Without the given Figure, precise answers can't be provided. Generally, acceleration is calculated as the slope of the velocity-time graph, and position is provided by the integral (or area under the graph) of the velocity-time graph. The object's speed is lowest when its velocity is minimal, and it is farthest from x= 0 when the accumulated area under the graph is maximum.
Explanation:Unfortunately, without the given Figure P2.50, it's impossible to accurately calculate the object's acceleration, position at different times, or specify when the object is moving with the lowest speed or is farthest from x = 0.
However, I can explain the general method to determine this information. Acceleration is calculated from the slope of the velocity-time graph. The position is generally obtained by calculating the area under the velocity-time graph from the beginning of the interval to the end. The object is moving with the lowest speed when the velocity is lowest (either positively or negatively). The object is farthest from x = 0 when the accumulated area under the velocity-time graph (counting areas below the time axis as negative) is a maximum. The total distance the object has moved is equal to the absolute sum of all the areas (both positive and negative) on the velocity-time graph.
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the blank secrets chemical messages that circulate through the blood and also communicates messages that influence behavior and many aspects of the biological functioning
(stagnation pressure) a hang glider soars through standard sea-level air with an airspeed of 11.8 m/s. what is the gage pressure at a stagnation point on the structure?
The pressure at a hang glider's stagnation point is the sum of atmospheric pressure and dynamic pressure due to airflow collision, resulting in higher than atmospheric pressure. Using Bernoulli's equation with provided values, the stagnation pressure is 95430 Pa, and at 150 m/s, the pressure is approximately 82920 Pa.
The calculation of the gage pressure at a stagnation point involves understanding the flow dynamics around the structure of a hang glider as it moves through air. The stagnation point is where the airflow comes to a complete stop at the leading edge of the wing, causing the pressure to be higher than the atmospheric pressure due to the dynamic pressure of the colliding air mass. Using the provided stagnation pressure of 95430 Pa and the density of air at sea-level conditions (1.14 kg/m³), and given that the airspeed is 11.8 m/s, we would apply Bernoulli's equation to confirm the dynamic pressure at the stagnation point. But for the second part of the question, we adjust the equation to account for the higher speed of 150 m/s over the wing and determine that the pressure is 82920 Pa, acknowledging that turbulent flow makes these results approximate.
The stagnation pressure on a hang glider's stagnation point at sea-level conditions and an airspeed of 11.8 m/s can be calculated as 95430 Pa using the sea-level air density (1.14 kg/m³) and standard atmospheric pressure (8.89 × 10⁴ N/m²).
The concept being discussed in the question is the stagnation pressure which is the pressure at a point on a structure where the fluid (air) velocity is zero because the fluid there has been brought to rest abruptly. This pressure is higher than the atmospheric pressure because it includes both the atmospheric pressure and the additional pressure resulting from the air hitting the structure. The stagnation pressure (Pstagnation) at a hang glider's stagnation point can be calculated using Bernoulli's equation.
Given that the airspeed is 11.8 m/s, the sea-level air density is 1.14 kg/m³, and standard atmospheric pressure is 8.89 × 10⁴ N/m², we can find the stagnation pressure on the structure. Using the provided values:
Pstagnation = Po + (1/2) ρV²
Pstagnation = 8.89 × 10⁴ Pa + (1/2)(1.14 kg/m³)(11.8 m/s)²
Pstagnation = 8.89 × 10⁴ Pa + (1/2)(1.14)(139.24)
Pstagnation = 8.89 × 10⁴ Pa + 79.37 Pa
Pstagnation = 95430 Pa
A river has a steady speed of vs. A student swims upstream a distance d and back to the starting point. (a) If the student can swim at a speed of v in still water, how much time tup does it take the student to swim upstream a distance d? Express the answer in terms of d, v, and vs. tup = Incorrect: Your answer is incorrect. (b) Using the same variables, how much time tdown does it takes to swim back downstream to the starting point? tdown = Incorrect: Your answer is incorrect. Your answer cannot be understood or graded. More Information
he speed of the student relative to shore is
v_ up = v- vs
v _down = v+ vs
The time required to travel distance d upstream
is
t_up = d/ v_up = d/ v- vs
(2)
The time required to swim the same distance d downstream is
t_down = d/ v_down = d/ v+ vs
find the velocity vb for the case where the acceleration is 7.66 m/s2. Anawer in units of m/s
"Giant Swing", the seat is connected to two cables as shown in the figure (Figure 1) , one of which is horizontal. The seat swings in a horizontal circle at a rate of 37.0 rev/min .
