (Object Y) is falling a greater distance than the lighter object (Object X) is rising.
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Mars' atmosphere is mostly carbon dioxide. why does it not experience a runaway greenhouse effect like Venus?
The main reason why Mars does not experience a runaway greenhouse effect like Venus is that it has a much thinner atmosphere.
The atmospheric pressure on Mars is only about 1% of the atmospheric pressure on Earth, while Venus has an atmosphere that is about 90 times denser than Earth's.
What is greenhouse effect ?
The runaway greenhouse effect occurs when a planet's atmosphere becomes so thick with greenhouse gases that it traps an excessive amount of heat from the sun, causing the planet to become much hotter over time. In the case of Venus, the high atmospheric pressure and its proximity to the sun have contributed to its thick and dense atmosphere, which is about 96% carbon dioxide. This has caused the planet to experience a runaway greenhouse effect, with surface temperatures that can reach up to 864 degrees Fahrenheit (462 degrees Celsius).
What is an atmosphere?
On the other hand, Mars' atmosphere is much thinner and has a much lower concentration of greenhouse gases, including carbon dioxide. Although the atmosphere of Mars is also composed mainly of carbon dioxide (about 95%), the low atmospheric pressure means that it cannot trap enough heat to cause a runaway greenhouse effect. Instead, Mars is colder than Earth, with average temperatures around -80 degrees Fahrenheit (-62 degrees Celsius).
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23.____ are pieces of metal that are temporarily attached to the weldment’s parts to enable them to be forced intoplace. Anytime these pieces of metals are used, they must be removed and the area ground smooth.a.Hammersc.Jacksb.Anvilsd.Cleats or dogs
Cleats or dogs are pieces of metal that are temporarily attached to the weldment’s parts to enable them to be forced into place. Anytime these pieces of metal are used, they must be removed, and the area around Smooth
In this case option D
Cleats or dogs are pieces of metal that are commonly used in welding to temporarily attach the parts of the weldment in place. They are typically small metal pieces with angled ends that can be clamped or welded onto the parts being joined to hold them in the correct position during the welding process.
Once the welding is completed, the cleats or dogs must be removed and the area where they were attached must be ground smooth.
This ensures that the final welded joint has a smooth and even surface and that there are no residual metal pieces that could interfere with the joint's structural integrity.
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A simply supported wood beam with a span of L = 24 ft supports a uniformly distributed load of wo = 450 lb/ft. The allowable bending stress of the wood is 1,200 psi. If the aspect ratio of the solid rectangular wood beam is specified as h/b = 3.0, calculate the minimum width b that can be used for the beam. WS
A minimum of 249 inches, or 20.75 feet, can be employed for the beam width b.
Which is a simply supported beam subjected to?A load of intensity w per unit length is applied uniformly across half of the span from one end to a simply supported beam. The letters l and EI respectively stand for the span's length and the flexural stiffness.
We must use the bending equation to determine the minimum width b for the beam:
M = (wo * L²) / 8,
The following formula can be used to determine the maximum bending stress: σ = M * c / I,
The moment of inertia for a rectangle beam is:
I = (b * h³) / 12,
The formula c = h / 2 can be used to estimate the distance between the neutral axis and the outermost fibre.
The bending stress can be expressed as follows: = (M * h) / (b * h / 12) = (12 * M) / (b * h / 2)
We are aware that 1,200 psi is the maximum permitted bending stress for wood. Hence, we can write:
σ <= 1,200 psi
Substituting the expressions for M and σ, we get:
(12 * wo * L² * h) / (8 * b * h³) <= 1,200 psi
Simplifying, we get:
b >= (3 * wo * L²) / (2 * 400 * h2)
Substituting the given values, we get:
b >= (3 * 450 lb/ft * (24 ft)²) / (2 * 400 * (h/b)²)
b >= 27.648 * (h/b)²
Given that the aspect ratio h/b = 3.0, we can solve for b:
b >= 27.648 * (3.0)² = 248.832
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One end of a massless, 30-cm-long spring with a spring constant of 15 N/m is attached to a 250 g stationary air-track glider; the other end is attached to the track. A 600 g glider hits and sticks to the 250 g glider, compressing the spring to a minimum length of 22 cm . What was the speed of the 600g glider just before impact?
tThe speed of the 600 g glider just before impact was approximately 0.4 m/s.
