NEED HELP URGENT!! A small, horizontal pipe with cross-sectional area A is joined to a large horizontal pipe with cross-sectional area 2A. Both pipes are at the same height, and completely filled with water (density 1000 kg/m^3). What is the speed of the water in the large pipe if the speed in the small pipe is 1 m/s?
Answers
v₁ = 1 m/s, the speed in the smaller pipe
ρ = 1000 kg/m³, the density of the water (constant)
A, m² = the cross-sectional area of the smaller pipe
2A, m² = the cross-sectional area of the larger pipe.
Let v₂ = the velocity of the water in the larger pipe.
The mass flow rate is constant, and it is
[tex]Q = (\rho \, \frac{kg}{m^{3}} )*(v_{1} \, \frac{m}{s} )*(A \, m^{2}) = (\rho \, \frac{kg}{m^{3}} )*(v_{2} \, \frac{m}{s} )*(2A \, m^{2})[/tex]
Because v₁ = 1 m/s, obtain
ρ*v₂*(2A) = ρA
2v₂ = 1
v₂ = 1/2 m/s
Answer: 0.5 m/s
Related Questions
If an object mechanical energy is equal to its potential energy how much Connecticut energy does the object have
Answers
(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?
Answers
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 rocket is launched at an angle of 56.0 degrees above the horizontal with an initial speed of 105 m/s. The rocket moves for 3.00 s along its initial line of motion with an acceleration of 28.0 m/s^2. At the time, it’s engines fail and the rocket proceeds to move as a projectile
Answers
= 100 + 30*3 = 190 m/s
lets call the maximum altitude of the rocket H and say that the height reached while the engines are running is h. Also we'll call the angle of the launch a.
we can get an expression for h as follows.
h = vsin(a)t + (1/2)at^2
h = 100*sin(53)*3 + 0.5*30*9
h = 374.6 m
and we can write an expression for the time it takes the rocket to reach maximum height after the engines have failed.
t = [0 - vsin(a)]/-g
= vsin(a)/g
then
H-h = vsin(a)t - 1/2*gt^2
if we sub in the expression for t and rearrange we get
H-h = [(vsin(a)^2]/2g
now add h
H = [(vsin(a)^2]/2g + h
H = 1549 m
b)
to obtain the total time of flight, first work out the time it takes for the rocket to fall from the top of its trajectory.
t = (v-u)/g
where v is now the final vertical velocity and u is the initial vertical velocity.
to get v
v^2 = u^2 + 2gH
so t = [Sqrt(2gH)]/g
now the total time is the falling time plus the time the engines are running, plus the time we calculated previously.
t = [Sqrt(2gH)]/g + 3 + vsin(a)/g
t = 36.72 s
c)
calculating the range is simple since the horizontal velocity stays the same for most of the flight.
We just need to work out the bit of distance traveled while the rocket accelerates for the first 3 seconds of flight.
r = ut + 1/2at^2
r = 100*3+0.5*30*9
r = 435 m
now we take the remaining flight time and put it in the following formula to work out the rest of the distance.
R-r = vcos(a)t
= 190*cos(53)*33.72
= 3856 m
add this to the previous distance we calculated to obtain the range
R = 3856 + 435
R = 4291 m
Hopefully that's right, I may have made an error in there so make sure you check through these calculations yourself to confirm. Hope this helps. :)
Questioning is the beginning of the scientific inquiry process. True or False
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Hope this helped! Please give me Brainliest!:)
Suppose you are a particle of water in a lake. Describe what happens to you when a motorboat passes by. Be sure to use words like vibration and crest in your description.
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Because not all stars are the same distance from the earth, it can be difficult to determine
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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?\
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M=mass=2250g=2.25Kg
C=specific heat capacity of water= 4.18KJ/Kg
ΔT= change in temp= ?
E=mcΔT
509=(2.25)x(4.18)xΔT
509=9.405ΔT
ΔT=509/9.405=54.1degrees
Initial temp = 100-54 = 46 degrees
Hope this helps :)
Which of the following is an appropriate unit for speed? *note: you can choose more than one.