Two passenger trains are passing each other on adjacent tracks. train a is moving east with a speed of 13 m/s, and train b is traveling west with a speed of 28 m/s.(a) what is the velocity (magnitude and direction) of train a as seen by the passengers in train b? (b) what is the velocity (magnitude anddirection) of train b as seen by the passengers in train a?
Since the two trains are passing in opposite directions, so this means that their relative velocities will be the sum of the two trains that is:
relative velocity = (13 + 28) = 41m/s
a. The passengers aboard on train B will see that train A is
moving at 41m/sec due east
b. The passengers aboard on train A will see that train B is moving at 41m/sec due west
Answer:
B and C
Explanation:
The first "lunar olympics" is to be held on the Moon inside a huge dome. Of the usual Olympic events-track and field, swimming, gymnastics, and so on- which would be drastically affected by the Moon's lower gravity? (Select all that apply.)
Jumping
Bicycling
Swimming (on a level track)
Throwing things
Running races (on a level track)
Weight lifting
Jumping, throwing, and weight lifting Olympic events would be drastically altered by the Moon's lower gravity in a hypothetical lunar Olympics. Athletes could jump higher and throw further due to gravity being 1/6th of earth's. Similarly, weights would feel lighter during weight lifting.
Explanation:In the context of the lunar olympics, it's critical to understand how the Moon's lower gravity would drastically impact certain sports events. The Moon's gravity is about 1/6th of Earth's gravity, which would significantly influence events involving lifting, jumping and throwing.
Jumping events would drastically change because the lower gravity would allow participants to jump about six times higher than on Earth. Throwing events would also see remarkable enhancements as objects would travel a greater distance before dropping.
Moreover, Weight lifting would be greatly impacted as the same weights would feel much lighter on the Moon. Interestingly, sports such as bicycling, swimming on a level track, and running races on a level track would not be significantly affected because these activities are less dependent on gravity relative to the others mentioned. Nonetheless, athletes might perform differently due to the reduced atmospheric pressure and possible difficulties in maintaining balance because of lower gravity.
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Final answer:
The lunar olympics would see drastic changes in events such as jumping, throwing, and weight lifting due to the Moon's 1/6th Earth gravity, while bicycling would be less affected. Swimming would also change, but it assumes the presence of water in a lunar environment.
Explanation:
The "lunar olympics" concept highlights how sports would be drastically affected by the Moon's reduced gravity. Given that gravity on the Moon is about 1/6th of that on Earth, events such as jumping, throwing things, and weight lifting would be greatly altered. With lower gravity, athletes could jump much higher, throw objects much farther, and lift heavier weights with less effort, compared to Earth. Running races might see a difference in athletes' strides and suspension period off the ground, potentially leading to faster times. However, activities like bicycling would be less affected as they rely more on the rider's strength and endurance than on gravitational forces. Swimming is not practical without substantial water, which is currently not present on the Moon, but if somehow managed in a dome, the reduced gravity could affect water dynamics and a swimmer's buoyancy, thereby affecting the execution of strokes and turns.
Water (2510 g ) is heated until it just begins to boil. if the water absorbs 5.01×105 j of heat in the process, what was the initial temperature of the water?\
Questioning is the beginning of the scientific inquiry process. True or False
A heavy 2.0Kg ball is moving at 10m/s when it is caught. A light 1.5Kilogram ball is travelling at 20m/s when it caught. Which ball required the greater impulse?