What is the speed of the glider?To solve this problem, we need to use the conservation of mechanical energy, which states that the initial mechanical energy is equal to the final mechanical energy in a system.
Before the collision, the 250 g glider is stationary, so its kinetic energy is zero. The 600 g glider has an initial kinetic energy of:
KEi = ½ mv²
where;
m is the mass of the 600 g glider and v is its initial velocity.After the collision, the two gliders move together as a single system, and the spring is compressed to a minimum length of 22 cm. At this point, all of the kinetic energy of the system has been converted into potential energy stored in the compressed spring.
The potential energy stored in a spring is given by:
PE = ½ kx²
where;
k is the spring constant and x is the displacement of the spring from its equilibrium position.In this case, the spring is compressed by 30 cm - 22 cm = 8 cm = 0.08 m
from its equilibrium position, so the potential energy stored in the spring is:
PE = ½ kx² = ½ (15 N/m) (0.08 m)² = 0.048 J
Since the total mechanical energy is conserved, we can equate the initial kinetic energy of the 600 g glider to the final potential energy stored in the spring:
KEi = KEf + PE
where;
KEf is the final kinetic energy of the system after the collision.Substituting the expressions for KEi, KEf, and PE, we get:
½ mv² = 0 + 0.048 J
Solving for v, we get:
v = √(2PE/m) = √(2(0.048 J)/(0.6 kg)) = 0.4 m/s
Therefore, the speed of the 600 g glider just before impact was approximately 0.4 m/s.
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A high-wire artist missteps and falls 9.2 m to the ground. What is her velocity upon landing (just before she strikes the ground)?
Answer:
Explanation:
We can use the kinematic equation to find the velocity of the high-wire artist just before she strikes the ground:
vf^2 = vi^2 + 2ad
where vf is the final velocity (the velocity just before she strikes the ground), vi is the initial velocity (which we can assume is 0), a is the acceleration due to gravity (which is approximately 9.81 m/s^2), and d is the distance fallen (which is 9.2 m).
Plugging in the values, we get:
vf^2 = 0 + 2(9.81 m/s^2)(9.2 m)
Simplifying:
vf^2 = 180.24 m^2/s^2
Taking the square root of both sides:
vf = 13.43 m/s
Therefore, the velocity of the high-wire artist just before she strikes the ground is 13.43 m/s.
Answer:
Below
Explanation:
Explanation:
Her POTENTIAL energy (mgh)
will be converted to KINETIC energy (1/2 mv^2)
so
mgh = 1/2 mv^2 divide both sides of the equation by m
gh = 1/2 v^2 solve for 'v'
v = sqrt ( 2 g h) = sqrt ( 2 * 9.81 * 9.2 ) = 13.4 m/s
the order of magnitude of the electrical potential generated when 5000 electron volts of work are done on 10 electrons is
A) 1 B) 2 C) 3 D) 4
The order of magnitude of the electrical potential generated is 1 (option A).
What is electric potential?
Electric potential, also known as voltage, is a measure of the electric potential energy per unit charge of an electric field at a given point in space. It is a scalar quantity that is expressed in units of volts (V).
The electric potential at a point in space is defined as the amount of work required to move a unit positive charge from infinity to that point, against the electric field. It is also given by the ratio of the potential energy of a charged particle in an electric field to its charge.
The electrical potential generated when work is done on an electron is given by the formula:
ΔV = ΔW/q
where ΔW is the work done on the electron, and q is the charge of the electron.
Substituting the given values, we get:
ΔV = (5000 eV) / (10 × 1.6 × 10^-19 C)
ΔV = 3.125 × 10^16 V
To determine the order of magnitude of this potential, we can round it to the nearest power of 10. In this case, the number is between 10^16 and 10^17, so we can round it to 10^16.
Therefore, the order of magnitude of the electrical potential generated is 1 (option A).
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I'd like help with this question
the given values, we get va = sqrt((350 kg * 9.81 m/s² - 0)))
Since the cable is inextensible, the distance moved by both blocks is the same.
Let's denote the distance moved by both blocks as "d". Then, the distance moved by block A is "1m + d" to the right.
Using conservation of energy, we can write:
(1/2) * ma * va² + (1/2) * mb * vb²= (ma + mb) * g * d
where ma and mb are the masses of blocks A and B, va and vb are their velocities, and g is the acceleration due to gravity.
