The total stopping distance required to bring it to a stop is 300 metres.
What exactly is stopping distance?
The stopping distance is the distance travelled between the time a moving vehicle's body decides to stop and the time the vehicle completely stops.
Bringing a moving vehicle to a stop involves three critical factors.
1.Distance of perception
2. Distance between reactions
3.Distance between the brakes
1. Perception distance is the amount of time that a vehicle travels while your brain perceives the need to apply the brakes.
An alert driver must react in about 0.75 seconds. This time represents approximately 60 feet of travel.
2. Reaction distance is the distance travelled before the driver applies the brakes. This takes about 0.75 seconds for a cautious
driver. This adds an additional 60 feet of travel.
3. Braking distance is the distance travelled by a vehicle before coming to a complete stop on dry ground. This distance is longer on slick ground (after snow or rain). The skidding distance on dry ground is approximately 170 feet.
This means that a cautious driver needs 60+60+170 = 290 feet (approximately 300 feet) to come to a complete stop.
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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
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 crane lifts an object weighing 25000N up with a constant speed of 0.8m/s. calculate the capacity of that crane
If the constant force F0 is applied to stretch a material described by the Maxwell model, what would be the initial elongation value at t=0.-0-F0/k-F0/(k+n)-F0/(k+(n/t))
The initial elongation of the material described by the Maxwell model is given by the formula, initial elongation = [tex]F_0[/tex]/k.
The initial elongation value at t=0 for the material described by the Maxwell model is given by the formula, initial elongation = [tex]F_0[/tex]/k.
Here, k represents the spring constant of the material.
Let's understand this in detail.
The Maxwell model is a type of viscoelastic model that is used to describe the behavior of certain materials. It is made up of a spring and a dashpot in series.
The spring represents the elastic component of the material and the dashpot represents the viscous component of the material.
In this model, the deformation of the material depends on the applied force as well as the time duration for which the force is applied.
The formula to calculate the initial elongation of the material is given by:
initial elongation = [tex]F_0[/tex]/k
where [tex]F_0[/tex] is the force applied to stretch the material and k is the spring constant of the material. The spring constant of a material is defined as the amount of force required to stretch the material by one unit.
The initial elongation of the material is calculated using the spring constant of the material. The spring constant represents the amount of force required to stretch the material by one unit.
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Topic: Rotational Motion
The motor in an electric saw brings the circular blade from rest up to the rated angular velocity of 80.0 rev/s in 240.0 rev. One type of blade has a moment of inertia of 1.41×10^-3 kg.m^2. Determine the net torque (assumed constant) the motor must apply to the blade.
Answer:
To solve this problem, we can use the equation for rotational motion:
Δθ = (1/2) α t^2 + ω0 t
where Δθ is the change in angle, α is the angular acceleration, t is the time, and ω0 is the initial angular velocity.
In this case, we know that the initial angular velocity is 0 (since the blade is at rest), the final angular velocity is 80.0 rev/s, and the number of revolutions is 240.0 rev. We can use these values to find the angular acceleration:
ωf = ω0 + αt
80.0 rev/s = 0 + α(240.0 rev)
α = 80.0 rev/s / 240.0 rev
α = 1/3 rev/s^2
Now that we know the angular acceleration, we can use the moment of inertia and the torque equation:
τ = Iα
where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
Substituting the given values, we get:
τ = (1.41×10^-3 kg.m^2)(1/3 rev/s^2)
τ = 4.70×10^-4 N.m
Therefore, the net torque the motor must apply to the blade is 4.70×10^-4 N.m.
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.
A 2000 kg truck moving at 10 m/sec to the right strikes a 500 kg car moving to the left with a velocity of 20 m/s. If the two bodies stick together and moving to the right after the collision, what is the final velocity of the combined mass?
The final velocity of the combined mass after the collision is 15 m/s to the right.
This can be calculated using the equation for Conservation of Momentum.
The momentum of the combined mass before the collision is the sum of the momentums of the truck and car individually:
Momentum of Truck = 2000kg x 10 m/s = 20000 kgm/s
Momentum of Car = 500 kg x 20 m/s = 10000 kgm/s
Total Momentum before Collision = 30000 kgm/s
The total momentum after the collision is the same as before, so the final velocity of the combined mass is:
Final Velocity = Total Momentum/Combined Mass
Final Velocity = 30000 kgm/s / 2500 kg
Final Velocity = 15 m/s
therefore, the final velocity is 15 m/s.
