Answer:
Case Study: General Andrew Jackson: Andrew Jackson's military career spanned several wars including the American Revolution, the Creek War, the War of 1812, and the Case Study: General Andrew Jackson: Andrew Jackson's military career spanned several wars including the American Revolution, the Creek War, the War of 1812, and the First Seminole War. After the Creek War, Jackson and the Creek Indians signed the
Explanation:
Case Study: General Andrew Jackson: Andrew Jackson's military career spanned several wars including the American Revolution, the Creek War, the War of 1812, and the First Seminole War. After the Creek War, Jackson and the Creek Indians signed the
Case Study: General Andrew Jackson: Andrew Jackson's military career spanned several wars including the American Revolution, the Creek War, the War of 1812, and the First Seminole War. After the Creek War, Jackson and the Creek Indians signed theCase Study: General Andrew Jackson: Andrew Jackson's military career spanned several wars including the American Revolution, the Creek War, the War of 1812, and the First Seminole War. After the Creek War, Jackson and the Creek Indians signed the
If a camera lens gives the proper exposure for a photograph at a shutter speed of 1/200 s at an f-stop of f/2.80, the proper shutter speed at f/7.92 is
Answer:
[tex]\dfrac{1}{28.15432}\ s[/tex]
Explanation:
The factor by which the shutter speed is increased is [tex]\dfrac{7.92}{2.8}[/tex]
Exposure time will be increased by
[tex]f=2^{\dfrac{7.92}{2.8}}=7.1037\ s[/tex]
The proper shutter speed is given by
[tex]\dfrac{1}{T}f=\dfrac{1}{200}\times 2^{\dfrac{7.92}{2.8}}=\dfrac{1}{\dfrac{200}{2^{\dfrac{7.92}{2.8}}}}\\ =\dfrac{1}{28.15432}\ s[/tex]
The proper shutter speed is [tex]\dfrac{1}{28.15432}\ s[/tex]
For small amplitudes of oscillation the motion of a pendulum is simple harmonic. Consider a pendulum with a period of 0.550 s Find the ground-level energy. Express your result in joules Find the ground-level energy. Express your result in election volts
To solve this problem we will use the concepts related to the expression of energy for harmonic oscillator. From our given values we have that the period is equivalent to
[tex]T = 0.55s[/tex]
Therefore the frequency will be the inverse of the period and would be given as
[tex]f= \frac{1}{T}[/tex]
[tex]f = \frac{1}{0.55}[/tex]
[tex]f = 1.82s^{-1}[/tex]
The ground state energy of the pendulum is,
[tex]E = \frac{1}{2} hv[/tex]
[tex]E = \frac{1}{2}(6.626*10^{-34}J\cdot s)(1.82s^{-1})[/tex]
[tex]E = 6.03*10^{-34}J[/tex]
The ground state energy in eV,
[tex]E = 6.03 * 10^{-34}J(\frac{1eV}{1.6*10^{-19}J})[/tex]
[tex]E = 3.8*10^{-15}eV[/tex]
The energy difference between adjacent energy levels,
[tex]\Delta E = hv[/tex]
[tex]\Delta E = (6.626*10^{-34}J\cdot s)(1.82s)[/tex]
[tex]\Delta E = 12.1*10^{-34}J[/tex]
As a parallel-plate capacitor with circular plates 24 cm in diameter is being charged, the current density of the displacement current in the region between the plates is uniform and has a magnitude of 20 A/m2.
(a) Calculate the magnitude B of the magnetic field at a distance r = 87 mm from the axis of symmetry of this region.
(b) Calculate dE/dt in this region.
Answer:
Explanation:
a.)
The magnitude field
[tex]B=\frac{\mu _0I_{enclosed}}{2\pi r}\\\\=\frac{\mu _0(J_ar^2)}{2\pi r}\\\\=\frac{1}{2}(\mu _0J_ar)\\\\(0.5)(4\pi \times 10^{-7})(20A/m^2)(87\times 10^{-3})\\\\=1.0933\times 10^{-6}T[/tex]
b.)
The displacement current
[tex]i_d=\epsilon_0A\frac{dE}{dt}[/tex]
then [tex]\frac{dE}{dt}=\frac{i_d}{\epsilon_0A}\\\\=\frac{J_d}{\epsilon_0}\\\\=\frac{20}{8.85\times 10^{-12}}\\\\=2.26\times 10^{12}V/ms[/tex]
A flat plate is oriented parallel to a 45 m/s airflow at 20°C and atmospheric pressure. The plate is L = 1 m in the flow direction and 0.5 m wide. On one side of the plate, the boundary layer is tripped at the leading edge, and on the other side there is no tripping device. Find the total drag force on the plate.
Answer:
4.192 N
Explanation:
Step 1: Identify the given parameters
Velocity of airflow = 45m/s
air temperature = 20⁰
plate length and width = 1m and 0.5m respectively.
Step 2: calculate drag force due to shear stress, [tex]F_{s}[/tex]
[tex]F_{s} = C_{f} \frac{1}{2} (\rho{U_{o}}WL)[/tex]
Note: The density and kinematic viscosity of air at 20⁰ at 1 atm, is 1.2 kg/m³ and 1.5 X 10⁻⁵ N.s/m²
⇒The Reynolds number ([tex]R_{eL}[/tex]) based on the length of the plate is
[tex]R_{eL} =\frac{VXL}{U}[/tex]
[tex]R_{eL} =\frac{45X1}{1.5 X 10^{-5}}[/tex]
[tex]R_{eL}[/tex] = 3 X10⁶ (flow is turbulent, Re ≥ 500,000)
⇒The average shear stress coefficient ([tex]C_{f}[/tex]) on the "tripped" side of the plate is
[tex]C_{f} = \frac{0.074}{(R_{e})^\frac{1}{5}}[/tex]
[tex]C_{f} = \frac{0.074}{(3 X10^6)^\frac{1}{5}}[/tex]
[tex]C_{f}[/tex] = 0.0038
⇒The average shear stress coefficient ([tex]C_{f}[/tex]) on the "untripped" side of the plate is
[tex]C_{f} = \frac{0.523}{in^2(0.06XR_{e})} -\frac{1520}{R_{e}}[/tex]
[tex]C_{f} = \frac{0.523}{in^2(0.06 X 3X10^6)} -\frac{1520}{3X10^6}[/tex]
[tex]C_{f}[/tex] = 0.0031
The total drag force = [tex]\frac{1}{2}(1.2 X 45^2 X 1 X 0.5 (0.0038 +0.0031)[/tex]
The total drag force is 4.192 N
Intelligent beings in a distant galaxy send a signal to earth in the form of an electromagnetic wave. The frequency of the signal observed on earth is 1.1% greater than the frequency emitted by the source in the distant galaxy. What is the speed vrel of the galaxy relative to the earth?
