Answer:
Option (2) is correct.
The energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.
Explanation:
An electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.
The energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields.
So, the correct option is (2).
The energy density is equally distributed among the magnetic field and electric field. Hence, option (2) is correct.
The given problem is based on the concept and fundamentals of electromagnetic waves. The waves created as a result of vibrations between an electric field and a magnetic field is known as Electromagnetic waves.
In other words, an electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.
Also, the energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields. So, the energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.
Thus, we can conclude that the energy density is equally distributed among the magnetic field and electric field.
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A load of 25 kg is applied to the lower end and of a steal wire of length 25 m and thickness 3.0mm .The other end of wire is suspeded from a rigid support calculate strain and stress produced in the wire
Answer:
the weight of the wire + 25kg
Explanation:
the rate of cooling determines ....... and ......
Answer:
freezing point and melting point
i don't understand this, can someone help please??
Explanation:
N2 + H2 --> NH3
balance them:
N2 + 3 H2 --> 2 NH3
so if 6 moles of N2 react, 12 moles of NH3 will form.
(you have to look at the big number in front, in this case its N2 and 2 NH3, therefore the amount of N2 will produce double the amount of NH3 )
Two forces are acting on a body. One acts east, the other at 35° north of east. If the
two forces are equal in magnitude of 50 N, find the resultant using the Law of Sines
and the Law of Cosines. Please answer with full solution. Thanks
A=B=50NAngle=theta=35°
We know
[tex]\boxed{\sf R=\sqrt{A^2+B^2+2ABcos\Theta}}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{50^2+50^2+2(50)(50)cos35}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{2500+2500+2(2500)\times (-0.9)}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{5000+5000(-0.9)}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{5000+(-4500)}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{5000-4500}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{-500}[/tex]
[tex]\\ \sf\longmapsto R=22.4i[/tex]
Resultant using the Law of Sines and the Law of Cosines will be R=95 N
What is force?Force is an external agent applied on any object to displace it from its position. Force is a vector quantity, so with magnitude it also requires direction. Direction is necessary to examine the effect of the force and to find the equilibrium of the force.
The Magnitude of two forces =50 N
Angle between the forces = 35
By using the resultant formula
[tex]\rm R=\sqrt{A^2+B^2+2ABCos\theta}[/tex]
[tex]\rm R=\sqrt{50^2+50^2+2(50)(50)Cos35}[/tex]
[tex]\rm R=\sqrt{5000+5000(0.81)}[/tex]
[tex]\rm R=\sqrt{5000+4500}[/tex]
[tex]\rm R=95\ N[/tex]
Hence the Resultant using the Law of Sines and the Law of Cosines will be R=95 N
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When the drag force on an object falling through the air equals the force of gravity, the object has reached
terminal force.
terminal acceleration,
terminal illness.
terminal velocity
A 20 N south magnetic force pushes a charged particle traveling with a velocity of 4 m/s west through a 5 T magnetic field pointing downwards . What is the charge of the particle ?
Answer:
Charge of the particle is 1 coulomb.
Explanation:
Force, F:
[tex]{ \bf{F=BeV}}[/tex]
F is magnetic force.
B is the magnetic flux density.
e is the charge of the particle.
V is the velocity
[tex]{ \sf{20 = (5 \times e \times 4)}} \\ { \sf{20e = 20}} \\ { \sf{e = 1 \: coulomb}}[/tex]
A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1° from the normal to the plane of the loop. The magnitude of the magnetic field is 3.35 T , and the radius of the wire loop is 0.240 m . Find the magnetic flux Φ through the loop.
Answer:
0.5849Weber
Explanation:
The formula for calculating the magnetic flus is expressed as:
[tex]\phi = BAcos \theta[/tex]
Given
The magnitude of the magnetic field B = 3.35T
Area of the loop = πr² = 3.14(0.24)² = 0.180864m²
angle of the wire loop θ = 15.1°
Substitute the given values into the formula:
[tex]\phi = 3.35(0.180864)cos15.1^0\\\phi =0.6058944cos15.1^0\\\phi =0.6058944(0.9655)\\\phi = 0.5849Wb[/tex]
Hence the magnetic flux Φ through the loop is 0.5849Weber
if C is The vector sum of A and B C = A + B What must be true about The directions and magnitudes of A and B if C=A+B? What must be tre about the directions and magnitudes of A and B if C=0?
