The common angular speed is 9090.91 rad/h.
Explanation:To find the common angular speed, we need to first find the initial and final radii of the full reel. The initial radius is 35 mm and the final radius is 11 mm. The angular speed can be found using the equation v = rω, where v is the linear speed, r is the radius, and ω is the angular speed. We can solve for ω by dividing the linear speed by the radius. Since both reels will have the same angular speed, we can use the same equation to find the angular speed when the tape has unspooled a distance of 191 m.
First, let's convert the radii from millimeters to meters:
Initial radius: 35 mm = 0.035 mFinal radius: 11 mm = 0.011 mNext, we'll find the linear speeds at the initial and final radii. Since the length of the tape is 191 m and it plays for 1.9 hours, the linear speed can be calculated by dividing the length of the tape by the playtime:
Linear speed at initial radius: (191 m)/(1.9 h) = 100 m/hLinear speed at final radius: (191 m)/(1.9 h) = 100 m/hFinally, we can find the common angular speed by dividing the linear speed by the radius:
Initial angular speed: (100 m/h)/(0.035 m) = 2857.14 rad/hFinal angular speed: (100 m/h)/(0.011 m) = 9090.91 rad/hTherefore, the common angular speed is 9090.91 rad/h.
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Which is true about the self-induced emf of an inductor?
It is a fixed value, depending on only the geometry of the device.
It depends on the amount of current through the inductor.
It depends on the rate of dissipation.
It depends on the rate at which the current through it is changing.
Answer:
It depends on the rate at which the current through it is changing.
Explanation:
As per the Faraday's law, the induced emf is given by :
[tex]\epsilon=-L\dfrac{di}{dt}[/tex]
Where
L is the inductance of the inductor
[tex]\dfrac{di}{dt}[/tex] is the rate of change of current
So, the self-induced emf of an inductor depends on the rate at which the current through it is changing. Hence, the correct option is (d).
A circuit with current increasing at a rate of 4 A/s contains an inductor, L. If the induced emf is -2 V, what is the inductance of the inductor?
Answer:
L=500 mH
Explanation:
Here di/dt = 4A/s, ε= -2V
Inductance of inductor, induced emf and rate of change of current have the following relation.
ε= [tex] - L\frac{di}{dt}[/tex]
⇒L= - ε/[tex]\frac{di}{dt}[/tex]
⇒L= -(-2)/ 4
⇒ L= 0.5 H or
⇒ L= 500 mH
The ballistic pendulum consists of a block of dense material suspended from a cord and allowed to swing freely. A bullet is fired into the block, lodging itself into the block in a __________________. The (block+bullet) having now acquired a velocity, swings upward to some maximum height. (See lab manual)
The subject of this question is Physics and it pertains to the concept of a ballistic pendulum.
Explanation:The subject of the question is Physics. Specifically, it is related to the concept of a ballistic pendulum.
A ballistic pendulum consists of a block of dense material suspended from a cord. When a bullet is fired into the block, it lodges itself into the block in a specific manner, either elastically or by sticking together. The (block+bullet) swings upward to some maximum height due to the acquired velocity.
For example, if the block and bullet stick together after the collision, the system forms a ballistic pendulum. The bullet's initial kinetic energy is converted into potential energy as the (block+bullet) rises.
The ballistic pendulum consists of a block of dense material suspended from a cord and allowed to swing freely. A bullet is fired into the block, lodging itself into the block in a sticky collision. The (block+bullet) having now acquired a velocity, swings upward to some maximum height.
A ballistic pendulum is a device that measures the speed of a projectile, such as a bullet. When a bullet of mass m is fired into a block of mass M suspended from a cord, it lodges itself into the block in a sticky collision. The combined mass then swings to a maximum height h, where the kinetic energy is converted into potential energy.
First, understand that the bullet embeds itself in the block. This scenario is a completely inelastic collision, also known as a sticky collision.Upon impact, the kinetic energy of the bullet and block system is high, which gets converted into potential energy at the maximum height of the swing.To find the initial speed of the bullet use the conservation of momentum and energy principles:1. Conservation of momentum: m[tex]v_{i}[/tex] = (m + M)[tex]v_{f}[/tex]
2. Conservation of energy: 1/2(m + M)[tex]v_{f}^{2}[/tex] = (m + M)gh.
From these equations, [tex]v_{i}[/tex] can be derived:
[tex]v_{i}[/tex] = (1 + M/m) √(2gh)
Thus, the ballistic pendulum helps measure the speed of the bullet before the collision.
A titanium (c = 520 J/kg*K) satellite of mass m = 500 kg at a temperature of 10 K is in geostationary orbit above the equator. It is impacted by a meteorite, which knocks it off course but does not affect its speed. Because of the change in trajectory, the satellite will now crash-land on Earth in a small pond containing 5 × 10^5 kg of water (c = 4186 J/kg*K) at 20°C. Ignoring air resistance, and assuming that the water stays in the pond somehow, what is the final temperature of the water? Please indicate if any water boils, and justify any simplifying assumptions you make in your solution.
A 2.0 kg block is held at rest against a spring with a force constant k = 264 N/m. The spring is compressed a certain distance and then the block is released. When the block is released, it slides across a surface that has no friction except for a 10.0 cm section that has a coefficient of friction μk = 0.54.
Find the distance in centimeters the spring was compressed such that the block's speed after crossing the rough area is 2.7 m/s.
Answer:
21.73 cm
Explanation:
We have given parameters:
Mass of block, m = 2.0 kg
Force constant of spring, k = 264 N/m
Length of rough area, L = 10 cm = 0.1 m
Co-efficient of kinetic friction , [tex]\mu_{k}[/tex] = 0.54
Block's speed after crossing rough area, v = 2.7 m/s
Block's initial speed ( when it was released from compressed spring), [tex]v_{0}[/tex] = 0 m/s
We need to find the distance that the spring was initially compressed, x = ?