Answer:
3900000 m/s
Explanation:
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
v = Speed of the galaxy relative to the earth
Observed frequency is
[tex]f'=(1+0.013)f\\\Rightarrow \dfrac{f'}{f}=1.013[/tex]
Here the Doppler relation must be used.
So, observed frequency is given by
[tex]f'=f\dfrac{v+c}{c}\\\Rightarrow \dfrac{f'}{f}=\dfrac{v+c}{c}\\\Rightarrow v=\dfrac{f'c}{f}-c\\\Rightarrow v=(1.013\times 3\times 10^8)-3\times 10^8\\\Rightarrow v=3900000\ m/s[/tex]
The speed of the galaxy relative to the earth is 3900000 m/s
A counter attendant in a diner shoves a ketchup bottle with a mass 0.30 kg along a smooth, level lunch counter. The bottle leaves her hand with an initial velocity 2.8 m/s. As it slides, it slows down because of the horizontal friction force exerted on it by the countertop. The bottle slides a distance of 1.0 m before
coming to rest. What are the magnitude and direction of the friction force acting on it?
The magnitude of the friction force acting on the ketchup bottle is 0.84 N, and it is directed opposite to the initial velocity of the bottle.
Explanation:The friction force acting on the ketchup bottle can be determined using the equation:
Friction force = Mass x Acceleration
Since the bottle comes to rest, its final velocity is 0 m/s and the acceleration can be calculated using the equation:
Acceleration = (Final Velocity - Initial Velocity) / Time
By substituting the given values, we find that the friction force acting on the bottle is 0.30 kg x (-2.8 m/s) / (1.0 m/s^2) = -0.84 N. The negative sign indicates that the friction force is acting in the opposite direction of the initial velocity, which is in the direction of the attendant's hand.
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An observer stands on the side of the front of a stationary train. When the train starts moving with constant acceleration, the observer manages to measure the time it.takes for the second cart to pass him to be 5.0 seconds How long will it take for the 10th car to pass him? Assume all cars to be of the same length.
a. 2.8 s
b. 2.4 s
c. 2.0 s
d. 1.5 s
e. 1.1 s
To solve this problem we will use the linear motion description kinematic equations. We will proceed to analyze the general case by which the analysis is taken for the second car and the tenth. So we have to:
[tex]x = v_0 t \frac{1}{2} at^2[/tex]
Where,
x= Displacement
[tex]v_0[/tex] = Initial velocity
a = Acceleration
t = time
Since there is no initial velocity, the same equation can be transformed in terms of length and time as:
[tex]L = \frac{1}{2} a t_1 ^2[/tex]
For the second cart
[tex]2L \frac{1}{2} at_2^2[/tex]
When the tenth car is aligned the length will be 9 times the initial therefore:
[tex]9L = \frac{1}{2} at_3^2[/tex]
When the tenth car has passed the length will be 10 times the initial therefore:
[tex]10L = \frac{1}{2}at_4^2[/tex]
The difference in time taken from the second car to pass it is 5 seconds, therefore:
[tex]t_2-t_1 = 5s[/tex]
From the first equation replacing it in the second one we will have that the relationship of the two times is equivalent to:
[tex]\frac{1}{2} = (\frac{t_1}{t_2})^2[/tex]
[tex]t_1 = \frac{t_2}{\sqrt{2}}[/tex]
From the relationship when the car has passed and the time difference we will have to:
[tex](t_2-\frac{t_2}{\sqrt{2}}) = 5[/tex]
[tex]t_2 (\sqrt{2}-1) = 3\sqrt{2}[/tex]
[tex]t_2= (\frac{5\sqrt{2}}{\sqrt{2}-1})^2[/tex]
Replacing the value found in the equation given for the second car equation we have to:
[tex]\frac{L}{a} = \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2[/tex]
Finally we will have the time when the cars are aligned is
[tex]18 \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_3^2[/tex]
[tex]t_3 = 36.213s[/tex]
The time when you have passed it would be:
[tex]20\frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_4^2[/tex]
[tex]t_4 = 38.172[/tex]
The difference between the two times would be:
[tex]t_4-t_3 = 38.172-36.213 \approx 2s[/tex]
Therefore the correct answer is C.
The passage time for the 10th car in a uniformly accelerating train would be less than 5.0 seconds but not as low as any of the options provided (a to e), as each car passes more quickly than the previous one due to constant acceleration.
Explanation:The question involves an observer timing how long it takes for individual cars of constant acceleration of a train to pass by. If it took 5.0 seconds for the second car to pass the observer, then the time it takes for the 10th car to pass him is merely the time it took for one car to pass because the train is accelerating at a constant rate, and all cars are of the same length. Considering the train is accelerating uniformly, the passage of each car happens more quickly over time.
Giving this knowledge, for the first car to pass, the train would have started at rest, so it would have taken longer than the second car which had the benefit of the train already moving at a certain velocity. Therefore, we can derive that the time for each subsequent car to pass will be less than 5.0 seconds, which questions the multiple-choice options. Based on this logic, we don't have to calculate the exact passing time for the 10th car, as there are no values given to calculate with, but we can infer that none of the times listed would be correct. Each car would be passing quicker than the last, but it would not be instantaneous; it would be a fraction of the 5 seconds but not as low as any of the times listed from a to e.
A loop-the-loop has a circular arc, with a marble that can run along a track and traverse the entire inside of the loop. When the marble is precisely at the top of the inside loop, it has its minimum speed. ~v What do we know about the direction of the net force on the marble at this point, the top of the loop? a. The net force is instantaneously in the direction of the marble’s velocity b. The net force is instantaneously in the opposite direction to the marble’s velocity ~v. c. The net force is vertically downward. d. The net force is vertically upward
Answer:
c. The net force is vertically downward.
Explanation:
At the top of the loop, the only external force that keeps the ball moving around the loop, is the centripetal force.
Now, this centripetal force, is not a " new" force, it's just the vector sum of the two external forces (neglecting friction) , that act simultaneously on the marble, making it to change its speed, in magnitude and direction: the gravity force (which it is always downward), and the normal force(which is always perpendicular to the contact surface, preventing that the marble comes trough the surface), in this case between the marble and the track, which, at the top of the loop, points down too.
So, the net force, exactly at the top of the loop, is vertically downward.
A rocket burns fuel at a rate of 264 kg/s and
exhausts the gas at a relative speed of 8 km/s.
Find the thrust of the rocket.
Answer in units of MN.
Answer:
2.112 MN
Explanation:
Force = mass × acceleration. Each second, 264 kg of gas is accelerated from 0 to 8 km/s.
F = ma
F = m Δv / Δt
F = 264 kg × (8000 m/s − 0 m/s) / 1 s
F = 2,112,000 kg m/s²
F = 2.112 MN
Round as needed.
The thrust of the rocket, calculated using the formula - Thrust = mass rate of exhaust x exhaust velocity, is 2.112 Mega Newtons.