Check attached photo
Check attached photo
Answer:
Explanation:
1. If C = A + B then the lines A and B may have the same magnitude or they may not. The direction of A for example may be northwest ↖️ and the direction of B must be south ⬇️ because the arrow of A and the point of B must connect. Then C’s direction is west ⬅️ because it shouldn’t be as equilibrium.
2. If C = 0 t means the force is at equilibrium. That means all forces add up to zero. A’s direction for example may be northeast ↗️ and the direction of B may be south ⬇️ and the direction of C must be west if it has to be at equilibrium.
The magnitude of A and B must be equal
A ball of mass 0.50 kg is rolling across a table top with a speed of 5.0 m/s. When the ball reaches the edge of the table, it rolls down an incline onto the floor 1.0 meter below (without bouncing). What is the speed of the ball when it reaches the floor?
PLEASE EXPLAIN HOW YOU GOT THE ANSWER THANK YOU SO MUCH
Answer:
0
Explanation:
The speed of the ball when it reaches the floor is 0 because when an object is at rest or in uniform motion, it has no speed/velocity
The final speed of the ball when it reaches the floor is 7.10 m/s.
What is the conservation of energy?The conservation of energy is a fundamental principle in physics that states that energy cannot be created or destroyed, but only converted from one form to another or transferred from one system to another. In other words, the total amount of energy in a closed system remains constant over time, even though it may be converted from one form to another.
This principle is based on the first law of thermodynamics, which states that the total energy of a closed system is always conserved, and can only be changed by the transfer of heat, work, or matter into or out of the system. The conservation of energy has important applications in various fields of physics, including mechanics, thermodynamics, and electromagnetism, and is a fundamental principle in the understanding of the natural world.
Here in the Question,
We can use the conservation of energy to solve this problem. Initially, the ball has kinetic energy due to its motion on the tabletop, but no potential energy since it is at a constant height. When the ball rolls off the edge of the table, it loses some kinetic energy due to friction but gains potential energy as it moves upward. When it reaches the floor, it has gained potential energy but lost kinetic energy due to friction. We can assume that the energy lost due to friction is converted to thermal energy, so the total energy of the system is conserved.
Let's start by calculating the potential energy gained by the ball as it moves from the edge of the table to the floor:
ΔPE = mgh
where ΔPE is the change in potential energy, m is the mass of the ball, g is the acceleration due to gravity, and h is the vertical distance traveled by the ball.
ΔPE = (0.50 kg)(9.81 m/s^2)(1.0 m) = 4.905 J
Now we can use the conservation of energy to find the final kinetic energy of the ball, which will allow us to calculate its final speed:
KEi + ΔPEi = KEf + ΔPEf
where KEi and ΔPEi are the initial kinetic and potential energies of the ball, respectively, and KEf and ΔPEf are the final kinetic and potential energies of the ball, respectively.
Since the ball is not bouncing, we can assume that its initial and final potential energies are zero. Therefore:
KEi = KEf + ΔKE
where ΔKE is the change in kinetic energy due to friction.
We can assume that the coefficient of kinetic friction between the ball and the incline is constant, and use the work-energy principle to find ΔKE:
Wfric = ΔKE
where Wfric is the work done by friction.
The work done by friction can be expressed as:
Wfric = ffricd
where ffric is the force of friction and d is the distance traveled by the ball on the incline.
The force of friction can be expressed as:
ffric = μmg
where μ is the coefficient of kinetic friction, and m and g have their usual meanings.
Putting it all together, we get:
KEi = KEf + ffricd
KEi = KEf + μmgd
(1/2)mv^2 = (1/2)mu^2 + μmgd
v^2 = u^2 + 2gd
where u is the initial speed of the ball on the tabletop, and v is the final speed of the ball on the floor.
Plugging in the given values, we get:
v^2 = (5.0 m/s)^2 + 2(9.81 m/s^2)(1.0 m)
v^2 = 50.405
v = 7.10 m/s
Therefore, the final speed of the ball when it reaches the floor is 7.10 m/s.