Hence, we well apply Work-Energy principle which indicates that,
Work done by the friction = Change in the total energy of block
[tex]- \mu_{k} * mg * L = (\frac{1}{2} * mv^{2} - \frac{1}{2} * mv_{0} ^{2} ) + (0 - \frac{1}{2} * kx^{2})[/tex]
-0.54 * 2 * 9.8 * 0.1 = (1/2 * 2 * [tex]2.7^{2}[/tex] - 1/2 * 2 * 0) + (0 - 1/2 * 264 * [tex]x^{2}[/tex])
x = 0.2173 m = 21.73 cm
8. A 40.0kg block of metal is suspended from a scale and is immersed in water. The dimensions of the block are 12.0cm x 10.0cm x 10.0cm. The 12.0cm dimension is vertical, and the top of the block is 5.00cm below the surface of the water. (a) What are the forces exerted by the water on the top and bottom of the block? (Take P0 = 1.0130 X 105 N/m2 .) (b) What is the reading of the spring scale? (c) Show that the buoyant force equals the difference between the forces at the top and bottom of the block.
Final answer:
The forces exerted by the water on the submerged block are due to pressure differences and can be calculated using the dimensions of the block, the density of water, and the depth of immersion. The scale reading reflects the difference between the block's weight and the buoyant force. Lastly, the buoyyant force is confirmed to be the difference between the top and bottom forces on the block, in alignment with Archimedes' principle.
Explanation:
The student's question involves calculating the forces exerted by water on a submerged block and the reading of a scale when the block is immersed in water. To calculate these forces, one must consider the forces exerted by the water on the top and bottom of the block, which are due to the differences in pressure at these points, as well as the buoyant force acting on the block. Using the principle of Archimedes, which states that the upthrust or buoyant force on a submerged object is equal to the weight of the fluid it displaces, we can find the buoyant force which in turn will help to determine the scale reading.
The weight of the water displaced can be calculated using the volume of the block and the density of the water. To answer part (a), the pressure difference at the top and bottom of the block due to the water column can be calculated using the given atmospheric pressure and depth of the block in the water. For part (b), the scale reading would be the weight of the block minus the buoyant force exerted by the water. Part (c) involves showing that the buoyant force is indeed the difference in forces at the top and bottom of the block, confirming Archimedes' principle.
A special system is set up in a lab that lets its user select any wavelength between 400nm and 700 nm with constant intensity. This light is directed at a thin glass film (n =1.53) with a thickness of 350 nm and that is surrounded by air. As one scans through these possible wavelengths, which wavelength of light reflected from the glass film will appear to be the brightest, if any?
a) 428 nm
b) 535 nm
c) 657 nm
d) 700 nm
e) Since the intensity of the light is constant,
all wavelengths of light reflected from the glass will appear to be the same.
Answer:
Explanation:
The case relates to interference in thin films , in which we study interference of light waves reflected by upper and lower surface of a medium or glass.
For constructive interference , the condition is
2μt = ( 2n+1)λ/2
μ is refractive index of glass , t is thickness , λ is wavelength of light.
putting the given values
2 x 1.53 x 350 x 10⁻⁹ = ( 2n+1) λ/2
λ = 2142nm / ( 2n+1)
For n = 2
λ = 428 nm
This wave length will have constructive interference making this light brightest of all .
For n = 1
λ = 714 nm
So second largest brightness will belong to 700 nm wavelength.
The second law of thermodynamics states that spontaneous processestend to be accompanied by entropy increase. Consider, however, thefollowing spontaneous processes:
the growth of plants from simple seeds to well-organizedsystems
the growth of a fertilized egg from a single cell to a complexadult organism
the formation of snowflakes from molecules of liquid water withrandom motion to a highly ordered crystal
the growth of organized knowledge over time
In all these cases, systems evolve to a state of less disorder andlower entropy, apparently violating the second law ofthermodynamics. Could we, then, consider them as processesoccurring in systems that are not isolated?
True or False?
Answer:
True
Explanation:
According to the definition of a closed system, It is true because it's not precisely a closed system. A closed system is a physical system that doesn't let certain types of transfers (such as transfer of mass in or out of the system), though the transfer of energy is allowed.
An average nerve axon is about 4x10-6m in radius, and the axoplasm that composes the interior of the axon has a resistivity of about 2.3 Ω ⋅m.
What is the resistance of just 2-cm length of this axon?
Provide your answer in mega-ohms (1 mega-ohm = 106 ohms or "millions of ohms").
(Note, this value is so large -- it corresponds to the resistance of tens of thousands of miles of the thinnest copper wire normally manufactured(!) -- that it explains why a nerve pulse traveling down an axon CANNOT simply be a current traveling along the axon. The voltage required to achieve a perceptible current in the axon would have to be gigantic! We will investigate how voltage pulses -- not current -- travel down axons in a future lab.)
For the calculation of resistance there are generally two paths. The first is through Ohm's law and the second is through the relationship
[tex]R = \frac{pL}{A}[/tex]
Where
p = Specific resistance of material
L = Length
A = Area
The area of nerve axon is given as
[tex]A = \pi r^2[/tex]
[tex]A = \pi (5*10^{-6})^2[/tex]
[tex]A = 7.854*10^{-11}m^2[/tex]
The rest of values are given as
[tex]p= 2 \Omega\cdot m[/tex]
[tex]L = 2cm = 0.02m[/tex]
Therefore the resistance is
[tex]R = \frac{pL}{A}[/tex]
[tex]R = \frac{2*0.02}{7.854*10^{-11}}[/tex]
[tex]R = 509.3*10^6\Omega[/tex]
[tex]R = 509.3M\Omega[/tex]
A silver bar of length 30 cm and cross-sectional area 1.0 cm2 is used to transfer heat from a 100°C reservoir to a 0°C block of ice.
How much ice is melted per second? (For silver, k = 427 J/s⋅m⋅°C. For ice, Lf = 334 000 J/kg.)
a. 4.2 g/s
b. 2.1 g/s
c. 0.80 g/s
d. 0.043 g/s
Answer:
d. 0.043 g/s
Explanation:
Formula for rate of conduction of heat through a bar per unit time is as
follows
Q = k A ( t₁ - t₂ ) / L
A is cross sectional area and L is length of rod ,( t₁ - t₂ ) is temperature
difference . Q is heat conducted per unit time
Putting the values in the equation
Q = (427 x 1 x 10⁻⁴ x 100 )/ 30 x 10⁻²
= 14.23 J/s
mass of ice melted per second
= Q / Latent heat of ice
= 14.23 / 334000
= 0.043 g/s
Final answer:
To find the mass of ice melted per second, the rate of heat transfer is calculated using the thermal conductivity of silver and the latent heat of fusion of ice. The calculated value of 0.426 g/s does not match the provided options, suggesting a possible typo in the options or the need for clarification.