Explanation:The thrust produced by a rocket is typically calculated using the formula: Thrust = mass rate of exhaust x exhaust velocity. In this scenario, the rocket is burning fuel at a rate of 264 kg/s (mass rate of exhaust), and the exhaust gas is ejected at a relative speed of 8 km/s.
However, to ensure the units are consistent, we have to convert the relative speed from km/s to m/s, which results in 8000 m/s (exhaust velocity).
Substituting these values into the formula, we get: Thrust = 264 kg/s x 8000 m/s = 2,112,000 N = 2.112 MN.
Therefore, the thrust of the rocket is 2.112 Mega Newtons.
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Stars form from clouds of gas and dust. As a protostar gravitationally contracts within its parent cloud, "conservation of angular momentum" says thatA) the protostar rotates more slowly.b) the protostar\'s rotation does not change.c) the protostar\'s axis of rotation will change direction.d) the protostar rotates more quickly.
Answer:
D) True. the protostar rotates more quickly.
Explanation:
If the system is isolated, the angular momentum must be retained.
Initial
L₀ = I w₀
Final
[tex]L_{f}[/tex] = [tex]I_{f}[/tex] [tex]w_{f}[/tex]
L₀ = [tex]L_{f}[/tex]
I w₀ = [tex]I_{f}[/tex][tex]w_{f}[/tex]
[tex]w_{f}[/tex] = I /[tex]I_{f}[/tex] w₀
In general, the radius of the cloud decreases significantly to form the star, the moment of inertia must decrease, so the angular velocity must increase
Let's examine the answers
A) False. The opposite happens
B) False. Speed changes
C) False. For this there must be an external force, which does not exist
D) True. You agree with the above
Final answer:
The correct answer to the question of what occurs to a protostar as it contracts due to the 'conservation of angular momentum' is that the protostar rotates more quickly (d). This is analogous to a figure skater pulling their arms in to spin faster and is supported by observations of star-forming regions like the Orion Nebula.
Explanation:
Stars form from clouds of gas and dust. As these clouds collapse under their own gravity, they form protostars, which spin due to the conservation of angular momentum. Conservation of angular momentum dictates that as the radius of a spinning object decreases, its rotation speed must increase to conserve angular momentum. This concept is similar to a figure skater who spins faster when they pull their arms in. Therefore, when a protostar gravitationally contracts within its parent cloud, it rotates more quickly because as it shrinks, the protostar's rate of spin increases to conserve angular momentum.
This increase in rotation speed results in the formation of a spinning accretion disk around the equator, which is easier to observe in some regions, such as the Orion Nebula or the Taurus star-forming region. Observations from telescopes like the Hubble Space Telescope support this understanding of the role of angular momentum in star formation. In conclusion, the correct answer to the question is (d) the protostar rotates more quickly.
A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnification of 400x with an objective lens that has a focal length of 0.40 cm. The distance between the eyepiece and objective lenses is 12 cm. 1) Find the focal length of the eyepiece lens assuming a near point of 25 cm (the closest an object can be and still be seen in focus). Do not neglect any values in your calculation. (Express your answer to two significant figures.)
Answer:
[tex]f_{e}[/tex] = 1.9 cm
Explanation:
The magnification of a microscope is the product of the magnification of the eyepiece by the magnifier with the objective
M = M₀ [tex]m_{e}[/tex]
Where M₀ is the magnification of the objective and [tex]m_{e}[/tex] is the magnification of the eyepiece.
The eyepiece is focused to the near vision point (d = 25 cm)
[tex]m_{e}[/tex] = 25 / [tex]f_{e}[/tex]
The objective is focused on the distances of the tube (L)
M₀ = -L / f₀
Substituting
M = - L/f₀ 25/[tex]f_{e}[/tex]
1) Let's look for the focal length of the eyepiece (faith)
[tex]f_{e}[/tex] = - L 25 / f₀ M
M = 400X = -400
[tex]f_{e}[/tex] = - 12 25 /0.40 (-400)
[tex]f_{e}[/tex] = 1.875 cm
Let's approximate two significant figures
[tex]f_{e}[/tex] = 1.9 cm
Final answer:
The focal length of the eyepiece lens in the microscope is approximately -0.01 mm.
Explanation:
To find the focal length of the eyepiece lens, we can use the formula for the magnification of a compound microscope, which is given by:
M = -fobjective/feyepiece
Where M is the overall magnification, fobjective is the focal length of the objective lens, and feyepiece is the focal length of the eyepiece lens.
Given that the overall magnification is 400x and the focal length of the objective lens is 0.40 cm, we can rearrange the formula to solve for the focal length of the eyepiece lens:
feyepiece = -fobjective/M = -0.40 cm/400 = -0.001 cm = -0.01 mm.
Therefore, the focal length of the eyepiece lens is approximately -0.01 mm.
will a flying bird have more kinetic energy than a sitting elephant
Answer:
Yes.
Explanation:
A sitting elephant has zero kinetic energy. A flying bird have some kinetic energy due to its motion. Regardless of their size, a moving object has always more kinetic energy than an object at rest.
You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the reading of 400. counts has diminished to 100. counts after 90.3 minutes, what is the half-life of this substance?
Answer : The half-life of this substance will be, 45 minutes.
Explanation :
First we have to calculate the value of rate constant.
Expression for rate law for first order kinetics is given by:
[tex]k=\frac{2.303}{t}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = ?
t = time passed by the sample = 90.3 min
a = initial amount of the reactant = 400
a - x = amount left after decay process = 100
Now put all the given values in above equation, we get
[tex]k=\frac{2.303}{90.3min}\log\frac{400}{100}[/tex]
[tex]k=1.54\times 10^{-2}\text{ min}^{-1}[/tex]
Now we have to calculate the half-life of substance, we use the formula :
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]1.54\times 10^{-2}\text{ min}^{-1}=\frac{0.693}{t_{1/2}}[/tex]
[tex]t_{1/2}=45min[/tex]
Therefore, the half-life of this substance will be, 45 minutes.
Consider two different rods. The greatest thermal conductivity will be in the rod with:
a. electrons that are freer to move from atom to atom.
b. the greater specific heat.
c. the greater cross-sectional area.
d. the greater length.
Answer:
Options A and D are correct
Explanation:
The thermal conductivity of a metal is the property of a metal to allow heat flow through it. conductivity is higher in conductors and low in insulators. Thermal conductivity is high in metals due to the metallic bonds that exist in metals and the presence of free electrons within the metal which allow easy flow of heat from one atom to another.From the problem the rod which contains freer electrons will allow more heat to flow easily hence have a higher thermal conductivity.