A. miles/hour
B. meters/second
C. blocks/min
D. newtons/sec
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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
Answers
Good luck! (:
if you were trying to locate a single star that was less luminous than the sun, but with hotter surface temperature, which group would you look in?
A. Super Giants
B. White Dwarfs
C. Main Sequence
D. Giants
Answer = ???
What do you guys think?
Answers
White Dwarfs are less luminous than the sun, but with hotter surface temperatures. Therefore, option (B) is correct.
What are White Dwarfs?A white dwarf can be described as a stellar core remnant made mostly of electron-degenerate matter. A white dwarf is dense because its mass is comparable to the Sun's and volume is comparable to the Earth's.
A white dwarf has faint luminosity which comes from the emission of residual thermal energy and no fusion occurs in a white dwarf. There are currently eight white dwarfs among the 100 star systems nearest the Sun.
The material in a white dwarf has no fusion reactions, so there is no source of energy. It cannot support itself by the energy generated by fusion but is supported by electron degeneracy pressure.
A carbon-oxygen white dwarf approaches this mass limit by mass transfer from a companion star and may explode as a kind Ia supernova via a process called carbon detonation.
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Which best describes the beginning of the Big Bang Theory?
A. All matter in the universe was compressed into a single point.
B. The universe will eventually collapse into a black hole.
C. Stars form from giant masses of gas and dust.
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A. All matter in the universe was compressed Into a single point.
(apex)
A. All matter in the universe was compressed Into a single point. this option describes the big bang theory.
What is the Big Bang theory?The Big Bang hypothesis states that all of the current and past matter in the Universe came into existence at the same time, roughly 13.8 billion years ago.
At this time, all matter was compacted into a very small ball with infinite density and intense heat called Singularity.
The universe begin;
The Big Bang was the moment 13.8 billion years ago when the universe began as a tiny, dense, fireball that exploded.
Here Most astronomers use the Big Bang theory to explain how the universe began, But what caused this explosion in the first place is still a mystery.
Thus matter was compressed into a single point and then exploded outward to form the universe describes the big bang theory.
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"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 .
Answers
Fx = [(233 + 840)/g]*v²/7.5
v = 32.3*2*π*7.5/60 m/s = 25.37 m/s
The horizontal component of force from the cables is Th + Ti*sin40º and the vertical component of force from the cable is Ta*cos40º
Thh horizontal and vertical forces must balance each other. First the vertical components:
233 + 840 = Ti*cos40º
solve for Ti. (This is the answer to the part b)
Horizontally
[(233 + 840)/g]*v²/7.5 = Th + Ti*sin40º
Solve for Th
Th = [(233 + 840)/g]*v²/7.5 - Ti*sin40º
using v and Ti computed above.
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
Answers
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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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.
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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.
Answers
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)
Y is a geometric factor related to the shape of the crack, which is assumed to be 1.0 for this problem
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.
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.
Answers
(85 kg)*(9.8 m/s²) = 833 N
Because the applied pressure is 9530 N/m², the weight is applied over an area, A, such that
(833 N)/(A m²) = 9530 N/m²
A = 833/9530 = 0.0874 m²
The total surface area of the human body ranges from 1.0 to 2.0 m².
Because the calculated area is very small compared to the surface area of the human body, Morgan is standing on the bed, possibly in high heels.
Answer: Standing 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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A box is at rest with respect to the surface of a flatbed truck. the coefficient of static friction between the box and the surface is μs. (a) find an expression for the maximum acceleration of the truck so that the box remains at rest with respect to the truck. your expression should be in terms of μs and g. how does your answer change if the mass of the box is doubled?
Answers
The maximum acceleration of the truck so that the box remains at rest can be determined using the coefficient of static friction and gravity, which gives us an expression of a_max = µ_s * g. Doubling the mass of the box does not change the maximum acceleration.
Explanation:The maximum acceleration of the truck such that the box remains at rest can be derived from Newton's Second Law, which shows that the force of friction must balance the force due to acceleration. The force of static friction can be calculated as the coefficient of static friction times the normal force, which for a box on a flatbed truck is equal to the mass of the box times gravity. In equation form, we have F_f = µ_s * m * g, where F_f is the static friction force, µ_s is the coefficient of static friction, m is the mass of the box, and g is the acceleration due to gravity.