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two bodies A and B with some asses 20 kg and 30 kg respectively above the ground which have greater potential
Answer:
B has greater potential
Explanation:
We know;
Potential Energy (PE) = mgh
where, m=mass of body
g=acceleration due to gravity
h=height of body
From the formula,
PE is directly proportional to the mass of the body
so the body with greater mass has greater potential.
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms. At what depth did this reflection occur? (The average propagation speed for sound in body tissue is 1540 m/s)
Answer:
10.01 cm
Explanation:
Given that,
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.
The average propagation speed for sound in body tissue is 1540 m/s.
We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,
[tex]v=\dfrac{2d}{t}\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{1540\times 0.13\times 10^{-3}}{2}\\\\d= $$0.1001\ m[/tex]
or
d = 10.01 cm
So, the reflection will occur at 10.01 cm.
The paper dielectric in a paper-and-foil capacitor is 0.0785 mm thick. Its dielectric constant is 2.35, and its dielectric strength is 49.5 MV/m. Assume that the geometry is that of a parallel-plate capacitor, with the metal foil serving as the plates.
Required:
a. What area of each plate is required for for a 0.300 uF capacitor?
b. If the electric field in the paper is not to exceed one-half the dielectric strength, what is the maximum potential difference that can be applied across the compactor?
Answer:
a) required area is 1.1318 m²
b) the maximum potential difference that can be applied across the compactor is 1931.1 V
Explanation:
Given the data in the question;
dielectric constant εr = 2.35
distance between plates ( thickness ) d = 0.0785 mm = 7.85 × 10⁻⁵ m
dielectric strength = 49.5 MV/m
a)
given that capacity capacitor C = 0.3 uF = 0.3 × 10⁻⁶ F
To find the Area, we use the following the expression.
C = ε₀εrA / d
we know that The permittivity of free space, ε₀ = 8.854 x 10⁻¹² (F/m)
we substitute
0.3 × 10⁻⁶ = [ (8.854 x 10⁻¹²) × 2.35 × A ] / 7.85 × 10⁻⁵
A = [ (0.3 × 10⁻⁶) × (7.85 × 10⁻⁵) ] / [ 2.35 × (8.854 x 10⁻¹²) ]
A = 2.355 × 10⁻¹¹ / 2.08069 × 10⁻¹¹
A = 1.1318 m²
Therefore, required area is 1.1318 m²
b)
the maximum potential difference that can be applied across the compactor.
We use the following expression;
⇒ 1/2 × dielectric strength × thickness d
we substitute
⇒ 1/2 × ( 49.5 × 10⁶ V/m ) × ( 7.85 × 10⁻⁵ m )
⇒ 1931.1 V
Therefore, the maximum potential difference that can be applied across the compactor is 1931.1 V
A 34-m length of wire is stretched horizontally between two vertical posts. The wire carries a current of 68 A and experiences a magnetic force of 0.16 N. Find the magnitude of the earth's magnetic field at the location of the wire, assuming the field makes an angle of 72.0° with respect to the wire.
Answer:
7.28×10⁻⁵ T
Explanation:
Applying,
F = BILsin∅............. Equation 1
Where F = magnetic force, B = earth's magnetic field, I = current flowing through the wire, L = Length of the wire, ∅ = angle between the field and the wire.
make B the subject of the equation
B = F/ILsin∅.................. Equation 2
From the question,
Given: F = 0.16 N, I = 68 A, L = 34 m, ∅ = 72°
Substitute these values into equation 2
B = 0.16/(68×34×sin72°)
B = 0.16/(68×34×0.95)
B = 0.16/2196.4
B = 7.28×10⁻⁵ T
distance of distinct vision.
is placed at a distance less than the distance of near point, its image o
will be blurred. Hence human eye can not see such object clearly.
ADDITIONAL INFORMATION
distance of distinct vision for a normal eye of different age groups
Babies = 7 cm
Adults = 25 cm
erson of age 55 years and above = 100 cm
ever, in our discussion we are concerned with a normal eye of an adult so least
The foulart position of an ahiect from a human eve so that the sh
The least distance up to which we can see the objects clearly without any strain is called least distance of distinct vision. Least distance of distinct vision for a normal human being is 25cm. For young people, the least distance of distant vision will be within 25cm which however it varies with age.