Thermal conductivity has the formula below;
[tex]k= \frac{QL}{AΔT}[/tex]
k is thermal conductivity,A is cross sectional area L is length of rod Q is quantity of heat transferred to material. ΔT is temperature change.From the above equation we can see that thermal conductivity is inversely proportional to A and directly proportional to L. This mean the rod with less area will have a higher thermal conductivity and the rod with a higher length will have higher k. Hence option C i wrong and option D is correct.
For specific heat, its very much different from thermal conductivity. Specific heat is the ability of a material to hold heat while thermal conductivity is the ability of heat to flow through a material.
In braking an automobile, the friction between the brake drums and brake shoes converts the car's kinetic energy into heat. If a 1 500-kg automobile traveling at 30 m/s brakes to a halt, how much does the temperature rise in each of the four 8.0-kg brake drums in °C? (The specific heat of each iron brake drum is 448 J/kg⋅°C).
To solve this problem it is necessary to apply the concepts related to energy conservation.
In this case the kinetic energy is given as
[tex]KE = \frac{1}{2} mv^2[/tex]
Where,
m = mass
v= Velocity
In the case of heat lost energy (for all 4 wheels) we have to
[tex]Q = mC_p \Delta T \rightarrow 4Q = 4mC_p \Delta T[/tex]
m = mass
[tex]C_p =[/tex] Specific Heat
[tex]\Delta T[/tex]= Change at temperature
For conservation we have to
[tex]KE = Q[/tex]
[tex]\frac{1}{2} mv^2 = 4mC_p \Delta T[/tex]
[tex]\Delta T = \frac{1}{2}\frac{mv^2}{4mC_p}[/tex]
[tex]\Delta T = \frac{1}{2}\frac{(1500)(30)^2}{4(8)(448)}[/tex]
[tex]\Delta T = 47.084\°C \approx 47\°C[/tex]
Therefore the temperature rises in each of the four brake drums around to 47°C
Halley's comet has an elliptical orbit with the sun at one focus. Its orbit shown below is given approximately by 10.71 r - 1 + 0.883 sin θIn the formula, r is measured in astronomical units. (One astronomical unit is the average distance from Earth to the sun, approximately 93 million miles.) Find the distance from Halley's comet to the sun at its greatest distance from the sun. Round to the nearest hundredth of an astronomical unit and the nearest million miles.
A. 12.13 astronomical units; 1128 million miles
B. 91.54 astronomical units; 8513 million miles
C. 5.69 astronomical units; 529 million miles
D. 6.06 astronomical units; 564 million miles
The formula of an elliptical orbit is given by
[tex]r = \frac{A}{1+Bsin\theta}[/tex]
Assuming the given expression that was wrongly typed and whose true function is
[tex]r = \frac{10.71}{1+0.883sin\theta}[/tex]
We could start by deducing that the greatest distance from the sun would be given at the angle
[tex]\theta = \frac{3\pi}{2}[/tex]
For that value the value of [tex]sin\theta=-1[/tex]
[tex]r = \frac{10.71}{1+0.883(-1)}[/tex]
[tex]r = 91.538 AU[/tex]
That is equal to
[tex]r = 91.54Au* (\frac{93*10^6milles}{1AU})[/tex]
[tex]r = 8513[/tex] million miles
Therefore the correct option is B.
A cylinder with a piston contains 0.300 mol of oxygen at 2.50×105 Pa and 360 K . The oxygen may be treated as an ideal gas. The gas first expands isobarically to twice its original volume. It is then compressed isothermally back to its original volume, and finally it is cooled isochorically to its original pressure.Find the work done by the gas during the initial expansion.Find the heat added to the gas during the initial expansion.Find internal-energy change of the gas during the initial expansion.Find the work done during the final cooling;Find the heat added during the final cooling;Find the internal-energy change during the final cooling;Find the internal-energy change during the isothermal compression
Answer:
a) W = 900 J. b) Q = 3142.8 J . c) ΔU = 2242.8 J. d) W = 0. e) Q = 2244.78 J. g) Δ U = 0.
Explanation:
(a) Work done by the gas during the initial expansion:
The work done W for a thermodynamic constant pressure process is given as;
W = p Δ V
where
p is the pressure and Δ V is the change in volume.
Here, Given;
P 1 = i n i t i a l p r e s s u r e = 2.5 × 10^ 5 P a
T 1 = i n i t i a l t e m p e r a t u r e = 360 K
n = n u m b er o f m o l e s = 0.300 m o l
The ideal gas equation is given by
P V = nRT
where ,
p = absolute pressure of the gas
V = volume of the gas
n = number of moles of the gas
R = universal gas constant = 8.314 K J / m o l K
T = absolute temperature of the gas
Now we will Calculate the initial volume of the gas using the above equation as follows;
PV = n R T
2.5 × 10 ^5 × V 1 = 0.3 × 8.314 × 360
V1 = 897.91 / 250000
V 1 = 0.0036 m ^3 = 3.6×10^-3 m^3
We are also given that
V 2 = 2× V 1
V2 = 2 × 0.0036
V2 = 0.0072 m^3
Thus, work done is calculated as;
W = p Δ V = p×(V2 - V1)
W = ( 2.5 × 10 ^5 ) ×( 0.0072 − 0.0036 )
W = 900 J.
(b) Heat added to the gas during the initial expansion:
For a diatomic gas,
C p = 7 /2 ×R
Cp = 7 /2 × 8.314
Cp = 29.1 J / mo l K
For a constant pressure process,
T 2 /T 1 = V 2 /V 1
T 2 = V 2 /V 1 × T 1
T 2 = 2 × T 1 = 2×360
T 2 = 720 K
Heat added (Q) can be calculated as;
Q = n C p Δ T = nC×(T2 - T1)
Q = 0.3 × 29.1 × ( 720 − 360 )
Q = 3142.8 J .
(c) Internal-energy change of the gas during the initial expansion:
From first law of thermodynamics ;
Q = Δ U + W
where ,
Q is the heat added or extracted,
Δ U is the change in internal energy,
W is the work done on or by the system.
Put the previously calculated values of Q and W in the above formula to calculate Δ U as;
Δ U = Q − W
ΔU = 3142.8 − 900
ΔU = 2242.8 J.
(d) The work done during the final cooling:
The final cooling is a constant volume or isochoric process. There is no change in volume and thus the work done is zero.
(e) Heat added during the final cooling:
The final process is a isochoric process and for this, the first law equation becomes ,
Q = Δ U
The molar specific heat at constant volume is given as;
C v = 5 /2 ×R
Cv = 5 /2 × 8.314
Cv = 20.785 J / m o l K
The change in internal energy and thus the heat added can be calculated as;
Q = Δ U = n C v Δ T
Q = 0.3 × 20.785 × ( 720 - 360 )
Q = 2244.78 J.
(f) Internal-energy change during the final cooling:
Internal-energy change during the final cooling is equal to the heat added during the final cooling Q = Δ U .
(g) The internal-energy change during the isothermal compression:
For isothermal compression,
Δ U = n C v Δ T
As their is no change in temperature for isothermal compression,
Δ T = 0 , then,
Δ U = 0.
