[tex]\rm 220^oC[/tex] is the surface temperature of the rod. The fuel must be kept below [tex]\rm 300^oC[/tex] to avoid reaching the cooling water from the critical heat flux and the temperature below that it is reasonable.
The average kinetic energy of particles in a medium, such as a gas, liquid, or solid, is measured as temperature. It is a basic physical characteristic that enables us to comprehend how hot or cold something is. Depending on the location, temperature is often measured using different scales, such as Celsius or Fahrenheit, or in scientific situations, Kelvin (K). Temperature, said simply, tells us how quickly the particles in a substance are moving. The substance is hotter or cooler depending on how quickly the particles are moving.
Heat generated = Heat transferred by convection + Heat conducted
[tex]\rm q_{gen }= q'' \times V\\\rm q_{gen} = 150 \times 10^{6 }W/m^3 \times \pi \times r^2\\\rm q_{conv} = h \times A \times (T_{s }- T )\\\rm q_{conv} = 5000 W/m^2\timesK \times2 \times \pi \times r \times L \times (T_s - T)\\\rm q_{cond} = k \times A \times (T_{s} - T) / L\rm 150 \times 10^6 \times \times \times r^2= 5000 \times 2 \times \pi \times r \timesL \times (T_{s} - T) + k \times 2 \times \times \times r \timesL \times (T_{s} - T) / L\\\rm T_{s }=220^oC[/tex]
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What equation gives the position at a specific time for an object with constant acceleration?
Answer:
[tex]x = x_{0}+v_{0}t+\frac{1}{2}at^{2}[/tex] equation gives the position at a specific time for an object with constant acceleration
Explanation:
[tex]x = x_{0}+v_{0}t+\frac{1}{2}at^{2}[/tex] equation gives the position at specific time for an object having constant acceleration. Constant acceleration is referred as the change of velocity with respect to the time is known as the acceleration, but when the velocity changes occurs at constant rate this rate is termed as the constant acceleration. The constant acceleration can never be zero. The velocity changes but that change occurs in the consistently. The acceleration is affected by the mass.
A ______ cell is the pattern mantle material takes as it rises when heated, then moves laterally as it cools, and sinks once its density is great enough.
Answer:
Convection cell
Explanation:
The convection current refers to the upward rising of the hot molten and less dense magma in the mantle. These currents are formed due to the heat supplied from the core of the earth. This upward movement of magma results in the movement of the lithospheric plates over the less dense viscous layer of the asthenosphere.
As the magma at the greater depth gets heated up, due to the increasing heat and pressure, it gets heated up and rises upward in the form of convection cells. As it rises upward, it cools and slowly moves laterally and eventually sinks after its density increases. This cycle repeats and forms the main mechanism for the plate tectonic movement.
A small ball is attached to one end of a spring that has an un- strained length of 0.200 m. The spring is held by the other end, and the ball is whirled around in a horizontal circle at a speed of 3.00 m/s. The spring remains nearly parallel to the ground during the motion and is observed to stretch by 0.010 m. By how much would the spring stretch if it were attached to the ceiling and the ball allowed to hang straight down, motionless?
Answer:
[tex]\Delta x=0.002287\ m=2.287\ mm[/tex]
Explanation:
Given:
un-stretched length of the spring, [tex]l=0.2\ m[/tex]speed of revolution of the ball in horizontal plane, [tex]v=3\ m.s^{-1}[/tex]length stretch during the motion, [tex]\Delta l=0.01\ m[/tex]Now the radius of revolution of the ball:
[tex]r=l+\Delta l[/tex]
[tex]r=0.2+0.01[/tex]
[tex]r=0.21\ m[/tex]
Now in this case the centrifugal force is equal to the spring force:
[tex]F_c=F_s[/tex]
[tex]m.\frac{v^2}{r} =k.\Delta l[/tex]
where:
m = mass of the ball
k = spring constant
[tex]m\times \frac{3^2}{0.21} =k\times 0.01[/tex]
[tex]k=(4285.714\times m)\ N.m^{-1}[/tex]
Now the extension in the spring upon hanging the ball motionless:
[tex]m.g=k.\Delta x[/tex]
[tex]9.8\times m=(4285.714\times m)\times \Delta x[/tex]
[tex]\Delta x=0.002287\ m=2.287\ mm[/tex]
You find a featureless black slab. There are two arms marked In and OUT. You find that moving the IN arm 20 inches causes the OUT arm to move 5 inches. You find that a 10 lb. pull on IN lifts 30 lbs. on OUT. What is the AMA?a. 1
b. 4
c. 3
d. 1/3
Answer:
c. 3
Explanation:
Given:
displacement in the input arm, [tex]d_i=20\ in[/tex]corresponding displacement in the output arm, [tex]d_o=5\ in[/tex]load on the output arm, [tex]F_o=30\ lbs[/tex]corresponding load on the input arm, [tex]F_i=10\ lbs[/tex]Since AMA i.e. actual mechanical advantage is defined as the ratio of the output force to the input force on a simple machine. This takes into account the losses occurred.