Answer:
25 you said ? thats incorecct
Explanation:
What is the meant of by renewable energy and non-renewrable with example of each.
Answer:
Renewable energy is a type of energy that can be renewed easily, such as sunlight. By using Solar panels to collect the suns energy, we are not depleting it, so this source is renewable.
Non-renewable energy is something that cannot easily be replenished. An example would be oil because oil takes millions of years to form and cannot be renewed easily.
Wind instruments like trumpets and saxophones work on the same principle as the "tube closed on one end" that we examined in our last experiment. What effect would it have on the pitch of a saxophone if you take it from inside your house (at 76 degrees F) to the outside on a cold day when the outside temperature is 45 degrees F ?
Answer:
The correct answer would be - Low pitch.
Explanation:
As it is known that if frequency increases then pitch will be increase as well as pitch depends on frequency, Now for the question it is mentioned that the tube closed on one end frequency is:
f = v/2l
Where,
l = length of the tube
v = velocity of longitudinal wave of gas filled in the tube
Now increase with the temperature the density of the gas decreases and velocity v is inversely proportional to density of gas so velocity increases. So if there is an increase in frequency so pitch also increases. As the temperature inside the house is at 750 F more than outsideat 450 Fso pitch is more inside and the pitch is low outside.
derive expression for pressure exerted by gas
1) Consider an electric power transmission line that carries a constant electric current of i = 500 A. The cylindrical copper cable used to transmit this current has a diameter o = 2.00 cm and a length L = 150 km. If there are 8.43x10^28 free electrons per cubic meter (m^3 ) in the cable, calculate how long it would take for an electron to cross the entire length of the transmitter line.
Answer:
t = 1.27 x 10⁹ s
Explanation:
First, we will find the volume of the wire:
Volume = V = AL
where,
A = Cross-sectional area of wire = πr² = π(1 cm)² = π(0.01 m)² = 3.14 x 10⁻⁴ m²
L = Length of wire = 150 km = 150000 m
Therefore,
V = 47.12 m³
Now, we will find the number of electrons in the wire:
No. of electrons = n = (Electrons per unit Volume)(V)
n = (8.43 x 10²⁸ electrons/m³)(47.12 m³)
n = 3.97 x 10³⁰ electrons
Now, we will use the formula of current to find out the time taken by each electron to cross the wire:
[tex]I =\frac{q}{t}[/tex]
where,
t = time = ?
I = current = 500 A
q = total charge = (n)(chareg on one electron)
q = (3.97 x 10³⁰ electrons)(1.6 x 10⁻¹⁹ C/electron)
q = 6.36 x 10¹¹ C
[tex]500\ A = \frac{6.36\ x\ 10^{11}\ C}{t}\\\\t = \frac{6.36\ x\ 10^{11}\ C}{500\ A}[/tex]
Therefore,
t = 1.27 x 10⁹ s
A bird has a kinetic energy of 3 J and a potential energy of 25 J. What is the mechanical energy of the bird?
Answer:
28 j
Explanation:
because when you add you get 28
An ideal parallel plate capacitor with a cross-sectional area of 0.4 cm2 contains a dielectric with a dielectric constant of 4 and a dielectric strength of 2 x 108 V/m. The separation between the plates of the capacitor is 5 mm. What is the maximum electric charge (in nC) that can be stored in the capacitor before dielectric breakdown
Answer: [tex]283.2\times 10^{-9}\ nC[/tex]
Explanation:
Given
Cross-sectional area [tex]A=0.4\ cm^2[/tex]
Dielectric constant [tex]k=4[/tex]
Dielectric strength [tex]E=2\times 10^8\ V/m[/tex]
Distance between capacitors [tex]d=5\ mm[/tex]
Maximum charge that can be stored before dielectric breakdown is given by
[tex]\Rightarrow Q=CV\\\\\Rightarrow Q=\dfrac{k\epsilon_oA}{d}\cdot (Ed)\quad\quad [V=E\cdot d]\\\\\Rightarrow Q=k\epsilon_oAE\\\\\Rightarrow Q=4\times 8.85\times 10^{-12}\times 0.4\times 10^{-4}\times 2\times 10^8\\\\\Rightarrow Q=28.32\times 10^{-8}\\\\\Rightarrow Q=283.2\times 10^{-9}\ nC[/tex]
Answer:
The maximum charge is 7.08 x 10^-8 C.