Jack tries to place magnets on the door of his refrigerator. He observes that the magnets don’t stick. He guesses that the door of the refrigerator is made of a nonmagnetic substance. What stage of scientific investigation does his guess represent?
Ike is cutting the grass using a human-powered lawn mower. he pushes the mower with a force of 45 n directed at an angle of 41° below the horizontal direction. calculate the work that mike does on the mower each time he pushes it 9.1 m across the yard.
Morgan has a mass of 85 kg and is on top of a bed in such a position that she can apply a pressure of 9530 n/m^2 on the mattress. would you calculate that morgan is standing sitting or lying on the bed.
To determine Morgan's position on the bed (whether standing, sitting, or lying), one must calculate the area of contact she has with the bed using her mass and the pressure she exerts. By comparing the calculated contact area with typical areas for different positions, one can infer her position.
Explanation:The question asks to determine whether Morgan is standing, sitting, or lying on the bed based on the pressure she exerts on the mattress and her mass of 85 kg. Pressure is defined as the force applied per unit area; in this case, force is due to Morgan's weight. Given that Morgan's weight (W) is the product of her mass (m) and Earth's gravity (g), we can write W = m * g = 85 kg * 9.81 m/s2. The weight finds us the force Morgan exerts downward.
To find the area of contact (A), we rearrange the pressure formula P = F/A, to A = F/P. Inserting the given pressure (P = 9530 N/m2) and her weight calculated earlier, we can find the area. A smaller contact area would suggest standing, while a larger area would suggest sitting or lying down. For example, if calculated area is around the size of feet, she's likely standing; if it's larger - similar to a body's contact area when sitting or much larger - she's sitting or lying down, respectively.
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Which will result in positive buoyancy and cause the object to float?
When the force of gravity is more than the buoyant force
When the buoyant force is greater than the force of gravity
When the buoyant force is equal to the force of gravity
When the buoyant force is less than the force of gravity
the answer should be:
When the buoyant force is equal to the force of gravity
You are an employee of the city of Chicago. A city official declares that all cables holding traffic lights in the city should be exactly horizontal and asks you to direct the project. What should be your reply?
A. That's impossible, because according to Newton's Second Law, there must be an upward component of the tension in the cable that balances the weight of the traffic light, and therefore, the cable can never be exactly horizontal.
B. No problem. All we have to do is increase the horizontal tension in the cable.
C. No problem. All it takes is a pay raise.
D. That's impossible, because by Newton's Third Law, the cable pulls horizontally on the poles with a force of the same magnitude force as the force with which the poles pull on the cable. Therefore, if the horizontal component of the tension is too large, it will pull the poles inward toward the cables.
The cables holding the traffic lights cannot be exactly horizontal due to Newton's Second Law. An upward component of tension is required to balance the weight of the traffic light, preventing a completely horizontal placement.
Explanation:Your objective to have the cables exactly horizontal is unfortunately not feasible due to the principles of physics, specifically Newton's Second Law. According to Newton's second law, there has to be an upward component of tension in the cable that counterbalances the weight of the traffic light. This means that the cable can't be placed horizontally as some part of the tension must always act vertically to balance the gravity acting on the traffic light. If we tried to set the cable perfectly horizontally, without the upward component of tension, the traffic light would not hold in place and would fall due to gravity.