Now, mathematically:
[tex]AMA=\frac{F_o}{F_i}[/tex]
[tex]AMA=\frac{30}{10}[/tex]
[tex]AMA=3[/tex]
Answer:
kuh
Explanation:
If the skater started from rest 4 m above the ground (instead of 7m), what would be the kinetic energy at the bottom of the ramp (which is still 1 m above the ground)?
--> mass of skater is 75 kg and the acceleration of gravity is 9.81 N/kg
If the skater started from rest 4 above the ground (instead of 7), what would be the kinetic energy at the bottom of the ramp (which is still 1 above the ground)?
a. 735 J
b. 2940 J
c. 2205 J
d. 4410 J
Answer:
C. 2205 J
Explanation:
First we need to find the final velocity of skater at the bottom.
We, have:
Height lost = h = 4 m - 1 m = 3 m
Initial Velocity = Vi = 0 m/s (since, it starts from rest)
acceleration due to gravity = g = 9.8 m/s²
using third equation of motion:
2gh = Vf² - Vi²
2(9.8 m/s²)(3 m) = Vf² - (0 m/s)²
Vf² = 58.8 m²/s²
Now, for kinetic energy at bottom:
K.E = (1/2) m Vf²
K.E = (1/2)(75 kg)(58.8 m²/s²)
K.E = 2205 J
Hence, the correct option is C. 2205 J
Jaclyn plays singles for South's varsity tennis team. During the match against North, Jaclyn won the sudden death tiebreaker point with a cross-court passing shot. The 57.5-gram ball hit her racket with a northward velocity of 26.7 m/s. Upon impact with her 331-gram racket, the ball rebounded in the exact opposite direction (and along the same general trajectory) with a speed of 29.5 m/s.
A. Determine the pre-collision momentum of the ball.
B. Determine the post-collision momentum of the ball.
C. Determine the momentum change of the ball.
D. Determine the velocity change of the racket.
Answer:
A) pbin = 1.535 Kgm/s (+)
B) pbf = 1.696 Kgm/s (-)
C) Δp = 3.3925 Kgm/s
D) Δvr = 10.249 m/s
Explanation:
Given
Mass of the ball: m = 57.5 g = 0.0575 Kg
Initial speed of the ball: vbi = 26.7 m/s
Mass of the racket: M = 331 g = 0.331 Kg
Final speed of the ball: vbf = 29.5 m/s
A) We use the formula
pbin = m*vbi = 0.0575 Kg*26.7 m/s = 1.535 Kgm/s (+)
B) pbf = m*vbf = 0.0575 Kg*29.5 m/s = 1.696 Kgm/s (-)
C) We use the equation
Δp = pbf - pbin = 1.696 Kgm/s - (-1.535 Kgm/s) = 3.3925 Kgm/s
D) Knowing that
Δp = 3.3925 Kgm/s
we can say that
Δp = M*Δvr
⇒ Δvr = Δp / M
⇒ Δvr = 3.3925 Kgm/s / 0.331 Kg
⇒ Δvr = 10.249 m/s
If you ignore air resistance after an initial force launches a projectile, name all forces acting on it as it moves through the air. What does this force cause the object to do?
Answer:
Neglecting air resistance, the only force acting on a projectile is gravity.
This force causes the object to accelerate.
Explanation:
As a projectile moves upward, there is a downward force and a downward acceleration due to force of gravity. That is, as the object is moving upward, force of gravity acting on the projectile is causing a steady slowing down of the projectile.
Hence, Gravity is the downward force upon a projectile that influences its vertical motion and causes the parabolic trajectory that is characteristic of projectiles.
From Newton's law of motion, it suggest that force is required to cause an acceleration and not motion. Therefore, force of gravity causes the object to accelerate downwards.
Which of the following was NOT considered a key characteristic of the genetic material even before nucleic acids were identified as the genetic material?
Genetic material must encode the phenotype.
Genetic material must encode complex information.
Genetic material must be copied accurately.
Genetic material must have a complex structure.
These are all key characteristics of genetic material.
Answer:
D:Genetic material must have a complex structure.
Before DNA was identified as the genetic material, a complex structure was not considered a key characteristic of genetic material. Proteins, with their diverse amino acids, were initially thought to be the genetic carriers due to their apparent structural complexity compared to DNA's four nucleotides.