Explanation:
Area, A = 0.4 cm^2
K = 4
Electric field, E = 2 x 10^8 V/m
separation, d = 5 mm = 0.005 m
Let the capacitance is C and the charge is q.
[tex]q = CV\\\\q=\frac{\varepsilon o A}{d}\times E d\\\\q = \varepsilon o A E\\\\q = 8.85\times 10^{-12}\times0.4\times 10^{-4}\times 2\times 10^8\\\\q = 7.08\times 10^{-8}C[/tex]
1. A turtle and a rabbit are to have a race. The turtle’s average speed is 0.9 m/s. The rabbit’s average speed is 9 m/s. The distance from the starting line to the finish line is 1500 m. The rabbit decides to let the turtle run before he starts running to give the turtle a head start. If the rabbit started to run 30 minutes after the turtle started, can he win the race? Explain.
Answer:no
Explanation:because 0.9*(30*60)=0.9*1800=1620
The turtle has already won the race
Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours
What will be the speed of the rabbit and the turtle?It is given
[tex]V_{t} = 0.9 \frac{m}{s}[/tex]
[tex]V_{r} = 9 \frac{m}{s}[/tex]
[tex]D=1500 m[/tex]
Time taken by turtle
[tex]T= \dfrac{D}{V_{t} }=\dfrac{1500}{0.9_{} }[/tex]
[tex]T=1666 minutes= 27 hours[/tex]
Time taken by rabbit
[tex]T= \dfrac{D}{V_{r} }=\dfrac{1500}{9_{} }[/tex]
[tex]T=166 minutes[/tex]
since rabbit started 30 minutes after turtle then
[tex]T= 136+30=196 minutes[/tex]
[tex]T= 3.2 hours[/tex]
Hence Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours
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A 1030 kg car has four 12.0 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles
Answer:
The required fraction is 0.023.
Explanation:
Given that
Mass of a car, m = 1030 kg
Mass of 4 wheels = 12 kg
We need to find the fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles.
The rotational kinetic energy due to four wheel is
[tex]=4\times \dfrac{1}{2}I\omega^2\\\\=4\times \dfrac{1}{2}\times \dfrac{1}{2}mR^2(\dfrac{v}{R})^2\\\\=mv^2[/tex]
Linear kinetic Energy of the car is:
[tex]=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times Mv^2[/tex]
Fraction,
[tex]f=\dfrac{mv^2}{\dfrac{1}{2}Mv^2}\\\\f=\dfrac{m}{\dfrac{1}{2}M}\\\\f=\dfrac{12}{\dfrac{1}{2}\times 1030}\\\\=0.023[/tex]
So, the required fraction is 0.023.
Two identical cars, each traveling at 16 m>s, slam into a concrete wall and come to rest. In car A the air bag does not deploy and the driver hits the steering wheel; in car B the driver contacts the deployed air bag. (a) Is the impulse delivered by the steering wheel to driver A greater than, less than, or equal to the impulse delivered by the air bag to driver B
Answer:
I = - m 16 the two impulses are the same,
Explanation:
The impulse is given by the relationship
I = Δp
I = p_f - p₀
in this case the final velocity is zero therefore p_f = 0
I = -p₀
For driver A the steering wheel impulse is
I = - m v₀
I = - m 16
For driver B, the airbag gives an impulse
I = - m 16
We can see that the two impulses are the same, the difference is that in the air bag more time is used to give this impulse therefore the force on the driver is less
Nhiệt dung riêng của một chất là ?
Answer:
enchantment table language
Explanation:
An infinite plane lies in the yz-plane and it has a uniform surface charge density.
The electric field at a distance x from the plane
a.) decreases as 1/x^2
b.) increases linearly with x
c.) is undertermined
d.) decreases linearly with x
e.) is constant and does not depend on x
Answer:
So the correct answer is letter e)
Explanation:
The electric field of an infinite yz-plane with a uniform surface charge density (σ) is given by:
[tex]E=\frac{\sigma }{2\epsilon_{0}}[/tex]
Where ε₀ is the electric permitivity.