Explanation:
The question is about calculating the amount of ice that is melted per second when a silver bar conducts heat from a hot reservoir to a block of ice. To answer this question, we will use the given thermal conductivity of silver (k = 427 J/s·m·°C), the cross-sectional area of the silver bar (1.0 cm2), and the latent heat of fusion for ice (Lf = 334,000 J/kg).
First, we need to calculate the rate of heat transfer (Q) from the 100°C reservoir through the silver bar to the 0°C ice using the formula:
Q = k · A · (ΔT/L),
where A is the cross-sectional area, ΔT is the temperature difference, and L is the length of the bar. Converting A to meters squared (A = 1.0 x 10-4 m2) and L to meters (L = 0.30 m), and plugging in the values:
Q = 427 J/s·m·°C · 1.0 x 10-4 m2 · 100°C / 0.30 m = 142.333 J/s.
Now, using the formula Q = mLf to find the mass of ice melted per second (m), where Q is the rate of heat transfer and Lf is the latent heat of fusion:
m = Q / Lf = 142.333 J/s / 334,000 J/kg = 0.000426 kg/s.
Converting this mass to grams (since 1 kg = 1000 g), we get:
m = 0.000426 kg/s · 1000 g/kg = 0.426 g/s.
This value is not exactly matching any of the options provided (a-d), but with rounding and considering significant figures, the closest answer would be 0.43 g/s, which is not listed amongst the options provided.
A 5 kg object near Earth's surface is released from rest such that it falls a distance of 10 m. After the object falls 10 m, it has a speed of 12 m/s. Which of the following correctly identifies whether the object-Earth system is open or closed and describes the net external force?
A. The system is closed, and the net external force is zero.
B. The system is open, and the net external force is zero.
C. The system is closed, and the net external force is nonzero.
D. The system is open, and the net external force is nonzero.
Answer:D
Explanation:
Given
mass of object [tex]m=5 kg[/tex]
Distance traveled [tex]h=10 m[/tex]
velocity acquired [tex]v=12 m/s[/tex]
conserving Energy at the moment when object start falling and when it gains 12 m/s velocity
Initial Energy[tex]=mgh=5\times 9.8\times 10=490 J[/tex]
Final Energy[tex]=\frac{1}{2}mv^2+W_{f}[/tex]
[tex]=\frac{1}{2}\cdot 5\cdot 12^2+W_{f}[/tex]
where [tex]W_{f}[/tex] is friction work if any
[tex]490=360+W_{f}[/tex]
[tex]W_{f}=130 J[/tex]
Since Friction is Present therefore it is a case of Open system and net external Force is zero
An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .
A tube has a length of 0.025 m and a cross-sectional area of 6.5 x 10-4 m2. The tube is filled with a solution of sucrose in water. The diffusion constant of sucrose in water is 5.0 x 10-10 m2/s. A difference in concentration of 5.2 x 10-3 kg/m3 is maintained between the ends of the tube. How much time is required for 5.7 x 10-13 kg of sucrose to be transported through the tube?
Answer:
The time required for sucrose transportation through the tube is 8.4319 sec.
Explanation:
Given:
L = 0.025 m
A = 6.5×10^-4 m^2
D = 5×10^-10 m^2/s
ΔC = 5.2 x 10^-3 kg/m^3
m = 5.7×10^-13 kg
Solution:
t = m×L / D×A×ΔC
t = (5.7×10^-13) × (0.025) / (5×10^-10)×(6.5×10^-4)×(5.2 x 10^-3)
t = 8.4319 sec.
The roads are icy, and you observe a head-on collision on Summit, at the corner with Rhodes: a 1ton car swerves out of his lane and slides through a stop sign at 41 mph straight into a 3 ton SUV traveling at 16mph in the other direction. The car and the SUV crumple from the collision, and stick together.
1. What is the final velocity, in MPH (you don't need to enter MPH in your answer) , of the SUV/car entanglement (the positive direction is the direction the car was initially going)
Answer:
final velocity = - 1.75 mph
Explanation:
given data
mass m1 = 1 ton
mass m2 = 3 ton
velocity v = 41 mph
velocity u = 16 mph
to find out
what is final velocity V
solution
we will apply here Conservation of momentum that is express as
mv + Mu = (m + M) × V ...........................1
put here value we get
1 × 41 - 3 × 16 = (1 + 3 ) × V
solve it we get
41 - 48 = 4 V
V = [tex]\frac{-7}{4}[/tex]
final velocity = - 1.75 mph
Final answer:
The final velocity of the SUV/car entanglement is 9.9 m/s.
Explanation:
In order to determine the final velocity of the SUV/car entanglement, we need to first calculate the momentum of each vehicle before the collision.
Momentum is determined by the product of an object's mass and velocity. So, the momentum of the 1 ton car before the collision is 1 ton (or 1000 kg) multiplied by its velocity of 41 mph (which is equivalent to 18.3 m/s). Therefore, the momentum of the car is 1000 kg imes 18.3 m/s = 18300 kg*m/s.
Similarly, the momentum of the 3 ton SUV before the collision is 3 ton (or 3000 kg) multiplied by its velocity of 16 mph (which is equivalent to 7.1 m/s). Therefore, the momentum of the SUV is 3000 kg imes 7.1 m/s = 21300 kg*m/s.
Since momentum is conserved in collisions, the total momentum before the collision is equal to the total momentum after the collision. This means that the final velocity of the SUV/car entanglement can be calculated by dividing the total momentum by the total mass. The total momentum is 18300 kg*m/s + 21300 kg*m/s = 39600 kg*m/s. The total mass is 1000 kg + 3000 kg = 4000 kg. So, the final velocity of the SUV/car entanglement is 39600 kg*m/s / 4000 kg = 9.9 m/s.
Light from a laser strikes a diffraction grating that has 5 300 grooves per centimeter. The central and first-order principal maxima are separated by 0.488 m on a wall 1.64 m from the grating. Determine the wavelength of the laser light. (In this problem, assume that the light is incident normally on the gratings.)