Explanation:The one characteristic that was NOT considered a key attribute of the genetic material before nucleic acids were identified is that the genetic material must have a complex structure. Historically, even though DNA does have a complex structure, early scientific thought leaned towards proteins as the biomolecules with sufficient complexity to encode genetic information. This belief was due to proteins being composed of 20 different amino acids, allowing for a high degree of variability and complexity, while DNA, composed of only four nucleotides, seemed too simple to account for the vast genetic diversity observed in living organisms.
It wasn't until the transformative experiments of Avery, McCarty, McLeod, and eventually the definitive work of Hershey and Chase, that DNA was confirmed to be the genetic material. The initial complexity perceived in proteins overshadowed the abstract concept of steep structural complexity in DNA. Afterward, it was understood that through the processes of replication and expression, DNA's arrangement of four bases could indeed encode and express the complexity necessary for life.
When two point charges are a distance dd part, the electric force that each one feels from the other has magnitude F.F . In order to make this force twice as strong, the distance would have to be changed to _____
Answer:
In order to make this force twice as strong, F' = 2 F, the distance would have to be changed to half i.e. r' = r/2.
Explanation :
The electric force between two point charges is directly proportional to the product of charges and inversely proportional to the square of the distance between charges. It is given by :
[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]
r is the separation between charges
[tex]F\propto \dfrac{1}{r^2}[/tex]
[tex]r=\sqrt{\dfrac{1}{F}}[/tex]
If F'= 2F
[tex]r'=\dfrac{1}{\sqrt{2F} }[/tex]
In order to make this force twice as strong, F' = 2 F, the distance would have to be changed to half i.e. [tex]r'=\dfrac{1}{\sqrt{2F} }[/tex]. Hence, this is the required solution.
A ball is thrown straight up, reaches a maximum height, then falls to its initial height. Make a statement about the direction of the velocity and acceleration as the ball is coming down.
Answer:
Explanation:
When the ball is thrown upwards it's velocity is pointing upwards while acceleration i.e. acceleration due to gravity is pointing downwards.
As the ball moves upwards it's velocity is decreasing and the ball reaches maximum height it's velocity becomes zero while the acceleration is still pointing downwards.
During downward motion, both velocity and acceleration are pointing downward.
The total electric flux through a closed cylindrical (length = 1.2 m , diameter = 0.20 m) surface is equal to −5 0. Nm / C^2. Determine the net charge within the cylinder in pC? (εo = 8.85z10^-12)
a) – 71
b) – 62
c) – 53
d) – 44
e) –16
Final answer:
The net charge within the cylinder, given the electric flux of -50 Nm²/C² and the permittivity of free space, is -44 picocoulombs (pC).
Explanation:
The question asks to determine the net charge within a cylindrical surface given the total electric flux and the electrical permittivity of free space. According to Gauss's law, the electric flux (φ) through a closed surface is equal to the net charge (Q) enclosed by the surface divided by the permittivity of free space (ε0). The formula is φ = Q / ε0. Given the electric flux is -50 Nm2/C2 and ε0 = 8.85 x 10-12 C2/Nm2, we can solve for Q:
Q = φ x ε0
Q = (-50 Nm2/C2)(8.85 x 10-12 C2/Nm2)
Q = -4.425 x 10-10 C
To convert this to picocoulombs (pC), we multiply by 1012 pC/C:
Q = -4.425 x 10-10 C x 1012 pC/C = -442.5 pC
The closest answer to -442.5 pC is -44 pC, so the correct answer is (d) -44.
What are we calculating when we calculate expected frequencies? What is the reason for calculating expected frequencies the way we do? In laymen’s terms, what do expected frequencies tell us?
Answer:
1. What we calculate when we calculate expected frequency is the predicted frequency that can be obtained from an experiment whose outcome is expected to be true.
2. The reason for using expected frequency is because it is a tool used for doing complex probability calculations and predictions. Also, it serves as a working tool that can be used in place of the observed frequency,when the observed frequency is not available.
3. In probability, expected frequency tell us the possible outcomes of an event.
Say, for example rolling of a fair die.
The probability of getting 1,2....,6 as the outcome is 1/6 but that doesn't mean that when we roll a fair die 6 times, we'll get outcomes of 1-6.
So, the essence of expected frequency is to tell us what to expect in an event (this may or may not turn out to be true).
Final answer:
Expected frequencies estimate how often an event should occur based on a probability distribution, providing a basis for comparing actual outcomes in a data set to determine consistency with expected probabilities.
Explanation:
When we calculate expected frequencies, we are determining how often we anticipate an event will occur based on a probability distribution. To find the expected frequency, one typically multiplies the expected percentage by the total number of observations or the sample size. For instance, if an event has a 10% chance of occurring out of 600 trials, the expected frequency would be 0.10 * 600 = 60.