During the initial expansion, the gas performed 900 J of work, absorbed 3142.8 J of heat, resulting in a change in internal energy of 2242.8 J. In subsequent processes, work and internal energy changes were determined, culminating in an isothermal compression with no internal energy change.
(a) Work done by the gas during the initial expansion:
The work done (W) for a thermodynamic constant pressure process is given by W = P ΔV, where P is the pressure and ΔV is the change in volume.
Given:
P₁ = initial pressure = 2.5 × 10^5 Pa
T₁ = initial temperature = 360 K
n = number of moles = 0.300 mol
The ideal gas equation is PV = nRT, where P is the absolute pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the universal gas constant (8.314 kJ/mol·K), and T is the absolute temperature of the gas.
Calculate the initial volume of the gas:
P₁V₁ = nRT₁
(2.5 × 10^5) × V₁ = 0.3 × 8.314 × 360
V₁ = (0.3 × 8.314 × 360) / (2.5 × 10^5)
V₁ = 0.0036 m³ = 3.6 × 10⁻³ m³
Given V₂ = 2 × V₁:
V₂ = 2 × 0.0036
V₂ = 0.0072 m³
Now, calculate the work done:
W = P ΔV = (2.5 × 10^5) × (0.0072 - 0.0036)
W = 900 J
(b) Heat added to the gas during the initial expansion:
For a diatomic gas, Cp = (7/2)R, where Cp is the molar heat capacity at constant pressure.
Cp = (7/2) × 8.314
Cp = 29.1 J/mol·K
For a constant pressure process, T₂/T₁ = V₂/V₁:
T₂ = (V₂/V₁) × T₁
T₂ = 2 × T₁ = 2 × 360 = 720 K
Heat added (Q) can be calculated as:
Q = nCpΔT = 0.3 × 29.1 × (720 - 360)
Q = 3142.8 J
(c) Internal-energy change of the gas during the initial expansion:
From the first law of thermodynamics Q = ΔU + W, where Q is the heat added or extracted, ΔU is the change in internal energy, and W is the work done on or by the system.
ΔU = Q - W = 3142.8 - 900
ΔU = 2242.8 J
(d) The work done during the final cooling:
The final cooling is a constant volume (isochoric) process, so there is no change in volume (ΔV = 0), and thus the work done is zero.
(e) Heat added during the final cooling:
For an isochoric process, Q = ΔU. The molar specific heat at constant volume is Cv = (5/2)R.
Cv = (5/2) × 8.314 = 20.785 J/mol·K
The change in internal energy and thus the heat added can be calculated as:
Q = ΔU = nCvΔT = 0.3 × 20.785 × (720 - 360)
Q = 2244.78 J
(f) Internal-energy change during the final cooling:
The internal-energy change during the final cooling is equal to the heat added during the final cooling, so ΔU = Q.
(g) The internal-energy change during the isothermal compression:
For isothermal compression, ΔU = 0 since there is no change in temperature (ΔT = 0).
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An object initially at rest falls from a height H until it reaches the ground. Two of the following energy bar charts represent the kinetic energy K and gravitational potential energy Ug of the object-Earth system at two positions. The first position is when the object is initially released, and the second position is when the object is halfway between its release point and the ground. Which two charts could represent the mechanical energy of the object-Earth system? Select two answers. by Ug K Ug к Mechanical Energy at Halfway Point Mechanical Energy at Release Point Mechanical Energy at Release Point Mechanical Energy at Halfway Point d. c. к Ug Ug K Mechanical Energy at Release Point Mechanical Energy at Halfway Point Mechanical Energy at Release Point Mechanical Energy at Halfway Point
Answer:
Initial: bar power U₀
Final: bar power U = U₀ / 2
bar Kinetic energy K = U₀ / 2
These two bars are half the height of the initial bar
Explanation:
To know which graph is correct, let's discuss the solution to the problem
Initial mechanical energy
Em₀ = U₀ = m g H
The mechanical energy at the midpoint
Em₂ = K + U₂
As there is no friction, mechanical energy is conserved
Em₀ = Em₂
U₀ = K + U₂
K = U₀ - U₂
K = m g (H - y₂)
Indicates that position 2 corresponds to y₂ = H / 2
K = m g (H –H / 2)
K = ½ m g H
K = ½ Uo
Therefore the graph must be
Initial: bar power U₀
Final: bar power U = U₀ / 2
bar Kinetic energy K = U₀ / 2
These two bars are half the height of the initial bar
Answer: B&C
Explanation:
Release point: Ug bar graph only K none
Halfway point: Ug and K are equal bar graph
second option,
Release point: empty graph
Halfway point: Ug down half K up half
If two electrons in the same atom have the same four quantum numbers, then they must have the same energy.a. trueb. falsec. They cannot both have the same four quantum numbers.
Final answer:
The statement that two electrons in the same atom with the same four quantum numbers must have the same energy is false.
Explanation:
The statement that two electrons in the same atom with the same four quantum numbers must have the same energy is false. The Pauli exclusion principle states that no two electrons in an atom can have the same set of four quantum numbers. The four quantum numbers are the principal quantum number (n), the angular momentum quantum number (l), the magnetic quantum number (m), and the spin quantum number (s).
Since the quantum numbers determine the energy and other properties of the electrons, if two electrons in the same atom have the same set of quantum numbers, they must have different spins in order to obey the Pauli exclusion principle. Therefore, they would have different energies.
The specific heat capacity of aluminum is about twice that of iron. Consider two blocks of equal mass, one made of aluminum and the other one made of iron, initially in thermal equilibrium.
Heat is added to each block at the same constant rate until it reaches a temperature of 500 K. Which of the following statements is true?
a. The iron takes less time than the aluminum to reach the final temperature.
b. The aluminum takes less time than the iron to reach the final temperature.
c. The two blocks take the same amount of time to reach the final temperature.
Answer:a
Explanation:
Given
Specific heat capacity of aluminium is twice that of iron.
[tex]c_{Al}=2c_{iron}[/tex]
Also mass of two blocks is equal
Rate of heat added is also same
[tex]\dot{Q_{Al}}=\dot{Q_{iron}}[/tex]
[tex]\frac{Q_{Al}}{t_{Al}}=\frac{Q_{iron}}{t_{iron}}[/tex]
[tex]\frac{mc_{Al}\Delta T}{t_{Al}}=\frac{mc_{iron}\Delta T}{t_{iron}}[/tex]
[tex]\frac{2c_{iron}}{t_{Al}}=\frac{c_{iron}}{t_{iron}}[/tex]
[tex]t_{Al}=2t_{iron}[/tex]
Thus Time taken by aluminium block will be more
The viewing screen in a double-slit experiment with monochromatic light. Fringe C is the central maximum. The fringe separation is ?y.