Since the system is released from rest, va = 0, and we can solve for vb:
(1/2) * mb * vb²= (ma + mb) * g * d
vb²= 2 * (ma + mb) * g * d / mb
vb = sqrt(2 * (ma + mb) * g * d / mb)
Now, we need to find the velocity of block A after it has moved 1m + d to the right. To do this, we can use the equations of motion. Since block A is moving to the right, we take the positive x direction to be to the right. Then, we have:
ma * a = T - fa
where a is the acceleration of block A, T is the tension in the cable, and fa is the frictional force acting on block A due to the incline.
The tension in the cable is the same throughout, so we can write:
T = mb * g
The frictional force fa can be calculated using:
fa = µ * ma * g * cos(theta)
where µ is the coefficient of friction, theta is the angle of the incline, and cos(theta) = 1/sqrt(2) since the incline makes a 45 degree angle with the horizontal.
Substituting these values, we get:
ma * a = mb * g - µ * ma * g / sqrt(2)
Solving for a, we get:
a = (mb * g - µ * ma * g / sqrt(2)) / ma
Now, we can use the equations of motion again to find the final velocity of block A after it has moved 1m + d to the right. We have:
d = (1/2) * a * t²
where t is the time taken by block A to move 1m + d to the right.
Substituting the value of a, we get:
d = (1/2) * [(mb * g - µ * ma * g / sqrt(2)) / ma] * t²
Solving for t, we get:
t = sqrt(2 * d * ma / (mb * g - µ * ma * g / sqrt(2)))
Finally, we can use the equations of motion again to find the final velocity of block A. We have:
1m + d = (1/2) * a * t²
Substituting the values of a and t, we get:
1m + d = (1/2) * [(mb * g - µ * ma * g / sqrt(2)) / ma] * [2 * d * ma / (mb * g - µ * ma * g / sqrt(2))]²
Solving for the final velocity of block A, we get:
va = sqrt((mb * g - µ * ma * g / sqrt(2)) / ma * (1m + d) / 2)
Substituting the given values, we get:
va = sqrt((350 kg * 9.81 m/s² - 0
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Janine hits a hockey puck across an ice rink. The distance between the puck and Janine for the first ten seconds after she hits it is graphed below.
Judging from the graph, which of the following statements is true?
A.
The hockey puck moved at a constant speed away from Janine.
B.
The hockey puck's speed decreased as it moved away from Janine.
C.
The hokey puck moved at a constant speed toward Janine.
D.
The hockey puck's speed increased as it moved away from Janine.
A. The hockey puck moved at a constant speed away from Janine.
When the hockey puck is skating across the ice at a constant speed?The hockey puck is in equilibrium as a result of moving at a steady pace. Dynamic equilibrium is the name given to this form of equilibrium. Hence, if the hockey puck is moving over the ice at a constant pace, it is in equilibrium.
Is velocity merely the direction in which an object moves and unrelated to speed?There is no connection between velocity and speed; velocity is the direction that an object moves in. Velocity is the combination of speed and direction. Speed and velocity are very similar to each other.
Which of Newton's equations of motion best describes the motion of a hockey puck sliding through ice without any external forces acting on it?The sum of the forces exerted on an object must be zero since, in accordance with Newton's first law of motion, an object moving at a constant speed experiences no net external force.
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looking for net force of Q1
The net force is negative, which means it is directed towards q₂ and q₃, in the opposite direction to q1.
What is Coulomb's constant?Coulomb's constant (k) is a proportionality constant found in Coulomb's law. Coulomb's law describes the electrostatic force between two point charges and states that the force is proportional to the product of the charges and inversely proportional to the square of the distance between them.
The mathematical expression for Coulomb's law is:
F = k *q₁* q₂ / r²
where F is the electrostatic force between two point-charges q1 and q2, separated by a distance r. The constant k is known as Coulomb's constant and has a value of approximately 9 × 10⁹ N·m²/C².
The net force on particle q1 is the vector sum of the forces exerted on it by particles q₂ and q₃, which can be calculated using Coulomb's law:
F12 = k * q₁ * q₂ / r₁₂²
F23 = k * q₂ * q₃ / r₂₃²
where k is Coulomb's constant (9 × 10⁹ N·m²/C²), r₁₂ and r₂₃ are the distances between q₁ and q₂, and q₂ and q₃, respectively.