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A hiker leaves her camp and walks 3.5 km in a direction of 55° south of west to the lake. After a short rest at the lake, she hikes 2.7 km in a direction of 16° east of south to the scenic overlook. What is the magnitude of the hiker’s resultant displacement? Round your answer to the nearest tenth. km What is the direction of the hiker’s resultant displacement? Round your answer to nearest whole degree. ° south of west
1) Magnitude
Let's take south as positive y-direction and east as positive x-direction. Then we have to resolve both displacements into their respective components:
[tex]d_{1x} = -(3.5 km) cos 55^{\circ}=-2.0 km[/tex]
[tex]d_{1y} = (3.5 km) sin 55^{\circ}=2.87 km[/tex]
[tex]d_{2x} = (2.7 km) sin 16^{\circ}=0.74 km[/tex]
[tex]d_{2y} = (2.7 km) cos 16^{\circ}=2.60 km[/tex]
So, the components of the total displacement are
[tex]d_x = d_{1x}+d_{2x}=-2.0 km +0.74 km=-1.26 km[/tex] east (so, 1.26 km west)
[tex]d_y=d_{1y}+d_{2y}=2.87 km + 2.60 km=5.47 km[/tex] south
So, the magnitude of the resultant displacement is
[tex]d=\sqrt{d_x^2+d_y^2}=\sqrt{(1.26)^2+(5.47)^2}=5.61 km[/tex]
2) Direction
the direction of the hiker's displacement is
[tex]\theta= arctan(\frac{d_y}{d_x})=arctan(\frac{5.47}{1.26})=arctan(4.34)=77.0^{\circ}[/tex] south of west.
The magnitude of the hiker's resultant displacement is 5.8 km south of west.
Explanation:To find the magnitude of the hiker's resultant displacement, we need to first find the x and y components of the displacement vectors.
Path 1, the hiker walks 3.5 km in a direction of 55° south of west. The x component of this displacement is -3.5 * cos(55°) = -2.503 km, and the y component is -3.5 * sin(55°) = -2.845 km.
Path 2, the hiker walks 2.7 km in a direction of 16° east of south. The x component of this displacement is 2.7 * sin(16°) = 0.732 km, and the y component is -2.7 * cos(16°) = -2.606 km.
The total x component is -2.503 + 0.732 = -1.771 km, and the total y component is -2.845 - 2.606 = -5.451 km. Using the Pythagorean theorem, the magnitude of the resultant displacement is sqrt((-1.771)^2 + (-5.451)^2) = 5.797 km.
The direction can be found using the inverse tangent function, atan((-5.451)/(-1.771)) = 73°. Since the displacement is south of west, the direction is 180 - 73 = 107° south of west.
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Find the speed at which Superman (mass=76.0 kg) must fly into a train (mass = 19969 kg) traveling at 65.0 km/hr to stop it.
Running into the train at that speed would severely damage both train and passengers. Calculate the minimum time Superman must take to stop the train, if the passengers experience an average horizontal force of 0.540 their own weight.
How far does the train then travel while being slowed to a stop?
If an object mechanical energy is equal to its potential energy how much Connecticut energy does the object have
What is an area of land in which all freshwater converges into the largest stream, ultimately draining into the largest body of water? hints what is an area of land in which all freshwater converges into the largest stream, ultimately draining into the largest body of water? estuary wetland spring watershed?
The answer is watershed. It is an area of land that drains all the rainfall and streams to a shared passage for instance, mouth of a bay, the outflow of a reservoir, or any point alongside a stream network. The word watershed is occasionally used interchangeably with catchment or drainage basin. This contains surface water like reservoirs, streams, lakes, and wetlands.
Answer:
THe answer is watershed
Explanation:
If a seagull drops a shell from rest at a height of 10 m , how fast is the shell moving when it hits the rocks?
The shell dropped by the seagull is moving at a speed of approximately 14 m/s when it hits the rocks. This is calculated through the conversion of potential energy to kinetic energy, using the relevant physics formula.
Explanation:Based on the principles of kinetic and potential energy transformation in Physics, we can estimate the speed at which the shell will hit the rocks. When the seagull drops the shell, it initially has potential energy which is converted into kinetic energy as it falls under the gravity. This can be calculated using the equation v² = 2gy, where v indicates the final velocity, g is the acceleration due to gravity approximated as 9.8 m/s² and y is the height from which the object is dropped.
Replacing the known values: v² = 2*9.8m/s²*10m. Therefore, v = √(2*9.8m/s²*10m). This calculation results with a final velocity of about 14 m/s. This means the shell is moving at a speed of 14 m/s when it hits the rocks.
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If the forces acting upon an object are balanced, then the object
A) All of these
B) Might be stopped.