The reason for calculating expected frequencies in this manner is to provide a basis for comparison with the actual frequencies that occur in a data set. This comparison allows us to determine if the outcomes observed are consistent with the expected probabilities, which can be particularly useful in statistical hypothesis testing, such as the chi-square test.
Expected frequencies give us insight into what we might predict to occur over a long period of trials or observations. They are theoretical estimates that can be appraised against actual outcomes to understand the probability distribution of a given scenario.
A cargo elevator on Earth (where g = 10 m/s2) lifts 3000 kg upwards by 20 m. 720 kJ of electrical energy is used up in the process. What is the efficiency of this process?
Answer: 83%
Explanation:
Efficiency of the process = work output/work input × 100%
Work input is the energy used up in the process = 720,000Joules
Work output = Force × distance
= (3000×10)× 20
= 600000 Joules
Efficiency= 600000/720000 × 100
= 0.83×100
= 83%
Finally, you are ready to answer the main question. Cheetahs, the fastest of the great cats, can reach 50.0 miles/hour in 2.22 s starting from rest. Assuming that they have constant acceleration throughout that time, find their acceleration in meters per second squared.
Explanation:
We have equation of motion v = u + at
Initial velocity, u = 0 m/s
Final velocity, v = 50 miles/hr = 80 km/hr = 22.22 m/s
Time, t = 2.22 s
Substituting
v = u + at
22.22 = 0 + a x 2.22
a = 10 m/s²
Acceleration is 10 m/s²
Acceleration of cheetah is 10 m/s²
A statement that the energy supplied to a system in the form of heat, minus the work done by the system, is equal to the change in internal energy represents the ___________
Answer:
First law of thermodynamics.
Explanation:
The first law of thermodynamics states that the change in internal energy of a system ΔE equals to the amount of heat energy supplied to the system H , minus the work done by the system on its surroundings. W.
The above definition can be expressed mathematically as: ΔE = H - W
This can also be restarted as energy of a close or isolated system is constant.
An energy storage system based on a flywheel (a rotating disk) can store a maximum of 4.1 MJ when the flywheel is rotating at 14000 revolutions per minute.
Incomplete question.The complete question is here
An energy storage system based on a flywheel (a rotating disk) can store a maximum of 4.1 MJ when the flywheel is rotating at 14000 revolutions per minute.What is the moment of inertia of the flywheel?
Answer:
Moment of inertia=3.82 kg.m²
Explanation:
Given data
Energy E=4.1 MJ=4.1×10⁶J
Angular speed ω= 14000 rev/min= 14000×2π/60=1465.3 rad/s
To find
Moment of Inertia
Solution
A rotating flywheel must store its energy as kinetic energy
Let I be moment of inertia of the flywheel.We Know
[tex]E(Energy)=\frac{1}{2}Iw^{2}\\ I=\frac{2E}{w^{2} }\\ I=\frac{2(4.1*10^{6}) }{(1465.3)^{2} }\\ I=3.82 kgm^{2}[/tex]
The subject of this question is Physics and it relates to an energy storage system based on a flywheel.
Explanation:The subject of this question is Physics. The question relates to an energy storage system based on a flywheel.
A flywheel is a rotating disk that can store energy in the form of rotational kinetic energy. The maximum energy that can be stored in the flywheel depends on its moment of inertia and the square of its angular velocity.
In this case, the flywheel can store a maximum of 4.1 MJ when it is rotating at 14000 revolutions per minute.
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A total resistance of 3.03 Ω is to be produced by connecting an unknown resistance to a 12.18 Ω resistance. (a) What must be the value of the unknown resistance, and
Answer:
(a) 4.0334Ω
(b)parallel
Explanation:
for resistors connected in parallel;
[tex]\frac{1}{R_{eq} } =\frac{1}{R1}+\frac{1}{R2}[/tex]
Req =3.03Ω , R1 =12.18Ω
[tex]\frac{1}{3.03 } =\frac{1}{12.18}+\frac{1}{R2}[/tex]
[tex]\frac{1}{R2}=\frac{1}{3.03 }-\frac{1}{12.18}[/tex]
[tex]\frac{1}{R2}=0.2479[/tex]
R2=1/0.2479
R2=4.0334Ω
(b)parallel connection is suitable for the desired total resistance. series connection can not be used to achieve a lower resistance as the equation for series connection is.
Req = R1+R2
To achieve a total resistance of 3.03 Ω when an unknown resistor is connected in parallel with a 12.18 Ω resistor, the unknown resistance must be 4.03 Ω.