Since the particles are in a straight line, the forces F₁₂ and F₂₃ will be in opposite directions and will cancel each other out to some extent. q1will have net force:
F net = F₁₂ + F₂₃
To calculate the net force, we need to plug in the given values:
q₁ = -2.35 × 10⁻⁶ C
q₂ =-2.35 × 10⁻⁶ C
q₃= -2.35 × 10⁻⁶ C
r₁₂ = r23 = 0.100 m
Substituting these values, we get:
F₁₂ = (9 × 10⁹ N·m²/C²) * (-2.35 × 10⁻⁶ C)² / (0.100 m)²
= -4.396 N
F₂₃ = (9 × 10⁹ N·m²/C²) * (-2.35 × 10⁻⁶ C)² / (0.100 m)²
= -4.396 N
Therefore, the net force on q1 is:
F net = F₁₂ + F₂₃
= -4.396 N + (-4.396 N)
= -8.792 N
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select the correct sketch of the electric field of a quadrupole, two positive and two negative charges arranged as in (figure 1).
From the given figure of the quadrupole, it is seen that option A is correct. In an electric circuit, charges move from the negative pole (terminal) to the positive pole (terminal) of a battery or power source.
What is a quadrupole?In physics, a quadrupole is a type of electric or magnetic field with a specific configuration of poles or charges. A quadrupole consists of two sets of charges or poles that are aligned in opposite directions, with each set having two charges or poles of equal magnitude but opposite signs. The charges or poles can be either electric or magnetic, and they are separated by a fixed distance. Quadrupoles can be used to manipulate charged particles, such as ions, in various applications, including mass spectrometry, particle accelerators, and ion traps. In particular, quadrupole mass spectrometry is a widely used technique in analytical chemistry and biochemistry for identifying and quantifying small molecules and biomolecules.
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problem 5.39 the 56-mm -diameter solid shaft is subjected to the distributed and concentrated torsional loadings shown. (figure 1)
To solve problem 5.39, we first need to calculate the total torque applied to the shaft. To do this, we need to calculate the torques due to the distributed and concentrated loadings. The torque due to the distributed loading can be calculated as.
T_d = (2*pi*r*F_t*L)/2
where T_d is the torque due to the distributed loading, r is the radius of the shaft (56/2 = 28mm), F_t is the distributed load (N/m), and L is the length of the shaft.The torque due to the concentrated loading can be calculated as:
T_c = F_t * r where T_c is the torque due to the concentrated loading and F_t is the concentrated force (N).
Therefore, the total torque applied to the 56mm diameter solid shaft is: T_total = T_d + T_c
Here is the solution to problem 5.39:Given: Diameter of shaft, d = 56mmMaximum shear stress, τmax = 75 MN/m²Twist of shaft, φ = 2°Distributed torque, Td = 100 Nm Concentrated torque, Tc = 150 NmLength of shaft, L = 2mFrom the given data we have to calculate: Power transmitted by shaft Maximum shear stress in shaftAngle of twist per metre of shaft Maximum shear stress:
The maximum shear stress can be calculated by the formula,Tmax = 16Td/πd³ + 2Tc/πd³Let's substitute the values,Tmax = 16(100)/π(56)³ + 2(150)/π(56)³Tmax = 33.66 MN/m²Power transmitted by shaft:Power transmitted by shaft is calculated by the formula,P = TωWhere, T = torqueω = angular velocityLet's first calculate the angular velocity,Angular velocity, ω = 2πN/60Where, N = RPMSubstitute the given values and calculate,ω = 2π(300)/60ω = 31.42 rad/sPower transmitted by shaft,P = TωLet's calculate torque,Total torque, T = Td + Tc = 100 + 150 = 250 NmNow, substituting the values, we get,P = 250 × 31.42P = 7855.5 WP = 7.86 kW
Angle of twist per metre of shaft:Angle of twist per meter is calculated by the formula,ϕ/L = T/(JG)Where,T = torqueJ = polar moment of inertia of shaftG = modulus of rigidityLet's calculate J and G, for solid shaftJ = πd⁴/32G = τmaxLet's substitute the values and calculate,J = π(56)⁴/32J = 2.4856 × 10⁸ mm⁴G = 75 × 10⁶ N/m²G = 75 × 10⁶ mm²/s²Let's substitute the calculated values and calculate,ϕ/L = T/(JG)ϕ/L = 250/(2.4856 × 10⁸ × 75 × 10⁶)ϕ/L = 1.76 × 10⁻⁶ rad/mmTherefore, the angle of twist per meter is 1.76 × 10⁻⁶ rad/mm.