C) Might be moving at a constant velocity
D) Is not accelerating
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 82.4 mpa1m (75.0 ksi1in.). if the plate is exposed to a tensile stress of 345 mpa (50,000 psi) during service use, determine the minimum length of a surface crack that will lead to fracture. assume a value of 1.0 for y.
Fracture toughness, minimum crack length calculation for fracture under stress.
Fracture toughness is the ability of a material to resist brittle fracture when a crack is present. In this scenario, we are given a fracture toughness value of 82.4 MPa√m.
To determine the minimum length of a surface crack that will lead to fracture when exposed to a tensile stress of 345 MPa, we can use the plane strain fracture toughness formula.
By plugging in the given values and solving the equation, we can find the minimum length of the surface crack that will cause fracture.
To find the minimum crack length that could lead to fracture in a steel plate, the plane strain fracture toughness equation is used, incorporating the values for fracture toughness, applied stress, and the geometric factor Y.
To determine the minimum length of a surface crack that will lead to fracture in a steel plate subjected to a tensile stress during service, we can use the plane strain fracture toughness equation:
\[a = \left(\frac{1}{\pi} \cdot \left(\frac{K_{IC}}{\sigma \cdot Y}\right)^{2}\right)
\]
Where:
a is the crack lengthKIC is the plane strain fracture toughness, which is 82.4 MPa√m (75.0 ksi√in.) in this casesigma is the stress applied during service, given as 345 MPa (50,000 psi)
Plugging in the values, the minimum crack length a can be calculated as follows:
\[a = \left(\frac{1}{\pi} \cdot \left(\frac{82.4 \, \text{MPa}\sqrt{m}}{345 \, \text{MPa} \cdot 1.0}\right)^{2}\right)
\]
After calculating the expression inside the parenthetical first and then squaring it, you would divide by π to get the minimum crack length a.
The height, s, of a ball thrown straight down with initial speed 64 ft/sec from a cliff 80 feet high is s(t) = â16t2 â 64t + 80, where t is the time elapsed that the ball is in the air. what is the instantaneous velocity of the ball when it hits the ground?
A thermos bottle has two layers, a glass layer and a vacuum layer. The vacuum layer is the better _____.
A 5000 kg truck is parked on a 7.0â slope. how big is the friction force on the truck?
For general projectile motion, which of the following best describes the horizontal component of a projectile's acceleration? Assume air resistance is negligible.
-The horizontal component of a projectile's acceleration continually decreases.
-The horizontal component of a projectile's acceleration is zero.
-The horizontal component of a projectile's acceleration remains a nonzero constant.
-The horizontal component of a projectile's acceleration continually decreases.
-The horizontal component of a projectile's acceleration initially decreases and then increases.
Answer:
The horizontal component of a projectile's acceleration is zero
Explanation:
When calculating all the different components of a projectile (acceleration, initial velocity, final velocity, etc.), you need to break it up into its X and Y directions.
During free fall, the only force acting on the projectile is gravity, and since gravity acts down (in the Y direction), there is no force acting in the X direction. Hence, the acceleration in the X direction is zero.
A 60-kg man uses an electric chair to carry him up the stairs. if the chair has a mass of 200 kg, what power must a motor generate to transport the man and chair 5 m vertically in 10 seconds, assuming a constant velocity and ignoring all forces except gravity? 900 w 1300 w 400 w 1200 w
The power must a motor generate to transport the man and chair is 1300 W.
What is gravitational potential energy?If an object is lifted, work is done against gravitational force. The object gains energy.
The equation for calculating gravitational potential energy is
GPE =mgh
where, GPE is the gravitational potential energy, m is the mass, h is the change in height and g is the acceleration due to gravity.
Given is the mass of chair =200kg and mass of man =60 kg, then the total mass to be lifted is
m = 200 +60 = 260 kg.
height to be raised h =5m and time taken t = 10s.
GPE = 260 x 10 x 5
GPE = 13000 J
Power to be generated P = GPE/t
Put the values, we get
P = 13000 / 10 = 1300 W
Thus, the power must be generated is 1300W.