Part A What will happen to the fringe spacing if the wavelength of the light is decreased?
a. Δy will decrease
b. Δy will increase
c. Δy will not change
Part B What will happen to the fringe spacing if the spacing between the slits is decreased?
a. Δy will decrease
b. Δy will increase
c. Δy will not change
Part C What will happen to the fringe spacing if the distance to the screen is decreased?
a. Δy will decrease
b. Δy will increase
c. Δy will not change
Part D Suppose the wavelength of the light is 460 nm . How much farther is it from the dot on the screen in the center of fringe E to the left slit than it is from the dot to the right slit?
Answer:
Part A:
a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.
Part B:
b) If the spacing between the slits is decreased the fringe spacing Δy will increase.
Part C:
a) If the distance to the screen is decreased the fringe spacing will decrease.
Part D:
The dot in the center of fringe E is [tex]920\ x\ 10^{-9} m[/tex] farther from the left slit than from the right slit.
Explanation:
In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.
The position of bright fringes in the screen where the pattern is formed can be calculated with
[tex]\vartriangle y =\frac{m \lambda L}{d} [/tex]
[tex]m=0,\pm 1,\pm 2,\pm 3,.....[/tex]
m is the order number.[tex]\lambda[/tex] is the wavelength of the monochromatic light.L is the distance between the screen and the two slits.d is the distance between the slits.Part A: a) In the above equation for the position of bright fringes we can see that if the wavelength of the light [tex]\lambda[/tex] is decreased the overall effect will be that the fringes are going to be closer. That means that the fringe spacing Δy will decrease.Part B: b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.Part C: a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at the center of the fringe E in the screen we use the condition for constructive interference. That says that the path length difference Δr between rays coming from the left and right slit must be [tex]\vartriangle r=m \lambda[/tex]We simply replace the values in that equation :
[tex]\vartriangle r= m \lambda =2.\ 460\ nm[/tex]
[tex]\vartriangle r= 920\ x\ 10^{-9} m[/tex]
The dot in the center of fringe E is [tex]920\ x\ 10^{-9}m[/tex] farther from the left slit than from the right slit.
By using the general equation for the double-slit maximums, we will get
A) aB) bC) aD) Can't be done completely, but below there is a approach to it.What is the distance between consecutive maximums?The equation for the constructive interference in a double-slit experiment is given by:
[tex]y = \frac{m*\lambda*D}{d} [/tex]
Where:
m is the number of the maximum.λ is the wavelength.D is the distance between the double-slit and the screen.d is the distance between the slits.Now let's answer:
A) If the wavelength is decreased, then the numerator is decreased, meaning the separation between consecutive fringes will also be decreased, so the correct option is a.
B) If d is decreased then the denominator decreases, meaning that the distance between consecutive fringes increases, so the correct option is b.
C) is the distance D is decreased, similar like in case A, the numerator decreases, meaning that the correct option is a again.
D) Sadly, as we do not know:
Which fringe is E.The value of DThe value of d.We can't answer this question.
What we should do here, is to compare the distance between the fringe and each slit, that distance will be the hypotenuse of a right triangle with one cathetus equal to D, and the other cathetus equal to y ± d/2, where each sign corresponds to each slit.
Then the difference in the distance will just be:
[tex]\sqrt{(y - d/2)^2 + D^2} - \sqrt{(y + d)^2 + D^2} [/tex]
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You need to determine the density of a ceramic statue. If you suspend it from a spring scale, the scale reads 28.4 N. If you then lower the statue into a tub of water so that it is completely submerged, the scale reads 17.0 N. For the steps and strategies involved in solving a similar problem, you may view a Video Tutor Solution.
Answer:
[tex]2491 kg/m^3[/tex]
Explanation:
Suppose g = 9.8 m/s2. When the statue is suspended from the spring scale, the scale reads 28.4 N. This means the mass of that statue is:
[tex]m = N/g = 28.4 / 9.8 = 2.9 kg[/tex]
When the tub is lowered and submerged in water, the scale reads 17N. So the statue is subjected to a force that make the difference of 28.4 - 17 = 11.4N. This equals to the gravity force of water displaced.
[tex]\rho_wVg = 11.4[/tex]
Let water density [tex]\rho_w = 1000kg/m^3[/tex], we can calculate the volume of the water displaced, which is also the volume of the statue:
[tex]V = \frac{11.4}{g\rho_w} = \frac{11.4}{9.8*1000} = 0.00116 m^3[/tex]
The density of the statue is mass divided by its volume:
[tex]\rho = \frac{m}{V} = \frac{2.9}{0.00116} = 2491 kg/m^3[/tex]
This question involves the concepts of density, weight, volume, and buoyant force.
The density of the ceramic statue is "2495.5 kg/m³".
First, we will find out the mass of the statue:
[tex]W = mg\\m=\frac{w}{g}[/tex]
where,
W = hanging weight of statue = 28.4 N
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]m =\frac{28.4\ N}{9.81\ m/s^2}\\[/tex]
m = 2.9 kg
Now, we will find out the volume of the statue. The difference, in weight of the statue upon submerging, must be equal to the buoyant force applied by the water. This buoyant force is equal to the weight of the volume of water displaced, which is equal to the volume of the statue.
[tex]Difference\ in\ weight\ of\ statue=(Density\ of\ water)(Volume\ of\ Statue)g\\28.4\ N-17\ N=(1000\ kg/m^3)(V)(9.81\ m/s^2)\\\\V=\frac{11.4\ N}{(1000\ kg/m^3)(9.81\ m/s^2)}[/tex]
V = 1.16 x 10⁻³ m³
Now, the density of the ceramic is given as follows:
[tex]\rho = \frac{m}{V} = \frac{2.9\ kg}{1.16\ x\ 10^{-3}\ m^3}\\\\\rho=2495.5\ kg/m^3[/tex]
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The earth has radius R. A satellite of mass 100 kg is in orbit at an altitude of 3R above the earth's surface. What is the satellite's weight at the altitude of its orbit
Answer:
W= 61.3 N
Explanation:
The only force acting on the satellite is the one due to the attraction from Earth, which obeys the Newton's Universal Law of Gravitation, as follows:
Fg =G*ms*me / (res)²
This force, also obeys the Newton's 2nd Law, so we can write the following equation:
G*ms*me*/ (res)² = ms* a = ms*g
We call to the product of the mass times the acceleration caused by gravity (g), the weight of this mass, so we can write as follows:
G*ms*me / (res)² = ms*g = W (1)
where G = 6.67*10⁻11 N*m²/kg², ms= 100 kg, me= 5.97*10²⁴ kg, and
res= 4 *re = 4*6.37*10⁶ m.