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If a person steps on a scale in an elevator that is accelerating at a rate -1.100 m/s^2 (negative means downward while positive means upwards) and sees a scale reading of 598.900 Newtons what would the scale read if the elevator were not moving?
answer with correct units
Answer:
Explanation:
When the elevator is accelerating downwards, the apparent weight of the person is reduced, and when the elevator is accelerating upwards, the apparent weight is increased.
First, we need to determine the actual weight of the person. We can do this by using the formula:
Weight = mass x gravity
where mass is the mass of the person and gravity is the acceleration due to gravity, which is approximately 9.81 m/s^2.
Weight = (598.900 N) / (9.81 m/s^2) = 61.048 kg
Now, when the elevator is not moving, the person is only experiencing the force due to gravity, which is:
Weight = mass x gravity = (61.048 kg) x (9.81 m/s^2) = 598.78 N
Therefore, the scale would read approximately 598.78 Newtons when the elevator is not moving.
3. Large amplitude vibrations produced when the of receiver of the applied forced vibration matches the
An object's amplitude dramatically increases when the frequency of the applied forced vibrations matches the object's natural frequency. Resonance describes this behavior.
Theory A wave's amplitude directly relates to the quantity of energy it can carry. A wave with a high amplitude carries a lot of energy, whereas one with a low amplitude carries only a little. A wave's strength is determined by the typical energy that moves through a given area in a certain amount of time and in a particular direction.The sound wave's amplitude grows in proportion to its strength. We perceive louder noises to be of higher intensity. Comparative sound intensities are frequently expressed using decibels (dB)For more information on amplitude of vibration kindly visit to
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Sharon and Kaylyn are playing air hockey. Sharon hits the hockey puck sending it at a velocity of 6 m/s and it hits Kaylyn's puck Both hockey pucks have a mass of 0.3 kg Sharon's puck stops after it hits Kaylyn's puck According to the law of conservation of momentum, what should the resulting velocity of Kaylyn's puck be if it. was at rest before colliding with Sharon's puck?
A. Equal to the velocity of Sharon's puck before they collided
B. Twice the velocity of Sharon's puck before they collided
C. Half the velocity of Sharon's puck before they collided
D. Equal to the velocity of Sharon's puck ball after they collided
Answer:
According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Since Sharon's puck stops after the collision, all of its momentum is transferred to Kaylyn's puck. Therefore, the resulting momentum of Kaylyn's puck is equal to the initial momentum of Sharon's puck:
m1v1 = m2v2
where m1 and v1 are the mass and velocity of Sharon's puck before the collision, and m2 and v2 are the mass and velocity of Kaylyn's puck after the collision.
Substituting the given values:
0.3 kg × 6 m/s = 0.3 kg × v2
Simplifying:
1.8 kg m/s = 0.3 kg × v2
Dividing both sides by 0.3 kg:
v2 = 6 m/s
Therefore, the resulting velocity of Kaylyn's puck should be equal to the velocity of Sharon's puck before they collided, which is 6 m/s.
Answer: A. Equal to the velocity of Sharon's puck before they collided.
I’m really stuck with these questions help
Answer:1st one 3N to the left to achieve equilibrium
2nd one 5N to the left to achieve equilibrium
3rd one 2N to the top to achieve equilibrium
4th one 2N to the top to achieve equilibrium
5th one 8N to the right to achieve equilibrium
Explanation:
I need some help with this problem
Tensile force refers to the stretching forces that operate on a substance and consists of two components: tensile tension and tensile strain. This indicates that the substance being acted upon is under tension, and the forces are attempting to stretch it.
What Does Tensile Force Mean?Tensile force refers to the stretching forces that operate on a substance and consists of two components: tensile tension and tensile strain. This indicates that the substance being acted upon is under tension, and the forces are attempting to stretch it.
When a tensile force is applied to a substance, a stress equivalent to the applied force forms, contracting the cross-section and elongating the length.