To solve this problem it is necessary to apply the concepts related to the principle of superposition, as well as to constructive interference. From the definition we know that this can be expressed mathematically as
[tex]sin\theta_m = \frac{m\lambda}{d}[/tex]
Where
m = Any integer which represent the number of repetition of spectrum
[tex]\lambda[/tex]= Wavelength
d = Distance between slits
From triangle (Watch image below)
[tex]tan\theta_1 = \frac{y_1}{L}[/tex]
[tex]tan\theta_1 = \frac{0.488}{1.64}[/tex]
[tex]\theta_1 = 16.57\°[/tex]
Replacing the angle at the first equation for m=1 we have
[tex]\lambda = d sin \theta_1[/tex]
Each of the distances (d) would be defined by
d = \frac{1}{5300} = 0.0001886
[tex]\lambda = (\frac{1}{5300}) sin(16.57)[/tex]
[tex]\lambda = 538nm[/tex]
Therefore the wavelength of the laser light is 538nm
The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This spitting ability is enabled by the presence of a groove in the roof of the mouth of the archerfish. The groove forms a long, narrow tube when the fish places its tongue against it and propels drops of water along the tube by compressing its gill covers.When an archerfish is hunting, its body shape allows it to swim very close to the water surface and look upward without creating a disturbance. The fish can then bring the tip of its mouth close to the surface and shoot the drops of water at the insects resting on overhead vegetation or floating on the water surface.At what speed v should an archerfish spit the water to shoot down an insect floating on the water surface located at a distance 0.800 m from the fish? Assume that the fish is located very close to the surface of the pond and spits the water at an angle 60∘ above the water surface.
Answer:
Explanation:
Here is the full question and answer,
The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This spitting ability is enabled by the presence of a groove in the roof of the mouth of the archerfish. The groove forms a long, narrow tube when the fish places its tongue against it and propels drops of water along the tube by compressing its gill covers.
When an archerfish is hunting, its body shape allows it to swim very close to the water surface and look upward without creating a disturbance. The fish can then bring the tip of its mouth close to the surface and shoot the drops of water at the insects resting on overhead vegetation or floating on the water surface.
Part A: At what speed v should an archerfish spit the water to shoot down a floating insect located at a distance 0.800 m from the fish? Assume that the fish is located very close to the surface of the pond and spits the water at an angle 60 degrees above the water surface.
Part B: Now assume that the insect, instead of floating on the surface, is resting on a leaf above the water surface at a horizontal distance 0.600 m away from the fish. The archerfish successfully shoots down the resting insect by spitting water drops at the same angle 60 degrees above the surface and with the same initial speed v as before. At what height h above the surface was the insect?
Answer
A.) The path of a projectile is horizontal and symmetrical ground. The time is taken to reach maximum height, the total time that the particle is in flight will be double that amount.
Calculate the speed of the archer fish.
The time of the flight of spitted water is,
[tex]t = \frac{{2v\sin \theta }}{g}[/tex]
Substitute [tex]9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}[/tex] for g and [tex]60^\circ[/tex] for [tex]\theta[/tex] in above equation.
[tex]t = \frac{{2v\sin 60^\circ }}{{9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}}}\\\\ = \left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\[/tex]
Spitted water will travel [tex]0.80{\rm{ m}}[/tex] horizontally.
Displacement of water in this time period is
[tex]x = vt\cos \theta[/tex]
Substitute [tex]\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}[/tex] for [tex]t\rm 60^\circ[tex] for [tex]\theta[/tex] and [tex]0.80{\rm{ m}}[/tex] for x in above equation.
[tex]\\0.80{\rm{ m}} = v\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\left( {\cos 60^\circ } \right)\\\\0.80{\rm{ m}} = {v^2}\left( {0.1767{\rm{ }}} \right)\frac{1}{2}{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\\\v = \sqrt {\frac{{2\left( {0.80{\rm{ m}}} \right)}}{{0.1767\;{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}}}} \\\\ = 3.01{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\[/tex]
B.) There are two component of velocity vertical and horizontal. Calculate vertical velocity and horizontal velocity when the angle is given than calculate the time of flight when the horizontal distance is given. Value of the horizontal distance, angle and velocity are given. Use the kinematic equation to solve the height of insect above the surface.
Calculate the height of insect above the surface.
Vertical component of the velocity is,
[tex]{v_v} = v\sin \theta[/tex]
Substitute [tex]3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}[/tex] for v and [tex]60^\circ[/tex] for [tex]\theta[/tex] in above equation.
[tex]\\{v_v} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\sin 60^\circ \\\\ = 2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\[/tex]
Horizontal component of the velocity is,
[tex]{v_h} = v\cos \theta[/tex]
Substitute [tex]3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}[/tex] for v and [tex]60^\circ[/tex] for [tex]\theta[/tex] in above equation.
[tex]\\{v_h} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\cos 60^\circ \\\\ = 1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\[/tex]
When horizontal [tex]({0.60\;{\rm{m}}}[/tex] distance away from the fish.
The time of flight for distance (d) is ,
[tex]t = \frac{d}{{{v_h}}}[/tex]
Substitute [tex]0.60\;{\rm{m}}[/tex] for d and [tex]1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}[/tex] for [tex]{v_h}[/tex] in equation [tex]t = \frac{d}{{{v_h}}}[/tex]
[tex]\\t = \frac{{0.60\;{\rm{m}}}}{{1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}}}\\\\ = 0.3987{\rm{ s}}\\[/tex]
Distance of the insect above the surface is,
[tex]s = {v_v}t + \frac{1}{2}g{t^2}[/tex]
Substitute [tex]2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}[/tex] for [tex]{v_v}[/tex] and [tex]0.3987{\rm{ s}}[/tex] for t and [tex]- 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}[/tex] for g in above equation.
[tex]\\s = \left( {2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right)\left( {0.3987{\rm{ s}}} \right) + \frac{1}{2}\left( { - 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right){\left( {0.3987{\rm{ s}}} \right)^2}\\\\ = 0.260{\rm{ m}}\\[/tex]
We had a homework problem in which the Arrhenius equation was applied to the blinking of fireflies. Several other natural phenomena also obey that equation, including the temperature dependent chirping of crickets. A particular species, the snowy tree cricket, has been widely studied. These crickets chirp at a rate of 178 times per minute at 25.0°C, and the activation energy for the chirping process is 53.9 kJ/mol. What is the temperature if the crickets chirp at a rate of 126 times per minute?