As we see, this electric field does not depend on distance, so the correct answer is letter e)
I hope it helps you!
The mass is released from the top of the incline and slides down the incline. The maximum velocity (taken the instant before the mass reaches the bottom of the incline) is 1.06 m/s. What is the kinetic energy at that time
Answer:
0.28 J
Explanation:
Let the mass of the object is 0.5 kg
The maximum velocity of the object is 1.06 m/s.
We need to find the kinetic energy at that time. It is given by :
[tex]K=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times 0.5\times (1.06)^2\\\\K=0.28\ J[/tex]
So, the required kinetic energy is equal to 0.28 J.
Place each description under the correct theory
Gravity is an attractive force.
Universal Law of Gravitation
General Theory of Relativity
Mass and distance affect force.
Time and space are absolute,
Time and space are relative.
Gravity is due to space-time curving.
Mass affects space-time curving.
Answer:
1) Law of Universal Gravitation Gravity is an attractive force
5) General relativity Gravity is due to the curvature of spacetime
Explanation:
In this exercise you are asked to relate the correct theory and its explanation
Theory Explanation
1) Law of Universal Gravitation Gravity is an attractive force
2) Law of universal gravitation Mass and distance affect force
3) Classical mechanics time and space are absolute
4) Special relativity Time and space are relative
5) General relativity Gravity is due to the curvature of
spacetime
6) General relativity Mass affects the curvature of space - time
Answer:
Explanation:
edge2022
During a particular thunderstorm, the electric potential difference between a cloud and the ground is Vcloud - Vground = 4.20 108 V, with the cloud being at the higher potential. What is the change in an electron's electric potential energy when the electron moves from the ground to the cloud?
Answer:
The electric potential energy is 6.72 x 10^-11 J.
Explanation:
Potential difference, V = 4.2 x 10^8 V
charge of electron, q = - 1.6 x 10^-19 C
Let the potential energy is U.
U = q V
U = 1.6 x 10^-19 x 4.2 x 10^8
U = 6.72 x 10^-11 J
In a large chemical factory, a feed pipe carries a liquid at a speed of 5.5 m/s. A pump pushes the liquid along at a gauge pressure of 140,000 Pa. The liquid travels upward 6.0 m and enters a tank at a gauge pressure of 2,000 Pa. The diameter of the pipe remains constant. At what speed does the liquid enter the tank
Answer:
v₂ = 15.24 m / s
Explanation:
This is an exercise in fluid mechanics
Let's write Bernoulli's equation, where the subscript 1 is for the factory pipe and the subscript 2 is for the tank.
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
They indicate the pressure in the factory P₁ = 140000 Pa, the velocity
v₁ = 5.5 m / s and the initial height is zero y₁ = 0
the tank is at a pressure of P2 = 2000 Pa and a height of y₂ = 6.0 m
P₁ -P₂ + ρ g (y₁ -y₂) + ½ ρ v₁² = ½ ρ v₂²
let's calculate
140,000 - 2000 + ρ 9.8 (0- 6) + ½ ρ 5.5² = ½ ρ v₂²
138000 - ρ 58.8 + ρ 15.125 = ½ ρ v2²
v₂² = 2 (138000 /ρ - 58.8 + 15.125)
v₂ = [tex]\sqrt{\frac{276000}{\rho } - 43.675 }[/tex]
In the exercise they do not indicate what type of liquid is being used, suppose it is water with
ρ = 1000 kg / m³
v₂ = [tex]\sqrt{\frac{276000}{1000} - 43.675}[/tex]
v₂ = 15.24 m / s
A 12.0 g sample of gas occupies 19.2 L at STP. what is the of moles and molecular weight of this gas?
At STP, 1 mole of an ideal gas occupies a volume of about 22.4 L. So if n is the number of moles of this gas, then
n / (19.2 L) = (1 mole) / (22.4 L) ==> n = (19.2 L•mole) / (22.4 L) ≈ 0.857 mol
If the sample has a mass of 12.0 g, then its molecular weight is
(12.0 g) / n ≈ 14.0 g/mol