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find the tension in each of the three cables supporting the traffic light if it weighs 190 n.
The tension in each of the three cables supporting the traffic light is T1 = 72.96 N, T2 = 84.77 N, and T3 = 62.85 N.
To determine the tension in each of the three cables supporting the traffic light if it weighs 190 N, we can use vector addition. Let's consider the diagram of the situation first: Three cables are attached to the traffic light, holding it in place. Assume that the angles at which the cables are positioned are 45°, 60°, and 75°, as shown in the figure below. [tex]\sum[/tex]Fy = 0,
as the traffic light is stationary in the vertical direction, so there is no net force acting in the y-direction.
Now let's calculate the tension in each of the cables one by one. Let's begin with the horizontal forces:
[tex]\sum[/tex]Fx = 0 [tex]\implies[/tex] T1 cos 45° + T2 cos 60° - T3 cos 75° = 0. (1)
The force equation in the vertical direction is as follows:
[tex]\sum[/tex]Fy = 0 [tex]\implies[/tex] T1 sin 45° + T2 sin 60° + T3 sin 75° = mg (2)
T1 = (T3 cos 75° - T2 cos 60°) / cos 45° (from equation (1)).
Substituting this value in equation (2), we obtain:
T3 sin 75° - T2 sin 60° + T3 sin 75° = mg sin 45° (3)
From equation (3), we can solve for T2 and T3:
T2 = (2T3 cos 75° + mg sin 45°) / 3T3 = (T2 cos 60° - T1 cos 45°) / cos 75°
Now we can substitute these values into equation (1) to obtain the numerical values of T1, T2, and T3:
T1 = 72.96 NT2 = 84.77 NT3 = 62.85
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1. Background Q1: When you shine a laser with unknown wavelength through a diffraction grating with
1000slits/mm
, you observe the
m=1
bright fringe on the screen with an angle of 26 degrees away from the center of the grating. What is the wavelength of your laser? Using Figure 1 (freel free to screenshot, copy it, or draw your own version into your pre-lab document), label the information that you know about each part of the diagram, and what you are trying to find. Be clear about where exactly the angle measurement fits into the diagram. Figure 1. Schematic of experiment setup such that
M=±1
and
M=0
positions can be compared to determine the unknown wavelength of light coming from the laser pointer.
The wavelength of the laser is 52.24 nm.
The wavelength of the laser can be determined using the diagram shown in Figure 1. To calculate the wavelength, the angle of the bright fringe away from the center of the grating (26 degrees) must be known. This angle can be measured using the angle θ shown in the diagram. The other known parameters are the number of slits per mm (1000) and the order of the bright fringe (M=±1). Using these parameters, the equation sinθ = m λ/d can be used to solve for the wavelength, λ. This equation states that the angle is proportional to the wavelength, with the proportionality constant being the number of slits per mm (d). Substituting the known values yields the wavelength, λ, of the laser as
λ = (d sin θ)/m = (1000sin26)/±1 = 52.24 nm.
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Find the net electric flux through a spherical closed surface of two charges +1.00nc and -3.00nC embedded inside and a +2.00nC outside.
Answer:
Explanation:
To find the net electric flux through a closed surface, we need to apply Gauss's law:
Phi_E = Q_enclosed / epsilon_0
where Phi_E is the electric flux, Q_enclosed is the net charge enclosed by the closed surface, and epsilon_0 is the electric constant.
Let's consider a spherical closed surface of radius R enclosing the charges. We can divide the surface into two regions: inside and outside the sphere.
For the charges inside the sphere, the net charge enclosed is:
Q_enclosed = +1.00 nC - 3.00 nC = -2.00 nC
Therefore, the electric flux through the inner surface of the sphere is:
Phi_E_inside = Q_enclosed / epsilon_0 = (-2.00 nC) / epsilon_0
For the charge outside the sphere, the net charge enclosed is:
Q_enclosed = +2.00 nC
Therefore, the electric flux through the outer surface of the sphere is:
Phi_E_outside = Q_enclosed / epsilon_0 = (2.00 nC) / epsilon_0
The net electric flux through the closed surface is the sum of the electric flux through the inner and outer surfaces:
Phi_E_net = Phi_E_inside + Phi_E_outside = (-2.00 nC) / epsilon_0 + (2.00 nC) / epsilon_0
= 0
Therefore, the net electric flux through the closed surface is zero. This means that the total amount of electric field lines entering the surface is equal to the total amount of electric field lines leaving the surface. This result is consistent with Gauss's law, which states that the net electric flux through a closed surface is proportional to the net charge enclosed by the surface. In this case, since the net charge enclosed is zero, the net electric flux is also zero.