Explanation:To determine the unknown resistance that needs to be connected to a 12.18 Ω resistor to produce a total resistance of 3.03 Ω, we can use the formula for resistors in parallel, since the total resistance is given and is less than the resistance of the known resistor. The formula is:
1 / Rtotal = 1 / R1 + 1 / R2
We can arrange the formula to solve for R2, the unknown resistance:
1 / R2 = 1 / Rtotal - 1 / R1
By plugging in the given values (Rtotal = 3.03 Ω, R1 = 12.18 Ω), we get:
1 / R2 = 1 / 3.03 - 1 / 12.18
Calculating the right side of the equation:
1 / R2 = 0.330 - 0.0821
1 / R2 = 0.2479
Finally, we take the reciprocal of both sides to find R2:
R2 = 1 / 0.2479
R2 = 4.03 Ω
So, the unknown resistance must be 4.03 Ω to achieve the specified total resistance when connected in parallel with a 12.18 Ω resistor.
A container of water is lifted vertically 3.0 m then returned to its original position. If the total weight is 30 N, how much work total was done?
Answer:
Work done, W = 0
Explanation:
Given that, a container of water is lifted vertically 3.0 m then returned to its original position.The total weight of the container is 30 N. We need to find the total work done by the container. We know that the work done by an object is given by :
W = F × d
Where
F is the applied force
d is the displacement of the object
It is mentioned that the container returns to its original position, the displacement of the container is equal to 0. Hence, no work was done by the container of water.
Answer: No work was done when a container of water is lifted vertically 3.0 m and then returned to its original position.
Explanation:
Work is defined as a force causing the movement or displacement of an object.
work=[tex]f\times s[/tex]
Given: f = force = 30N
s = displacement = 0.0 m (as the initial and final position is same, there is no displacement)
Putting in the values we get,
work=[tex]30N\times 0m=0J[/tex]
Thus amount of work will be 0 J.
An airplane weighing 5000 lb is flying at standard sea level with a velocity of 200 mi/h. At this velocity the L/D ratio is a maximum. The wing area and aspect ratio are 200 ft2 and 8.5, respectively. The Oswald efficiency factor is 0.93. Calculate the total drag on the airplane.
Answer:
98.15 lb
Explanation:
weight of plane (W) = 5,000 lb
velocity (v) = 200 m/h =200 x 88/60 = 293.3 ft/s
wing area (A) = 200 ft^{2}
aspect ratio (AR) = 8.5
Oswald efficiency factor (E) = 0.93
density of air (ρ) = 1.225 kg/m^{3} = 0.002377 slugs/ft^{3}
Drag = 0.5 x ρ x [tex]v^{2}[/tex] x A x Cd
we need to get the drag coefficient (Cd) before we can solve for the drag
Drag coefficient (Cd) = induced drag coefficient (Cdi) + drag coefficient at zero lift (Cdo)
where
induced drag coefficient (Cdi) = [tex]\frac{Cl^{2} }{n.E.AR}[/tex] (take note that π is shown as n and ρ is shown as [tex]p[/tex])where lift coefficient (Cl)= [tex]\frac{2W}{pAv^{2} }[/tex]=[tex]\frac{2x5000}{0.002377x200x293.3^{2} }[/tex] = 0.245
therefore
induced drag coefficient (Cdi) = [tex]\frac{Cl^{2} }{n.E.AR}[/tex] = [tex]\frac{0.245^{2} }{3.14x0.93x8.5}[/tex] = 0.0024
since the airplane flies at maximum L/D ratio, minimum lift is required and hence induced drag coefficient (Cdi) = drag coefficient at zero lift (Cdo)Cd = 0.0024 + 0.0024 = 0.0048Now that we have the coefficient of drag (Cd) we can substitute it into the formula for drag.
Drag = 0.5 x ρ x [tex]v^{2}[/tex] x A x Cd
Drag = 0.5 x 0.002377 x (293.3 x 293.3) x 200 x 0.0048 = 98.15 lb
The total drag on the airplane is approximately 376.4 pounds (lb).
To calculate the total drag on the airplane, we'll use the following formula for total drag:
[tex]\[D = \frac{W^2}{q \cdot S \cdot \pi \cdot e \cdot AR}\][/tex]
Where:
- \(D\) is the total drag in pounds (lb).
- \(W\) is the weight of the airplane in pounds (lb), which is 5000 lb in this case.
- \(q\) is the dynamic pressure in lb/ft^2, given by [tex]\(q = 0.5 \cdot \rho \cdot V^2\)[/tex], where[tex]\(\rho\)[/tex] is the air density at standard sea level (0.002377 lb/ft^3) and \(V\) is the velocity in ft/s (200 mi/h converted to ft/s).
- S is the wing area in square feet (ft^2), which is 200 ft^2.
- pi is approximately 3.1416.
- e is the Oswald efficiency factor, which is 0.93 in this case.
- AR is the aspect ratio, which is 8.5 in this case.