Replacing all these known values in (1), we get the value of W:
W =(( 6.67*5.97/(4*6.37)²) *( 10⁻¹¹ * 10²⁴ /10¹²) )* 100 N = 61.3 N
The satellite's weight at the altitude of its orbit is;
61.31 N
Formula for gravitational force is;
F = GMm/R²
Where;
G is gravitational constant = 6.67 × 10⁻¹¹ N.m²/kg²
m is mass of earth = 5.97 × 10²⁴ kg
M is mass of satellite = 100 kg
Now, we are told that the altitude is 3R above the Earth's surface.
At the Earth's surface, the distance from the Earth's center is R where R is radius of earth.
Thus, total altitude from the Earth's center to the satellite it (3R + R) = 4R
Thus;
F = GMm/(4R)²
Where R is radius of earth = 6371 × 10⁶ m
Thus;
F = (6.67 × 10⁻¹¹ × 100 × 5.97 × 10²⁴)/(4 × 6371 × 10⁶)
F = 61.31 N
Now, from Newton's second law of motion, we know that the force is equal to the weight.
Thus;
Weight = 61.31 N
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A monochromatic light passes through a narrow slit and forms a diffraction pattern on a screen behind the slit. As the wavelength of the light decreases, the diffraction pattern
a. spreads out with all the fringes getting wider.
b. becomes dimmer.
c. spreads out with all the fringes getting alternately wider and then narrower.
d. shrinks with all the fringes getting narrower.
e. remains unchanged.
We need to apply the definition of Young's double slit experiment. For fringe width of bright and dark fringe we have that
[tex]\beta = \frac{D\lambda}{d}[/tex]
Or expressed in terms of the wavelength we have that
[tex]\lambda = \frac{\beta d}{D}[/tex]
Where,
[tex]\lambda[/tex]= Wavelength
[tex]\beta[/tex]= Fringe width
d = Slit separation
D = Distance between slit and screen
From the ratios given in the equation, we have that as the wavelength decreases, the pattern determined for the diffraction pattern shrinks, which therefore causes all fringes to get narrower.
Final answer:
When the wavelength of monochromatic light decreases, the single-slit diffraction pattern shrinks with fringes getting narrower (option d). A decrease in slit width results in a wider diffraction pattern. More lines per centimeter on a diffraction grating cause bands to spread farther from the central maximum.
Explanation:
When a monochromatic light passes through a narrow slit, it forms a diffraction pattern due to the phenomenon of diffraction. The pattern consists of a series of bright and dark bands. The bright areas are known as maxima, while the dark areas are known as minima.
If the wavelength of the light decreases, the diffraction pattern shrinks, with all the fringes getting narrower (answer choice d). This is because the angle of diffraction is directly related to the wavelength, and as the wavelength becomes shorter, the angle at which light is bent decreases, causing the fringes to become narrower and the overall pattern to shrink.
Similarly, if the width of the slit producing a single-slit diffraction pattern is reduced, the pattern produced changes as well, with the bands spreading out and becoming wider. This is due to the inverse relationship between slit width and the angle of diffraction. A smaller slit width results in a larger spread of the diffraction pattern.
Lastly, if pure-wavelength light falls on a diffraction grating that has more lines per centimeter, the interference pattern will have bands that spread farther from the central maximum because a higher number of slits per unit area increases the diffractive effects.
Water is pumped from a lake to a storage tank 15 m above at a rate of 70 L/s while consuming 15.4 kW of electric power. Disregarding any frictional losses in the pipes and any changes in kinetic energy, determine (a) the overall efficiency of the pump-motor unit and (b) the pressure difference between the inlet and the exit of the pump.
Answer:
(a) 66.9%
(b) 147.14 kPa
Explanation:
Given:
Elevation of the water tank z = 15 m
Water volume flow rate V = 70 L/s = 0.07 m³/s
Input electric power consumption by the pump, Welec, in = 15.4 kW
Assuming there are no frictional losses in the pipes and changes in kinetic energy, the efficiency of the pump-motor will be;
η = ΔEmech ÷ Welec,
Where;
η is the overall efficiency
ΔEmech is the workdone to move the water pumped from the lake to a storage tank 15m above
Welec is the Input electric power consumption by the pump
Solving for ΔEmech ;
ΔEmech = mgh
mass, m = density × volume
Density of water = 1000 kg/m³
m = 1000 kg/m³ × 0.07 m³/s
m = 70 kg/s
∴ ΔEmech = 70 × 9.8 × 15
= 10.3 kW
Substituting the values of ΔEmech and Welec to calculate the overall efficiency
η = (10.3 kW ÷ 15.4 kW) × 100 %
= 0.6688 × 100 %
= 66.88 %
= 66.9 %
The overall efficiency of the pump-motor unit is = 66.9 %
(b) The pressure difference between the inlet and the exit of the pump is calculated to be;
Pressure = ΔEmech ÷ V
= 10.3 ÷ 0.07
= 147.14 kPa
This question involves the concept of potential energy, pressure difference, and electrical work.
(a) Efficiency of pump-motor unit is "66.9 %".
(b) The pressure difference between the inlet and the exit of the pump is "147.15 KPa".
(a) Pump EfficiencyThe efficiency of the pump-motor unit can be given by the following formula:
[tex]\eta = \frac{W_{P.E}}{W_{elect}}[/tex]
where,
[tex]\eta[/tex] = efficiency = ?P.E = power due to potential energy = [tex]\frac{mgh}{t}=\frac{\rho Vgh}{t}[/tex][tex]\rho[/tex] = density of water = 1000 kg/m³g = 9.81 m/s²h = height = 15 m[tex]\frac{V}{t}[/tex] = volume flow rate = 70 L/s = 0.07 m³/s[tex]W_{elect}[/tex] = electrical power input = 15.4 KW = 15400 WTherefore,
[tex]\eta = \frac{\rho Vgh}{t\ W_{elect}}\\\\\eta=\frac{(1000\ kg/m^3)(0.07\ m^3/s)(9.81\ m/s^2)(15\ m)}{15400\ W}\\\\\eta =0.669 = 66.9\ \%[/tex]
(b) PRESSURE DIFFERENCEThe pressure difference between inlet and outlet of the pump can be found using the following equation:
[tex]\Delta P = \rho gh\\\\\Delta P = (1000\ kg/m^3)(9.81\ m/s^2)(15\ m)[/tex]
[tex]\Delta P = 147150\ Pa = 147.15\ KPa[/tex]
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A flight attendant pulls her 70.0 N flight bag a distance of 253 m along a level airport floor at a constant velocity. The force she exerts is 40.0 N at an angle of 52 degrees above the horizontal. Find the following: a) the work she does on the flight bag b) the work done on the flight bag c) the coefficient of kinetic friction between the flight bag and the floor.