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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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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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A beam consisting of five types of ions labeled A, B, C, D, and E enters a region that contains a uniform magnetic field as shown in the figure below. The field is perpendicular to the plane of the paper, but its precise direction is not given. All ions in the beam travel with the same speed. The table below gives the masses and charges of the ions. Note: 1 mass unit = 1.67 x 10â€"27 kg and e = 1.6 x 10â€"19 C
Which ion falls at position 2?
At position 2, ion B falls. It is less deflected because it has a lesser mass than ions C, D, and E and the same charge as ion A.
A force perpendicular to the charged particle's velocity and the magnetic field's direction is applied when it reaches the magnetic field. The right-hand rule asserts that the palm will face the direction of the force if the thumb of the right hand points in the direction of the particle's velocity and the fingers point in the direction of the magnetic field. The particle's charge, velocity, and magnetic field intensity all affect how much force is generated.
Since all ions are moving at the same speed in this scenario, the force exerted on each ion is proportional to its charge to mass ratio. Ion B has the smallest mass of all the ions, so the least force and is least deflected of the ions, falling at position 2.
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A 30.0-kg box is being pulled across a carpeted floor by a horizontal force of 230 N , against a friction force of 210 N . What is the acceleration of the box?
Please answer only part E
A 30.0-kg box is being pulled across a carpeted floor by a horizontal force of 230 N , against a friction force of 210 N . The acceleration of the box is 0.667 m/s².
The acceleration of the box can be calculated using the formula:
acceleration = (Net force) / (mass)
The given values in the question are:
mass of the box = 30.0 kg
force applied on the box = 230 N
friction force acting on the box = 210 N
Now, let's calculate the net force acting on the box:
Net force = (force applied) - (friction force)= 230 N - 210 N= 20 N
Thus, the net force acting on the box is 20 N.
Using the formula mentioned above, the acceleration of the box can be calculated as:
acceleration = (Net force) / (mass)
= 20 N / 30.0 kg
= 0.667 m/s²
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The outer edge of a rotating Frisbee with a diameter of 30 cm has a linear speed of 3.2 m/s
What is the angular speed of the Frisbee? In rad/s
The angular speed of the Frisbee is 21.33 rad/s.
What is Angular speed ?
Angular speed is a measure of how fast an object is rotating or moving around a central point or axis. It is a scalar quantity, which is defined as the rate of change of the object's angular displacement over time, expressed in units of radians per second (rad/s).
Angular speed is calculated using the formula:
Angular speed (ω) = Δθ / Δt
The linear speed of the outer edge of the Frisbee is given by:
v = 3.2 m/s
The diameter of the Frisbee is given by:
d = 30 cm = 0.3 m
The radius of the Frisbee is half the diameter:
r = d/2 = 0.15 m
The linear speed of a point on the edge of a rotating object is related to the angular speed by the formula:
v = ωr
where ω is the angular speed in radians per second.
Substituting the given values, we can solve for ω:
ω = v/r
ω = 3.2 m/s / 0.15 m
ω = 21.33 rad/s (rounded to two decimal places)
Therefore, the angular speed of the Frisbee is 21.33 rad/s.
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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
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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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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A crate remains stationary after it has been placed on a ramp inclined at an angle with the horizontal. Which of the following statements must be true about the magnitude of the frictional force that acts on the crate?(a) It is larger than the weight of the crate.(b) It is at least equal to the weight of the crate.(c) It is equal to μsn.(d) It is greater than the component of gravitational force acting down on the ramp.(e) It is equal to the component of gravitational force acting down on the ramp.
Answer:
(D) the frictional force must be greater than the component of the gravitational force that acts down the ramp
Which of the substances listed in the table on the first page would heat up more quickly if an equal amount
of heat energy were applied to all of the substances at the same time? Explain your answer.
If an equal amount of heat energy is applied to all the substances at the same time, aluminum will heat up more quickly compared to the other substances.
What is Specific Heat ?
Specific heat is the amount of heat energy required to raise the temperature of a substance by a certain amount, usually one degree Celsius or one Kelvin, per unit mass of the substance. It is a physical property that determines how much energy is needed to change the temperature of a material. Different materials have different specific heat values, which can be used to predict how they will behave when heated or cooled.
Looking at the table, we can see that water has the highest specific heat capacity of all the substances listed, at 4.18 J/g°C. This means that it requires more energy to raise the temperature of water compared to the other substances. Therefore, if an equal amount of heat energy is applied to all the substances at the same time, water will heat up more slowly compared to the other substances.
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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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what type of tide occurs when the moon is in the positions marked by the straight arrows in this image? please give one word as your answer.
The positions marked by the straight arrows in the image are the first quarter and third quarter phases of the Moon the tide that occurs is called a "neap tide".