Answer:
Temperature = 20.35 °C
Explanation:
Arrhenius equation is as follows:
k = A*exp(-Ea/(R*T)) , where
k = rate of chirps
Ea = Activation Energy
R = Universal Gas Constant
T = Temperature (in Kelvin)
A = Constant
Given Data
Ea = 53.9*10^3 J/mol
R = 8.3145 J/(mol.K)
T = 273.15 + 25 K
k = 178 chirps per minutes
Calculation
Using the Arrhenius equation, we can find A,
A= 4.935x10^11
Now we can apply the same equation with the data below to find T at k=126,
k = A*exp(-Ea/(R*T))
Ea = 53.9*10^3
R = 8.3145
k = 126
T = 20.35 °C
In a football game, a 90 kg receiver leaps straight up in the air to catch the 0.42 kg ball the quarterback threw to him at a vigorous 21 m/s, catching the ball at the highest point in his jump. Right after catching the ball, how fast is the receiver moving?
To solve this problem it is necessary to apply the equations related to the conservation of momentum. Mathematically this can be expressed as
[tex]m_1v_1+m_2v_2 = (m_1+m_2)v_f[/tex]
Where,
[tex]m_{1,2}[/tex]= Mass of each object
[tex]v_{1,2}[/tex] = Initial velocity of each object
[tex]v_f[/tex]= Final Velocity
Since the receiver's body is static for the initial velocity we have that the equation would become
[tex]m_2v_2 = (m_1+m_2)v_f[/tex]
[tex](0.42)(21) = (90+0.42)v_f[/tex]
[tex]v_f = 0.0975m/s[/tex]
Therefore the velocity right after catching the ball is 0.0975m/s
The receiver's final velocity is approximately 0.098 m/s.
Explanation:First, we need to find the initial momentum of the ball by multiplying its mass (0.42 kg) by its velocity (21 m/s). The initial momentum of the ball is therefore 8.82 kg·m/s. Next, we need to find the final momentum of the receiver by multiplying his mass (90 kg) by his final velocity. Since he catches the ball at the highest point in his jump, his final velocity is 0 m/s. So the final momentum of the receiver is 0 kg·m/s. According to the law of conservation of momentum, the initial momentum of the ball must be equal to the final momentum of the receiver. Therefore, the final velocity of the receiver is 8.82 kg·m/s divided by his mass (90 kg), which is approximately 0.098 m/s.
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An engineer has an odd-shaped 13.5 kg object and needs to find its rotational inertia about an axis through its center of mass. The object is supported on a wire stretched along the desired axis. The wire has a torsion constant κ = 0.618 N·m. If this torsion pendulum oscillates through 28 cycles in 58.1 s, what is the rotational inertia of the object?
Answer:
I = 0.0674 kg.m²
Explanation:
given,
mass = 13.5 Kg
torsion constant = k = 0.618 N.m
number of cycle = 28
time = 58.1 s
Time of one cycle
[tex]T = \dfrac{58.1}{28}[/tex]
[tex]T =2.075\ s[/tex]
we know,
[tex]T = 2\pi\sqrt{\dfrac{I}{k}}[/tex]
[tex]I = k (\dfrac{T}{2\pi})^2[/tex]
[tex]I =0.618\times \dfrac{T^2}{4\pi^2}[/tex]
[tex]I =0.618\times \dfrac{2.075^2}{4\pi^2}[/tex]
I = 0.0674 kg.m²
the rotational inertia of the object is equal to I = 0.0674 kg.m²
A parallel plate capacitor is being charged by a constant current i. During the charging, the electric field within the plates is increasing with time.
Which one of the following statements concerning the magnetic field between the plates is true?
A) The magnetic field within a parallel plate capacitor is always equal to zero teslas.
B) The induced magnetic field is directed antiparallel to the increasing electric field.
C) The induced magnetic field strength has its largest value at the center of the plates and decreases linearly toward the edges of the plates.
D) At a given moment, the induced magnetic field strength has the same magnitude everywhere within the plates of the capacitor, except near the edges.
E) The induced magnetic field strength is zero teslas near the center of the plates and increases as r increases toward the edges of the plates.
Answer:
E
Explanation:
The parallel plate capacitor is being charged by a steady current i. If the constant current is flowing along a straight wire, what happens inside the capacitor, between the plates is that the induced magnetic field strength is zero Tesla near the center of the plates and increases as r increases toward the edges of the plates.
From Ampère–Maxwell law, the magnetic field between the capacitor plates assuming that the capacitor is being charged at a constant rate by a steady current is
B = μ₀r /2A [tex]\frac{dQ}{dt} e_{o}[/tex]
where;
μ₀ is permeability
Q(t) is the instantaneous charge on the positive plate,
A is the cross-sectional area of a plate and
e₀ is a unit vector
what consistent physiological pattern are more common in women's teeth? A. Women's canines are generally sharper. B. Women have straighter teeth. C. Women have 1 less set of molar teeth. D. Women normally have longer pre-molars.
Answer:
B
Explanation:
B. Women have straighter teeth.
Bandar Industries Berhad of Malaysia manufactures sporting equipment. One of the company’s products, a football helmet for the North American market, requires a special plastic. During the quarter ending June 30, the company manufactured 3,200 helmets, using 2,368 kilograms of plastic. The plastic cost the company $15,629. According to the standard cost card, each helmet should require 0.68 kilograms of plastic, at a cost of $7.00 per kilogram. Required: 1. What is the standard quantity of kilograms of plastic (SQ) that is allowed to make 3,200 helmets? 2. What is the standard materials cost allowed (SQ × SP) to make 3,200 helmets? 3. What is the materials spending variance? 4. What is the materials price variance and the materials quantity variance? (For requirements 3 and 4, indicate the effect of each variance by selecting "F" for favorable, "U" for unfavorable, and "None" for no effect (i.e., zero variance). Input all amounts as positive values. Do not round intermediate calculations.)
Answer:
1. 2176 kilograms of plastic 2. $15,232 3. $397 (U) 4. $947 (F) $1,344 (U)
Explanation:
Generally, cost variance analysis can be used to estimate the difference between the actual cost and the expected costs. However, if the actual cost is more than the expected cost, then the variance is said to be unfavorable and vice versa.
1. The standard quantity of kilograms of plastic (SQ) that is allowed to make 3,200 helmets?
We know that each helmet requires 0.68 kilograms of plastic. Thus, to make 3200 helmets, we will need 0.68*3200 = 2176 kilograms of plastic
2. The standard materials cost allowed (SQ × SP) to make 3,200 helmets.
We also know that each helmet costs $7.00 per kilogram. Therefore, to make 3200 helmets, the standard materials cost = $7.00*2176 = $15,232
3. The material spending variance = difference between the actual cost and the standard cost = $15,629 - $15,232 = $397 U
4. The materials price variance and the materials quantity variance?
The materials price variance is the actual cost- (standard cost per kilogram x actual number of plastic used) . Therefore:
Materials price variance = $15,629 - ($7 x 2,368 kg)
Materials price variance = $15,629 - $16,576 = ($947) F
Since the budgeted cost is relatively higher than the actual cost, the materials price variance is favorable (F) by $947.