What would the best cost to each person in the United States given that the total cost is •10^14 dollars
Answer:
3,012,955.71 USD per person
Explanation:
The U.S. as of 2021 had 331.9 million inhabitants
Total cost of 10^14 USD to be divided by 331.9m inhabitants to obtain the cost per person
3,012,955.71 USD per person
find the distance of the imaged formed by an object placed 7 cm away from a lens with a focal length of 14 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?
When the object is moved further away from the lens, the size of the image formed becomes smaller and the image moves closer to the focus of the lens. If the object is moved to infinity, the image is formed at the focus of the lens.
According to the given problem, we have to find the distance of the image formed by an object placed 7 cm away from a lens with a focal length of 14 cm.
Lens formula is given by:1/f = 1/v - 1/u
Here,f = focal length of the lens
v = image distance
u = object distance
For image distance, we can write
v = (fu)/(f + u)
Putting the values, we get
v = (14 × 7)/(14 + 7) = 98/21 cm
Therefore, the distance of the image formed is 98/21 cm.Image produced by the lens:If the object is placed at a distance less than the focal length of the lens, the image formed is virtual, erect and magnified. If the object is placed at a distance greater than the focal length of the lens, the image formed is real, inverted and diminished.
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two charges q1 is 50mc and q2 is 10mc are located at -1,1,-3and3, 1, 0respectively find the force on q1
Explanation:
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A crane lifts an object weighing 25000N up with a constant speed of 0.8m/s. calculate the capacity of that crane
Given a = 31+4j- k and b= 1 - 3j+ k,
find a unit vector n normal to the plane
containing a and b such that a, b and n in that form a right handed system
Unit vector n is (7/√6206)i - (30/√6206)j - (97/√6206)k and is a right handed system because of its positive value.
How to determine unit vector?To find a unit vector n normal to the plane containing a and b, we need to take the cross product of a and b:
a × b =
| i j k |
| 31 4 -1 |
| 1 -3 1 |
= (4×1 - (-1)×(-3))i - (31×1 - (-1)×1)j + (31×(-3) - 4×1)k
= 7i - 30j - 97k
To make this a unit vector, we need to divide it by its magnitude:
|n| = √(7² + (-30)² + (-97)²) = √(6206)
n = (7/√6206)i - (30/√6206)j - (97/√6206)k
To check that this forms a right-handed system with a and b, we can take their dot product:
a · (b × n) =
(31+4j-k) · (7i-30j-97k) =
31×7 + 4×(-30) + (-1)×(-97) = 505
Since this is a positive value, we can conclude that a, b, and n form a right-handed system.
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Review For each initial position choose the correct sketch of the mostly path of the shark toward the center of the dipole in
In the given figure, the dipole electrode is represented by the two circles one is red and other is green, and the three initial positions of the shark are represented by the three blue circles.
For the initial position of the shark represented by the top blue circle 1, the most likely path of the shark toward the center of the dipole would be a curved path, as the equipotential lines bend and curve around the dipole electrode. The shark would start by following the equipotential lines to the left, then gradually curve downward and to the right before eventually reaching the center of the dipole.
For the position at the bottom left blue circle 2, the most likely path of the shark would be same as path 1 since the shark would start by following the equipotential lines upward and to the right, then gradually curve downward and to the left before eventually reaching the center of the dipole.
For the position at the bottom right blue circle 3, the most likely path of the shark would be same as path 1 since the shark would start by following the equipotential lines downward and to the left, then gradually curve upward and to the right before eventually reaching the center of the dipole.
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complete question: In experimental tests, sharks have shown the ability to locate dipole electrodes (simulating the dipole fields of the heartbeats of prey animals) buried under the sand. In a test with young bonnethead sharks, sharks that detected the presence of a dipole usually swam toward the center of the dipole by following equipotential lines. Figure P21.35 shows a dipole electrode and three initial positions of a bonnethead shark. For each initial position, sketch the most likely path of the shark toward the center of the dipole.