Let's calculate it step by step:
1. Convert the velocity from miles per hour (mi/h) to feet per second (ft/s):
[tex]\[V = 200 \, \text{mi/h} \times \frac{5280 \, \text{ft/mi}}{3600 \, \text{s/h}} = 293.33 \, \text{ft/s}\][/tex]
2. Calculate the dynamic pressure (\(q\)):
[tex]\[q = 0.5 \cdot \rho \cdot V^2 = 0.5 \cdot 0.002377 \, \text{lb/ft}^3 \cdot (293.33 \, \text{ft/s})^2 = 21.81 \, \text{lb/ft}^2\][/tex]
3. Now, plug the values into the total drag formula:
[tex]\[D = \frac{5000 \, \text{lb}^2}{21.81 \, \text{lb/ft}^2 \cdot 200 \, \text{ft}^2 \cdot 3.1416 \cdot 0.93 \cdot 8.5} \approx 376.4 \, \text{lb}\][/tex]
So, the total drag on the airplane is approximately 376.4 pounds (lb).
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When Earth catches up to a slower-moving outer planet and passes it in its orbit the planet
a.exhibits retrograde motion.
b. slows down because it feels Earth's gravitational pull.
c. decreases in brightness as it passes through Earth's shadow.
d. moves into a more elliptical orbit.
Answer:
a. exhibits retrograde motion.
Explanation:
When earth catches up to a slower moving outer planets like for example Saturn and jupitar they seems to show the retrograde motion which is actually an illusion but there exits the real retrograde motion too in some cases. In retrograde motion planets appears to move in the backward direction like from east to west rather then from west to east in the sky.
A 6.0-cm-diameter horizontal pipe gradually narrows to 4.5 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 31.0 kPa and 24.0 kPa, respectively. What is the volume rate of flow?
Answer:
0.0072 m³/s
Explanation:
Using Bernoulli's law
P₁ + 1/2ρv₁² = P₂ + 1/2ρv₂ since the pipe is horizontal
1/2ρv₂² - 1/2ρv₁² = P₁ - P₂
flow rate is constant
A₁v₁ = A₂v₂
A₁ = πr₁² = π (0.06/2)² = 0.0028278 m²
A₂ = πr₂² = π (0.0225)² = 0.00159 m²
v₁ = (A₂ / A₁)v₂
v₁ = (0.00159 m²/ 0.0028278 m²) v₂ = 0.562 v₂
substitute v₁ into the Bernoulli's equation
1/2ρv₂² - 1/2ρv₁² = P₁ - P₂
500 ( 1 - 0.3161 ) v₂² = (31.0 - 24 ) × 10³ Pa
341.924 v₂² = 7000
v₂² = 20.472
v₂ = √ 20.472 = 4.525 m/s
volume follow rate = 0.00159 m² × 4.525 m/s = 0.0072 m³/s
[tex]0.0072 \;\rm m^{3}/s[/tex]The volume flow rate at the exit of the pipe is [tex]0.0072 \;\rm m^{3}/s[/tex].
Given data:
The initial diameter of horizontal pipe is, d = 6.0 cm.
The final diameter of pipe is, d' = 4.5 cm.
The gauge pressure at inner section is, P = 31.0 kPa.
The gauge pressure at outer section is, P' = 24.0 kPa.
Applying the Bernoulli's concept, which says the total pressure energy and kinetic energy throughout the flow remains constant .
So, for the horizontal pipe, the expression is,
[tex]P + \dfrac{1}{2} \rho v^{2} = P' + \dfrac{1}{2} \rho v'^{2}[/tex] ............................................(1)
Here, [tex]\rho[/tex] is the density of water throughout the flow, which remains constant.
Now, apply the continuity equation as,
[tex]A\times v = A' \times v'\\\\(\pi/4 \times d^{2}) \times v = (\pi/4 \times d'^{2}) \times v'\\\\ d^{2} \times v = d'^{2} \times v'\\\\v/v' = 0.045^{2}/0.006^{2}\\\\v = 0.5625\times v'[/tex]
Now substitute the value in equation (1) as,
[tex]P -P' = \dfrac{1}{2} \rho v'^{2} - \dfrac{1}{2} \rho v^{2}\\\\(31 -24) \times 10^{3} \;\rm Pa = \dfrac{1}{2} \times 1000 v'^{2} - \dfrac{1}{2} \times 1000 (0.5625 v')^{2}\\v' = 4.52 \;\rm m/s[/tex]
Then the flow rate is calculated as,
[tex]Q' = A' \times v'\\\\Q' = (\pi /4) d'^{2} \times v'\\Q' = (\pi /4) \times 0.06'^{2} \times 4.52\\\\Q' = 0.0072 \;\rm m^{3}/s[/tex]
Thus, the required value of volume flow rate is, [tex]0.0072 \;\rm m^{3}/s[/tex].