The materials quantity variance = (Actual number of plastic used x Standard cost per kilogram) - Standard cost
Materials quantity variance = ($7 x 2,368 kg) - $15,232
Materials quantity variance = $16,576 - $15,232 = $1,344 U
Since the budgeted cost is relatively higher than the standard cost, the materials quantity variance is unfavorable (U) by $1,344.
To find the standard quantity of plastic and standard materials cost allowed, multiply the standard quantity per helmet by the number of helmets produced and multiply the standard quantity by the standard price per kilogram, respectively. The materials spending variance is the difference between the actual cost and the standard cost that should have been incurred. The materials price variance is the difference between the actual quantity of plastic used multiplied by the standard price per kilogram and the actual cost, while the materials quantity variance is the difference between the standard quantity of plastic allowed multiplied by the standard price per kilogram and the actual cost.
Explanation:To calculate the standard quantity of plastic (SQ) allowed to make 3,200 helmets, multiply the standard quantity per helmet by the number of helmets produced. In this case, SQ = 0.68 kg/helmet * 3,200 helmets = 2,176 kg. To calculate the standard materials cost allowed (SQ × SP), multiply the standard quantity by the standard price per kilogram. In this case, the materials cost allowed is $15,232 (2,176 kg * $7.00/kg).
To calculate the materials spending variance, subtract the actual cost from the standard cost that should have been incurred. In this case, the materials spending variance is $397 (Standard Cost - Actual Cost = $15,232 - $15,629). The materials price variance is calculated by subtracting the actual quantity of plastic used multiplied by the standard price per kilogram from the actual cost.
The materials quantity variance is calculated by subtracting the standard quantity of plastic allowed multiplied by the standard price per kilogram from the actual cost. In this case, the materials price variance is $7,774 (Actual Cost - Actual Quantity * Standard Price = $15,629 - 2,368 kg * $7.00/kg) and the materials quantity variance is -$2,145 (Actual Quantity * Standard Price - Standard Cost = 2,368 kg * $7.00/kg - $15,232).
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A continuous and aligned fiber reinforced composite having a cross-sectional area of 1130 mm^2 (1.75 in.^2) is subjected to an external tensile load. If the stresses sustained by the fiber and matrix phases are 156 MPa (22,600 psi) and 2.75 MPa (400 psi), respectively; the force sustained by the fiber phase is 74,000 N (16,600 lbf); and the total longitudinal strain is 1.25 10^ -3, determine: (a)the force sustained by the matrix phase, (b)the modulus of elasticity of the compos- ite material in the longitudinal direction, and (c) the moduli of elasticity for fiber and matrix phases.
Answer:
a) [tex]F_m=1803.013\ N[/tex]
b) [tex]E=53665.84\ MPa[/tex]
c) [tex]E_f=124800\ MPa[/tex]
[tex]E_m=2200\ MPa[/tex]
Explanation:
Given:
cross-sectional area of reinforced composite, [tex]A=1130\ mm^2[/tex]stress sustained by the fiber phase, [tex]\sigma_f=156\ MPa[/tex]force sustained by the fiber phase, [tex]F_f=74000\ N[/tex]Total strain on the composite, [tex]\epsilon=1.25\times 10^{-3}[/tex]stress sustained in the matrix phase, [tex]\sigma_m=2.75\ MPa[/tex]Now, the area of fiber phase:
[tex]A_f=\frac{F_f}{\sigma_f}[/tex]
[tex]A_f=\frac{74000}{156}[/tex]
[tex]A_f=474.359\ mm^2[/tex]
∴Area of matrix phase:
[tex]A_m=A-A_f[/tex]
[tex]A_m=1130-474.359[/tex]
[tex]A_m=655.641\ mm^2[/tex]
(a)
Now the force sustained by the matrix phase:
[tex]F_m=\sigma_m\times A_m[/tex]
[tex]F_m=2.75\times 655.641[/tex]
[tex]F_m=1803.013\ N[/tex]
(b)
Total stress on the composite:
[tex]\sigma=\frac{(F_f+F_m)}{A}[/tex]
[tex]\sigma=\frac{(74000+1803.013)}{1130}[/tex]
[tex]\sigma=67.082\ MPa[/tex]
Now,Modulus of elasticity of the composite:
[tex]E=\frac{\sigma}{\epsilon}[/tex]
[tex]E=\frac{67.082}{1.25\times 10^{-3}}[/tex]
[tex]E=53665.84\ MPa[/tex]
(c)
Since, strain will be same in this case throughout the material.
Now the modulus of elasticity of fiber phase:
[tex]E_f=\frac{\sigma_f}{\epsilon}[/tex]
[tex]E_f=\frac{156}{1.25\times 10^{-3}}[/tex]
[tex]E_f=124800\ MPa[/tex]
Now the modulus of elasticity of matrix phase:
[tex]E_m=\frac{\sigma_m}{\epsilon}[/tex]
[tex]E_m=\frac{2.75}{1.25\times 10^{-3}}[/tex]
[tex]E_m=2200\ MPa[/tex]
The force sustained by the matrix phase is 3125 N, the modulus of elasticity of the composite material in the longitudinal direction is 124800 MPa, and the moduli of elasticity for the fiber and matrix phases cannot be calculated without information about the volume fractions.
Explanation:To determine the force sustained by the matrix phase, we can use the equation:
Force sustained by matrix phase = Stress in matrix phase * Cross-sectional area
Using given values, the force sustained by the matrix phase is:
Force sustained by matrix phase = 2.75 MPa * 1130 mm2 = 3125 N
To determine the modulus of elasticity of the composite material in the longitudinal direction, we can use the equation:
Modulus of elasticity = Stress / Strain
Using given values, the modulus of elasticity of the composite material in the longitudinal direction is:
Modulus of elasticity = 156 MPa / (1.25 x 10-3) = 124800 MPa
To determine the moduli of elasticity for the fiber and matrix phases, we can multiply the modulus of elasticity of the composite material by the volume fractions of the fiber and matrix phases. In this case, we don't have information about the volume fractions, so we cannot calculate the moduli of elasticity for the fiber and matrix phases.