To calculate the maximum acceleration of the truck, we can equate the force of static friction to the force due to acceleration (F = m * a), which gives us the equation µ_s * m * g = m * a_max, where a_max is the maximum acceleration. Simplifying the equation gives us a_max = µ_s * g as the maximum acceleration of the truck such that the box remains at rest.
If the mass of the box is doubled, the maximum acceleration of the truck required for the box to remain at rest would remain the same. This is because while the force of friction would double due to the increased mass, the force needed to accelerate the box would also double, keeping the acceleration unchanged.
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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.
Answers
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 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?
Answers
The momentum of ball 1 should be:
momentum 1= mass1 * velocity1
momentum 1= 2kg * 10m/s= 20kg m/s
The momentum of ball 2 should be:
momentum 2= mass2 * velocity2
momentum 2= 1.5kg * 20 m/s= 30kg m/s
The answer would be: 1.5kg ball because it was travelling faster
A 5000 kg truck is parked on a 7.0â slope. how big is the friction force on the truck?
Answers
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
Answers
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.
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?
Answers
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:
parallel and series circuits
I am quite confused as to how I can tackle this question
Answers
Read it from the top down.
If you ever come back again, with one that has
an AC source and an L or a C in the circuit, I'll
be looking for more than 5 points for that one. :-)
During a softball game, a shortstop catch a ground ball. The action forces is the ball pushing on the glove. What is the reaction?
A. The fielder pushing on the ball
B. The ball pushing on the ground
C. The glove pushing on the ball
D. The ground pushing on the ball
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Answer:
The answer is C. The glove pushing on the ball. I took a test that had this exact question and it was correct.
Why do ultraviolet rays from the sun reach earth
Answers
learned this in my class.. Hope it helps. :)
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?
Answers
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:
Problem 2.1 (closed system) a piston-cylinder device contains 25 g of saturated water vapor that is maintained at a constant pressure of 300 kpa. a resistance heater within the cylinder is turned on and passes a current i) of 0.2 a for 5 min for a 120 v source. at the same time, a heat loss of 3.7 kj occurs. determine the final temperature of the steam.
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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?
Answers
Hope it helps :)
A fly has a mass of 1 gram at rest. how fast would it have to be traveling to have the mass of a large suv, which is about 3000 kilograms?
Answers
We solve this using special
relativity. Special relativity actually places the relativistic mass to be the
rest mass factored by a constant "gamma". The gamma is equal to 1/sqrt
(1 - (v/c)^2).
We want a ratio of 3000000 to 1, or 3 million to 1.
Therefore:
3E6 = 1/sqrt (1 - (v/c)^2)
1 - (v/c)^2 = (0.000000333)^2
0.99999999999999 = (v/c)^2
0.99999999999999 = v/c
v= 99.999999999999% of the speed of light ~ speed of light
v = 3 x 10^8 m/s
Final answer:
A fly must travel at speeds approaching the speed of light to have the mass of an SUV according to the theory of relativity. However, such speeds are practically impossible to achieve due to energy limitations and physical laws.
Explanation:
To understand how fast a fly must travel to have the mass of a large SUV, we first need to consider the concept of relativistic mass.
Relativistic mass is an object's mass when it is moving and is given by the equation m = m0 / sqrt(1 - v^2/c^2), where m0 is the rest mass, v is the velocity of the object, and c is the speed of light in a vacuum (approximately 3 x 108 m/s).
Setting the relativistic mass m equal to the mass of an SUV (3000 kg) and solving for v theoretically reveals the speed. However, the answer lies in the realm of speeds very close to the speed of light, indicating an exponential increase in mass as speed approaches c.
This illustrates a principle of Einstein's theory of relativity: as an object's speed approaches the speed of light, its mass approaches infinity, requiring infinite energy to reach c.
Practically, even for an immense acceleration, a fly cannot reach such speeds due to the limitations of energy sources and the law of physics as we understand them. Thus, the question is more theoretical than practical, highlighting the fascinating insights of relativistic physics rather than providing a scenario that could occur in the real world.
the blank secrets chemical messages that circulate through the blood and also communicates messages that influence behavior and many aspects of the biological functioning
Answers
Hope this helps!
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