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Two steamrollers begin 105 m apart and head toward each other, each at a constant speed of 1.30 m/s . At the same instant, a fly that travels at a constant speed of 2.40 m/s starts from the front roller of the southbound steamroller and flies to the front roller of the northbound one, then turns around and flies to the front roller of the southbound once again, and continues in this way until it is crushed between the steamrollers in a collision.What distance does the fly travel?
According to the question,
Object's relative speed,
[tex]v_{rel} = 1.30+1.30[/tex]
[tex]= 2.60 \ m/s[/tex]
Time taken by the two steamrollers,
[tex]t = \frac{d}{v_{rel}}[/tex]
[tex]= \frac{105}{2.60}[/tex]
[tex]= 40.39 \ seconds[/tex]
hence,
The distance of fly travel will be:
→ [tex]s = vt[/tex]
By putting the values,
[tex]= 2.40\times 40.39[/tex]
[tex]= 96.94 \ m[/tex]
Thus the above response is right.
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Both a gage and a manometer are attached to a gas tank to measure its pressure. If the reading on the pressure gage is 80 kPa, determine the distance between the two fluid levels of the manometer if the fluid is
(a) mercury (r 5 13,600 kg/m3) or
(b) water (r 5 1000 kg/m3).
To determine the distance between the two fluid levels of the manometer, we use the equation P = ρgh. For mercury, the density is 13,600 kg/m3, and for water, the density is 1000 kg/m3.
Explanation:To determine the distance between the two fluid levels of the manometer, we need to consider two cases: one with mercury and one with water.
(a) For mercury, we can use the equation P = ρgh, where P is the pressure difference, ρ is the density of mercury, g is the acceleration due to gravity, and h is the height difference between the two fluid levels. Rearranging the equation, we have h = P / (ρg). Plugging in the values of P = 80 kPa, ρ = 13,600 kg/m3, and g = 9.8 m/s2, we can calculate the value of h in meters.
(b) For water, we can use the same equation P = ρgh, but with the density of water. Plugging in the values of P = 80 kPa, ρ = 1000 kg/m3, and g = 9.8 m/s2, we can calculate the value of h in meters.
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Michael opens his eyes in the morning to see the alarm clock on his dresser. At that point, his eyes are receiving light energy, which they change into neural messages for the brain. This conversion of one form of energy into another is called _____.
Answer:
Transduction.
Explanation:
Transduction is actually a process which involves the conversion of energies or messages from one kind to another. So in this case the transfer of light energy received by the eyes into the message for the brain is also called as transduction.
explain application of bernoulli's principle in lift of aeroplane, spinning ball, heart attack
The lift in the wing is similar to the Bernoulli's equation. The pressure on the one side of the ball makes it to mover faster than the other side. In heart attack As the blood rushes through the constricted artery, the internal pressure drops and again the artery closes.
Explanation:
The determination of the relation between the pressure, density and velocity at each point of a fluid can be done with Bernoulli's principle.The lift of the aircraft makes it to move forward and upward. This is achieved based on relationship between the angle of the wing and the turbulent flow that helps whether to increase or decrease the lift of the wing. This is similar to the Bernoulli's equation and hence they are associated with each other.
A ball contains stitches on the middle. While a ball spins the pressure is induced by the ball stitches. This will increase the pressure to decrease one side when compared to the pressure of the other side of the ball. This enables it to move faster. The pressure must be increased for maintaining a constant rate flow of blood in artery. Heart muscle is very essential factor for increasing this pressure. The person will have heart attach when there is dislodging of the plaque occurs.
Networks that are designed to connect similar computers that share data and software with each other are called: client/server networks peer-to-peer networks host networks client networks local area networks?
Answer:
Peer to peer network
Explanation:
Peer-to-peer networks are designed to connect similar computers allowing them to share data and software. These networks have played a crucial role in the formation of the internet as we know it today. AOL is an example of a commercial service provider that adopted this networking model.
Explanation:The networks that are designed to connect similar computers that share data and software with each other are called peer-to-peer networks. Peer-to-peer networks allow all computers, termed nodes, to share resources amongst each other without the need for a central server. Every computer on a peer-to-peer network can function as both a server and a client, sharing and receiving files simultaneously.
Examples of this can be found in early internet developments like the Advanced Research Projects Agency Network (ARPANET). As technologies improved, commercial online service providers like America Online (AOL) adopted this networking model, serving as gateways to the burgeoning internet.
Prior to the existence of more sophisticated network infrastructure like cable and fiber optics, these peer-to-peer networks provided a means for computers to communicate and share data, laying the groundwork for the commercial internet as we understand it today.