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You look towards a traffic light and see a yellow light. If you were to drive towards it at near the speed of light, what color would it appear?
A. Red
B. Green
Answer:
B. Green
Explanation:
The change in wavelength that is caused when one object is moving towards another object from the perspective of the viewer is called the Doppler effect.
When objects move close to one another then wavelength reduces which is called blue shift while the opposite case causes the wavelength to increase which is called red shift.
Here, the color of the traffic light is yellow and you are getting closer to it so the wavelength should blue shift and green should appear.
A bowling ball of mass 5.8 kg moves in a
straight line at 1.59 m/s.
How fast must a Ping-Pong ball of mass
1.528 g move in a straight line so that the two
balls have the same momentum?
Answer in units of m/s.
Answer:
Velocity of the ping pong ball must be = V2= 6,035.34m/s
Explanation:
M1= momentum of the bowling ball
m1 = mass of the bowling ball= 5.8kg
v1= velocity of the bowling ball= 1.59m/s
M2= momentum of the ping pong ball
m2= mass of the ping pong ball= 1.528 g/1000= 0.001528kg
v2= velocity of the ping pong ball
Momentum of the bowling ball= M1= m1v1= 5.8* 1.59= 9.222 kg-m/s
Momentum of the ping pong ball = M2= M1= m2v2
= 0.001528 *v2= 9.222
v2= 9.222/0.001528= 6,035.34 m/s
NASA operates a 2.2-second drop tower at the Glenn Research Centre in Cleveland, Ohio. At this facility, experimental packages are dropped from the top of the tower, on the 8th floor of the building. During their 2.2 seconds of free fall, experiments experience a microgravity environment similar to that of a spacecraft in orbit,
(a) What is the drop distance of a 2.2-s tower?
(b) How fast are the experiments travelling when they hit the air bags at the bottom of the tower?
(c) If the experimental package comes to rest over a distance of 0.75 m upon hitting the air bags, what is the average stopping acceleration?
Answer:
23.7402 m
21.582 m/s
310.521816 m/s²
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
g = Acceleration due to gravity = 9.81 m/s² = a
Equation of motion
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 2.2^2\\\Rightarrow s=23.7402\ m[/tex]
The drop distance is 23.7402 m
[tex]v=u+at\\\Rightarrow v=0+9.81\times 2.2\\\Rightarrow v=21.582\ m/s[/tex]
When they hit the air bags at the bottom of the tower the speed of the experiments is 21.582 m/s
The final speed of the fall will be the initial velocity of stopping
[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-21.582^2}{2\times 0.75}\\\Rightarrow a=-310.521816\ m/s^2[/tex]
The average stopping acceleration is 310.521816 m/s²
Final answer:
Calculations reveal a drop distance of 23.65 meters, a final velocity of 21.56 m/s, and an average stopping acceleration of 311.43 m/s² for the experimental package in the 2.2-second drop tower scenario.
Explanation:
The question encompasses the calculation of drop distance, final velocity, and average stopping acceleration of an experiment package in a microgravity test environment. To solve these, we apply the equations of motion under uniform acceleration.
To find the drop distance (d) for a 2.2-second fall, we use the equation d = 0.5 * g * t, where g is the acceleration due to gravity (9.8 m/s²) and t is time in seconds. This gives us d = 0.5 * 9.8 * 2.22 = 23.65 meters.To calculate the final velocity (v) when the experiments hit the air bags, we employ v = g * t, resulting in v = 9.8 * 2.2 = 21.56 m/s.The average stopping acceleration (a) can be derived from the equation v² = 2 * a * d, rearranged to a = v² / (2*d), where d is the stopping distance of 0.75 m, leading to a = (21.562) / (2*0.75) = 311.43 m/s².This analysis presupposes a vacuum environment in the drop tower to negate air resistance, thereby approximating conditions of free fall and achieving microgravity for the duration of the drop.
A miniature spring-loaded, radio-controlled gun is mounted on an air puck. The gun's bullet has a mass of 7.00 g, and the gun and puck have a combined mass of 150 g. With the system initially at rest, the radio controlled trigger releases the bullet causing the puck and empty gun to move with a speed of 0.530 m/s. What is the bullet's speed?
Answer:
Explanation:
Given
mass of bullet [tex]m=7 gm[/tex]
mass of gun and Puck [tex]M=150 gm[/tex]
speed of gun and Puck is [tex]v=0.53 m/s[/tex]
Let speed of bullet be u
conserving Momentum
initial momentum=Final Momentum
[tex]0=Mv+mu[/tex]
[tex]u=-\frac{M}{m}v[/tex]
[tex]u=-\frac{150}{7}\times 0.53=-11.35 m/s[/tex]
negative sign indicates that bullet is moving in opposite direction
Using the law of conservation of momentum, it's possible to calculate that the bullet moves in the opposite direction of the gun and puck at a speed of approximately 113.6 m/s.
Explanation:This question represents a practical example of the conservation of momentum. According to the law of conservation of momentum, the total momentum of an isolated system remains constant if no external forces act on it. As the system (gun and bullet) is initially at rest, the total momentum is zero, and it remains zero even after the bullet is fired.
Let's denote the bullet's velocity as v. The momentum of the bullet after firing is its mass times its velocity, or 0.007kg * v. The momentum of the gun and puck is their combined mass times their velocity, or 0.150kg * 0.530m/s. As the total momentum should remain zero, these two quantities must be equal and opposite. Thus, 0.007kg * v = - 0.150kg * 0.530m/s. Solving for v, we find that the bullet's speed is approximately -113.6 m/s. The negative sign indicates that the bullet moves in the opposite direction of the gun and puck.
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Choose the scenario under which each of the given Doppler shift effects will be seen.a. The source and observer are approaching one anotherb. The source and observer are moving away from one anotherc. The source and observer are stationary relative to one another
Answer:
scenario A and B
Explanation:
The Doppler effect is the change in frequency of a wave by the relative movement of the source and the observer. It is described by the expression
f ’= (v + v₀) / (v - [tex]v_{s}[/tex]) f₀
Where f₀ is the emitted frequency, v the speed of the wave, v₀ and [tex]v_{s}[/tex] the speed of the observer and the source, respectively, the signs are for when they are approaching and in the case of being away the signs are changed.
Consequently, from the above for the Doppler effect to exist there must be a relative movement of the source and the observer.