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What’s the velocity of an 11-kilogram object with 792 joules of kinetic energy?
7 m/s
8 m/s
9 m/s
11 m/s
12 m/s
Answer:
12 m/s bye
Explanation:
The velocity of the object is 12 m/s. The correct answer is 12 m/s. The correct option is (e).
The kinetic energy (KE) of an object can be calculated using the formula:
KE = 0.5 × m × v²
Where:
KE is the kinetic energy
m is the mass of the object
v is the velocity of the object
The object has a mass of 11 kilograms and kinetic energy of 792 joules, we can rearrange the formula to solve for velocity:
v² = (2 × KE) / m
v = √((2 × KE) / m)
v = √((2 × 792) / 11)
v = 12 m/s
So, the correct answer is 12 m/s. The correct option is (e).
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You plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, you will need to leave the earth with enough speed to make it to the moon. Some information that will help during this problem: mearth = 5.9742 x 1024 kg rearth = 6.3781 x 106 m mmoon = 7.36 x 1022 kg rmoon = 1.7374 x 106 m dearth to moon = 3.844 x 108 m (center to center) G = 6.67428 x 10-11 N-m2/kg2 1) On your first attempt you leave the surface of the earth at v = 5534 m/s. How far from the center of the earth will you get?
The distance from the center of the earth that you will get, given that you leave the surface of the earth with a velocity of 5534 m/s, is [tex]6.5099\times10^6\ m[/tex]
How to calculate the distance from the earth you will get to?
The distance from the center of the earth that you will get, knowing that you leave the surface of the earth with a velocity of 5534 m/s can be calculated as shown below:
Velocity of takeoff (v) = 5534 m/sMass of earth (M) = 5.9742×10²⁴ KgGravitational constant (G) = 6.67428×10¯¹¹ Nm²/Kg²Distance you will get considering your takeoff velocity (d) =?[tex]d = \frac{GM}{2v^2} \\\\d = \frac{6.67428\times10^{-11}\ \times\ 5.9742\times10^{24}}{2\ \times\ 5534^2} \\\\d = 6.5099\times10^6\ m[/tex]
From the above calculation, we can conclude that the distance you will get is [tex]6.5099\times10^6\ m[/tex]
How does the law of conservation of mass apply to this reaction: C2H4+O2-->2H2O+2CO2?
The equation needs to be balanced. There are fewer oxygen atoms in the equation than hydrogen or carbon.
Only the oxygen needs to be balanced. There are equal numbers of hydrogen and carbon.
The law of conservation of mass has already been applied. There is an equal number of each element on both sides of the equation.
Each element needs to be balanced
Regarding the equation C₂H₄ + O₂ ⇒ 2 H₂O + 2 CO₂, it is true that only the oxygen needs to be balanced. There are equal numbers of hydrogen and carbon.
According to the law of conservation of mass, matter is not created nor destroyed over the course of a chemical reaction. Thus, equations need to be balanced so the same number of atoms are on both sides.
How does the law of conservation of mass apply to this reaction:
C₂H₄ + O₂ ⇒ 2 H₂O + 2 CO₂?
The equation needs to be balanced. There are fewer oxygen atoms in the equation than hydrogen or carbon. NO. We just compare oxygen atoms in the products with oxygen atoms in the reactants. Only the oxygen needs to be balanced. There are equal numbers of hydrogen and carbon. YES. Carbon and hydrogen atoms are already balanced. The law of conservation of mass has already been applied. There is an equal number of each element on both sides of the equation. NO. There is not the same number of oxygen atoms on both sides. Each element needs to be balanced. NO. In this equation, carbon and oxygen have already been balanced.Regarding the equation C₂H₄ + O₂ ⇒ 2 H₂O + 2 CO₂, it is true that only the oxygen needs to be balanced. There are equal numbers of hydrogen and carbon.
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Answer: the answer is The law of conservation of mass has already been applied. There is an equal number of each element on both sides of the equation.
Explanation: i hope this helps:)