What is neap tide?
During these phases, the Moon and the Sun are at right angles to each other with respect to the Earth, which causes the gravitational forces of the Moon and the Sun to partially cancel out. As a result, the tidal range is at its minimum, and the tide that occurs is called a "neap tide". Therefore, the answer is "neap".
What is gravitational forces?
Gravitational forces refer to the attractive force that exists between any two objects in the universe that have mass. This force is governed by Newton's law of universal gravitation, which states that the gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
This means that the larger the mass of the objects, the stronger the gravitational force between them, and the farther apart they are, the weaker the gravitational force. Gravitational forces are responsible for keeping celestial bodies, such as planets, moons, and stars, in their orbits, and for the phenomena of tides on Earth caused by the gravitational pull of the Moon and the Sun.
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Complete question is: "neap tide" type of tide occurs when the moon is in the positions marked by the straight arrows in this image.
two charges q1 is 50mc and q2 is 10mc are located at -1,1,-3and3, 1, 0respectively find the force on q1
Explanation:
dhifgjrgrkfgd jsjhdhdj
shdjdjdj
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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
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.
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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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:
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:
Explanation:
According to the law of conservation of momentum, the total momentum before the collision must be equal to the total momentum after the collision. We can use this principle to determine the resulting velocity of Kaylyn's puck after the collision.
Before the collision, Sharon's puck has a momentum of:
p1 = m1 * v1 = 0.3 kg * 6 m/s = 1.8 kg m/s
where m1 is the mass of Sharon's puck and v1 is its velocity.
Kaylyn's puck is at rest before the collision, so its momentum is:
p2 = m2 * v2 = 0.3 kg * 0 m/s = 0
where m2 is the mass of Kaylyn's puck and v2 is its velocity.
After the collision, Sharon's puck is at rest, so its momentum is:
p1' = m1 * v1' = 0.3 kg * 0 m/s = 0
where v1' is the velocity of Sharon's puck after the collision.
The total momentum after the collision is the momentum of Kaylyn's puck:
p2' = m2 * v2'
where v2' is the resulting velocity of Kaylyn's puck after the collision.
Using the conservation of momentum principle, we can write:
p1 + p2 = p1' + p2'
Substituting the values we have calculated:
1.8 kg m/s + 0 = 0 + 0.3 kg * v2'
Solving for v2':
v2' = (1.8 kg m/s) / (0.3 kg) = 6 m/s
Therefore, the resulting velocity of Kaylyn's puck after the collision is 6 m/s, which is equal to the velocity of Sharon's puck before they collided. The correct answer is option A.
calculate T1 , T2 and T3
Using the triangle of forces to get the system of the forces;
T1 = 866 NT2 = 500 NT3 = 1000 NWhat is the triangle of forces theorem?If three forces acting on a body are in equilibrium, then they can be represented in magnitude and direction by the three sides of a triangle taken in order.
In other words, the three forces can be drawn as vectors, and these vectors can be arranged to form a closed triangle.
We know that we have the other end of the triangle to be;
100 Kg * 10 m/s^2 = 1000 N
The missing angle is;
180 - (30 + 60)
= 90 degrees
Thus;
1000/Sin 90 = T1/Sin 60
T1 = 100 Sin 60/Sin 90
T1 = 866/1
T1 = 866 N
1000/Sin 90 = T2/Sin 30
T2 = 1000 Sin 30/Sin 90
T2 = 500 N
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How many Earths could fit inside Jupiter (assuming you could fill up all the volume)? The equation for the volume of a sphere is V=(43)πr3V=(43)πr3. (The radius of Jupiter is RJupiterRJupiter = 71492 kmkm.)Express your answer using three significant figures.
The number of earths that could fit inside Jupiter is 6000 where the radius of Jupiter is 71492 km.
Given the volume of a sphere is (V) =[tex](4/3)*\pi*R^3[/tex]
The radius of Jupiter is (RJupiter) = 71492 km
Then the volume of Jupiter is (VJupiter) = [tex](4/3)*\pi*(71492)^3[/tex]
VJupiter = [tex]6.6 * 10^{15} km^3.[/tex]
We know the radius of earth = Re = 6371km
Then the volume of earth (Ve) =[tex]4/3 * \pi * (6371)^3[/tex]
Ve = [tex]1.08 * 10^{12} km^3.[/tex]
Let the number of earths that could fit inside Jupiter = n
Therefore, n = VJupiter/VEarth
n = [tex]6.6 * 10^{15} km^3/1.08 * 10^{12} km^3 = 6 * 10^3[/tex], = 6,000 Earths.
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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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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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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).