Let's examine the scenarios
A) True. You agree with the equation shown
B) True. Only the signs should be changed and it is described by the equation shown
C) False. If there is no relative movement there is no Doppler effect
You are lowering two boxes, one on top of the other, down a ramp by pulling on a rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 15.0 cm/s. The coefficient of kinetic friction between the ramp and the lower box is 0.444, and the coefficient of static friction between the two boxes is 0.800.
(a) What force do you need to exert to accomplish this? (b) What are the magnitude and direc- tion of the friction force on the upper box?
Answer:
(a) 57.17 N
(b) 146.21 N up the ramp
Explanation:
Assume the figure attached
(a)
The angle of ramp, [tex]\theta=tan ^{-1} \frac {2.5}{4.75}=27.75854^{\circ}\approx 27.76^{\circ}[/tex]
[tex]N=(32+48)*9.81*cos 27.76^{\circ}=694.4838 N[/tex]
m=32+48=80 kg
[tex]T+ \mu N= mg sin \theta[/tex]
[tex]T=mg sin \theta - \mu N= mg sin \theta- \mu mg cos \theta= mg (sin \theta - \mu cos \theta) [/tex] where [tex]\mu[/tex] is coefficient of kinetic friction
[tex]T=80*9.81(sin 27.76^{\circ} -0.444 cos 27.76^{\circ})=57.16698 N \approx 57.17 N[/tex]
(b)
Upper box doesn’t accelerate
[tex]F_r= mgsin\theta= 32*9.81sin 27.76^{\circ}=146.2071\approx 146.21 N[/tex]
The direction will be up the ramp
To lower the boxes down the ramp, a force is needed that is equal to the force of kinetic friction on the lower box. The magnitude and direction of the friction force on the upper box is the same as the force of kinetic friction on the lower box.
Explanation:To determine the force required to lower the two boxes down the ramp, we can use the equation: force = friction force + weight force. Since the boxes are moving at a constant speed, the force of kinetic friction on the lower box is equal to the force applied to it. Therefore, the force needed to lower the boxes is equal to the force of kinetic friction on the lower box which can be calculated using the equation: force = coefficient of kinetic friction × normal force.
For the magnitude and direction of the friction force on the upper box, we can use the equation: friction force = coefficient of static friction × normal force. Since the boxes are moving together, we can assume that the friction force on the upper box is equal to the force of kinetic friction on the lower box. Therefore, the magnitude and direction of the friction force on the upper box is the same as the force of kinetic friction on the lower box.
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A 0.6-m3 rigid tank contains 0.6 kg of N2 and 0.4 kg of O2 at 300 K. Determine the partial pressure of each gas and the total pressure of the mixture. The gas constant for N2 is 0.2968 kPa·m3/kg·K and the gas constant for O2 is 0.2598 kPa·m3/kg·K.
Answer:
Pnitrogen=3.18 kPa, Poxygen=1.62 kPa , Ptotal= 4.80 kPa
Explanation:
partial pressure equation becomes Ptotal = Pnitrogen + Poxygen
Partial pressure of Nitrogen
Pnitrogen= nRT/V
n=no of moles =mass/molar mass
mass of nitrogen=0.6kg
Molar mass of nitrogen gas=28gmol^-1
n=0.6/28=0.0214moles
R=0.2968 kPa·m3/kg·K
T=300k
V=0.6m^3
Pnitrogen=(0.0214 * 0.2968 * 300)/0.6
Pnitrogen=3.18 kPa
Likewise
Poxygen=nRT/V
n=0.4/32=0.0125moles
R=0.2598 kPa·m3/kg·K
T=300k
V=0.6m^3
Poxygen=(0.0125 * 0.2598 * 300)/0.6
Poxygen=1.62 kPa
Ptotal= 3.18+1.62= 4.80 kPa
Using the ideal gas law, the partial pressure for Nitrogen ([tex]N_{2}[/tex]) and Oxygen ([tex]O_{2}[/tex]) is 148.4 kPa and 86.6 kPa respectively. The total pressure of the mixture is the sum of these two partial pressures, equaling 235 kPa.
Explanation:The pressure exerted by individual gases in a mixture is known as partial pressure. The calculation of the partial pressure of each gas, and the total pressure of the mixture involves using the ideal gas law. In this case, the ideal gas equation is P = mRT/V, where P represents the pressure, m is the mass of gas, R is the gas constant, T is the temperature and V is the volume. Thus, for Nitrogen ([tex]N_{2}[/tex]), the partial pressure is (0.6 kg × 0.2968 kPa·m3/kg·K × 300K) / 0.6 m3 = 148.4 kPa, and for Oxygen ([tex]O_{2}[/tex]), the partial pressure is (0.4 kg× 0.2598 kPa·m3/kg·K × 300K) / 0.6 m3 = 86.6 kPa. The total pressure, then, is the sum of these partial pressures, which equals 148.4 kPa + 86.6 kPa = 235 kPa.
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The system consists of a 20-lb disk A, 4-lb slender rod BC, and a 1-lb smooth collar C. If the disk rolls without slipping, determine the velocity of the collar at the instant the rod becomes horizontal. The system is released from rest when != 45°.
The velocity of the collar at the instant the rod becomes horizontal can be determined using the conservation of angular momentum principle. By calculating the moment of inertia of the disk and equating the initial and final angular momenta, we can solve for the final angular velocity. Since the disk rolls without slipping, the final linear velocity of the disk can be determined using the equation v' = Rω'.
Explanation:To determine the velocity of the collar at the instant the rod becomes horizontal, we can use conservation of angular momentum. Since the disk rolls without slipping, its angular momentum is conserved. The angular momentum is given by the product of the moment of inertia and the angular velocity. When the rod becomes horizontal, the angular velocity of the disk is equal to the velocity of the collar.
We can calculate the moment of inertia of the disk using the formula I = (1/2)MR^2, where M is the mass of the disk and R is its radius. Substituting the values, we get I = (1/2)(20 lb)(2 ft)^2. We also know that the angular momentum is conserved, so the initial angular momentum is equal to the final angular momentum. The initial momentum is Iω, where ω is the initial angular velocity. The final momentum is Iω', where ω' is the final angular velocity.
Since the disk rolls without slipping, the linear velocity of the disk is equal to the radius times the angular velocity, v = Rω. So, the final linear velocity of the disk is equal to the final angular velocity times the radius, v' = Rω'. Substituting the values, we can solve for ω'.
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