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A rotating flywheel can be used as a method to store energy. If it is required that such a device be able to store up to a maximum of 1.00 ×106J when rotating at 400 rad/s, what moment of inertia is required? a. 50 kg⋅m2b. 25 kg⋅m2c. 12.5 kg⋅m2d. 6.3 kg⋅m2
Answer:
[tex]I=12.5\ kg.m^2[/tex]
Explanation:
Given that
Stored energy ,[tex]E=1\times 10^6[/tex]
Angular speed ,[tex]\omega =400\ rad/s[/tex]
Lets take moment of inertia of the flywheel = I
As we know that stored energy in the flywheel is given as
[tex]E=\dfrac{1}{2}\omega^2 I[/tex]
[tex]I=\dfrac{2\ttimes E}{\omega^2}[/tex]
Now by putting the values in the above equation we get
[tex]I=\dfrac{2\times 1\times 10^6}{400^2}\ kg.m^2[/tex]
[tex]I=12.5\ kg.m^2[/tex]
Therefore the moment of inertia of the flywheel will be [tex]I=12.5\ kg.m^2[/tex]
The answer will be C.
[tex]I=12.5\ kg.m^2[/tex]
A positively charged particle initially at rest on the ground accelerates upward to 140 m/s in 2.10 s. The particle has a charge-to-mass ratio of 0.100 C/kg and the electric field in this region is constant and uniform.What are the magnitude and direction of the electric field?
Answer:
764.7 N/C, the direction is upward
Explanation:
E electric field of the electron = F / charge
F force of the electron = Eq
acceleration of the body = change in velocity / time = 140 m / 2.10 s = 66.67 m/s² since the electron is from rest.
F upward = F of electron - force of gravity
F upward = Eq - mg
ma = Eq -mg
divide through with m
a = E (q/m) - g
(a + g) / (q/m) = E
(66.67 + 9.8) / 0.100 = 764.7 N/C, the direction is upward
g and the electric field in this region is constant and uniform.What are the magnitude and direction of the electric field?
If three uncharged styrofoam balls are placed together and agitated so that one gains +3 C of charge and another gains +4 C of charge, how much charge must there be on the third one?If three uncharged styrofoam balls are placed together and agitated so that one gains 3 of charge and another gains 4 of charge, how much charge must there be on the third one?a.+7 Cb.+1 Cc.0 Cd.−7 C
Answer:
(d) −7 C
Explanation:
Like charges attracts and unlike charges repel. When the three charges are are agitated, the +3 C and +4 C will repel to give of +7 C. To create a balance in the system, the third charge will -7 C to give an algebraic sum of zero in the system.
A tennis player swings her 1000 g racket with a speed of 15.0 m/s. She hits a 60 g tennis ball that was approaching her at a speed of 18.0 m/s. The ball rebounds at 40.0 m/s.
Part A) How fast is her racket moving immediately after the impact? You can ignore the interaction of the racket with her hand for the brief duration of the collision.
_________m/s
Part B) If the tennis ball and racket are in contact for 7.00 , what is the average force that the racket exerts on the ball?
_________N
Final answer:
The racket is moving at a speed of 15 m/s immediately after the impact. The average force that the racket exerts on the ball is 0.189 N.
Explanation:
Part A:
To calculate the racket's speed immediately after the impact, we can use the principle of conservation of momentum. The momentum before the impact is equal to the momentum after the impact.
The momentum before the impact is given by the product of the racket's mass and its initial speed:
Momentum before = (0.001 kg)(15.0 m/s) = 0.015 kg·m/s
The momentum after the impact is equal to the product of the racket's final speed (which is what we need to find) and its mass:
Momentum after = (0.001 kg)(v)
Since the momentum is conserved, we can equate the two expressions:
0.015 kg·m/s = (0.001 kg)(v)
Simplifying this equation gives:
v = 15 m/s
Therefore, the racket is moving at a speed of 15 m/s immediately after the impact.
Part B:
To find the average force that the racket exerts on the ball, we can use the impulse-momentum principle. The impulse exerted on the ball is equal to the change in momentum of the ball:
Impulse = change in momentum
The change in momentum can be calculated using the formula:
change in momentum = (final momentum) - (initial momentum)
The initial momentum of the ball is given by the product of its mass and its initial speed:
Initial momentum = (0.060 kg)(18.0 m/s) = 1.08 kg·m/s
The final momentum of the ball is given by the product of its mass and its final speed:
Final momentum = (0.060 kg)(40.0 m/s) = 2.4 kg·m/s
Substituting these values into the formula, we have:
Impulse = 2.4 kg·m/s - 1.08 kg·m/s
Impulse = 1.32 kg·m/s
Since impulse is equal to the average force multiplied by the time of contact, we can rearrange the formula to solve for the average force:
Average force = Impulse / Time of contact
Substituting the values for impulse and the given time of contact, we have:
Average force = 1.32 kg·m/s / 7.00 s
Average force = 0.189 N