Answer:
6230.49413 J
0 J
0.63998
Explanation:
F = Force = 40 N
[tex]\theta[/tex] = Angle = 52°
Work done is given by
[tex]W=Fscos50\\\Rightarrow W=40\times 253\times cos52\\\Rightarrow W=6230.49413\ J[/tex]
The work she does on the flight bag is 6230.49413 J
The work done on the flight bag will be the opposite of the work done by the flight attendant
[tex]W=-6230.49413\ J[/tex]
So net work will be
[tex]W_n=6230.49413-6230.49413\\\Rightarrow W_n=0[/tex]
The net work done on the flight bag is 0 J
Coefficient of friction is given by
[tex]\mu=\dfrac{F_f}{F_N}\\\Rightarrow \mu=\dfrac{F_f}{F_g-F_{app}}\\\Rightarrow \mu=\dfrac{40cos52}{70-40sin52}\\\Rightarrow \mu=0.63998[/tex]
The coefficient of friction is 0.63998
(a) The work done by the attendant on the flight bag is 6230.49 J.
(b) The work on the flight bag is 0 J .
(c) The coefficient of kinetic friction between the flight bag and the floor is 0.35.
Given data:
The weight of flight bag is, W = 70.0 N.
The distance covered by the bag is, d = 253 m.
The magnitude of force exerted on bag is, F = 40.0 N.
The angle of inclination with horizontal is, [tex]\theta =52^{\circ}[/tex].
Work done defined as the product of force and distance covered due to applied force.
(a)
The work done on the flight bag is given as,
[tex]W'=F \times dcos\theta[/tex]
Solving as,
[tex]W'=40 \times 253cos52\\W'=6230.49 \;\rm J[/tex]
Thus, the work done by the attendant on the flight bag is 6230.49 J.
(b)
The work done on the flight bag will be the opposite of the work done by the flight attendant. So,
W'' = - W
W'' = - 6230.49 J
Then net work done on the flight bag is,
[tex]W_{net}=W'+W''\\W_{net}=6230.49 - 6230.49\\W_{net}=0[/tex]
Thus, the net work on the flight bag is 0 J .
(c)
The expression for the frictional force is given as,
[tex]f = \mu \times W[/tex]
[tex]\mu[/tex] is the coefficient of kinetic friction. And the frictional force is due to the horizontal component of applied force. Then,
[tex]f=Fcos\theta[/tex]
So,
[tex]Fcos\theta= \mu \times W\\\\40 \times cos52=\mu \times 70\\\\\mu=\dfrac{40 \times cos52}{70} \\\\\mu = 0.35[/tex]
Thus, the coefficient of kinetic friction between the flight bag and the floor is 0.35.
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A 30 gram bullet is shot upward at a wooden block. The bullet is launched at the speed vi. It travels up 0.40 m to strike the wooden block. The wooden block is 20 cm wide and 10 cm high and its thickness gives it a mass of 500 g. The center of mass of the wooden block with the bullet in it travels up a distance of 0.60 m before reaching its maximum height.
Answer
Mass of bullet (m) = .03 kg
Mass of wooden block M = 0.5 kg
Since the center of mass of the wooden block with the bullet in it travels up a distance of 0.60 m before reaching its maximum height
finding the launch speed of bullet
Velocity of wooden block + bullet just after impact
= [\tex]\sqrt{2gh}[/tex]
=[\tex]\sqrt{2\times 9.8 \times 0.6}[/tex]
= 3.43 m/s
v₁ be the launch velocity
Applying law of conservation of momentum
0.03 x v₂ = 0.530 x 3.43
v₂ = 60.6 m /s
if v₁ be initial velocity
v₂² = v₁² + 2 g h
v₁² = v₂² - 2 gh
v₁² = 60.6 ² - 2 x (-9.8 )x 0.4
v₁ = 60.65 m /s this is launch speed
A water wave traveling in a straight line on a lake is described by the equation:y(x,t)=(2.75cm)cos(0.410rad/cm x+6.20rad/s t)Where y is the displacement perpendicular to the undisturbed surface of the lake. a. How much time does it take for one complete wave pattern to go past a fisherman in a boat at anchor, and what horizontal distance does the wave crest travel in that time? b. What are the wave number and the number of waves per second that pass the fisherman? c. How fast does a wave crest travel past the fisherman, and what is the maximum speed of his cork floater as the wave causes it to bob up and down?
Answer:
A) The wave equation is defined as
[tex]y(x,t) = A\cos(kx + \omega t)=0.0275\cos(0.0041x + 6.2t)\\[/tex]
Using the wave equation we can deduce the wave number and the angular velocity. k = 0.0041 and ω = 6.2.
The time it takes for one complete wave pattern to go past a fisherman is period.
[tex]\omega = 2\pi f\\ f = 1/ T[/tex]
T = 1.01 s.
The horizontal distance the wave crest traveled in one period is
[tex]\lambda = 2\pi / k = 2\pi / 0.0041 = 1.53\times 10^3~m[/tex]
[tex]y(x = \lambda,t = T) = 0.0275\cos(0.0041*1.53*\10^3 + 6.2*1.01) = 0.0275~m[/tex]
B) The wave number, k = 0.0041 . The number of waves per second is the frequency, so f = 0.987.
C) A wave crest travels past the fisherman with the following speed
[tex]v = \lambda f = 1.53\times 10^3 * 0.987 = 1.51\times 10^3~m/s[/tex]
The maximum speed of the cork floater can be calculated as follows.
The velocity of the wave crest is the derivative of the position with respect to time.
[tex]v(x,t) = \frac{dy(x,t)}{dt} = -(6.2\times 0.0275)\sin(0.0041x + 6.2t)[/tex]
The maximum velocity can be found by setting the derivative of the velocity to zero.
[tex]\frac{dv_y(x,t)}{dt} = -(6.2)^2(0.0275)\cos(0.0041*1.53\times 10^3 + 6.2t) = 0[/tex]
In order this to be zero, cosine term must be equal to zero.
[tex]0.0041*1.53\times 10^3 + 6.2t = 5\pi /2\\t = 0.255~s[/tex]
The reason that cosine term is set to be 5π/2 is that time cannot be zero. For π/2 and 3π/2, t<0.
[tex]v(x=\lambda, t = 0.255) = -(6.2\times0.0275)\sin(0.0041\times 1.53\times 10^3 + 6.2\times 0.255) = -0.17~m/s[/tex]
(a) The time taken "1.013 s".
(b) Number of waves "0.987 Hz".
(c) Maximum speed "0.1750 m/s".
A further explanation is below.
Given:
[tex]y(x,t) = (2.75 \ cm) Cos [(0.41 \ rad/cm)x+(6.20 \ rad/s)t][/tex](a)
The time taken will be:
→ [tex]T = \frac{2 \pi}{W}[/tex]
[tex]= \frac{2 \pi}{6.20}[/tex]
[tex]= 1.013 \ s[/tex]
The covered horizontal distance will be:
→ [tex]\lambda = \frac{2 \pi}{K}[/tex]
[tex]= \frac{2 \pi}{0.410}[/tex]
[tex]= 15.3 \ cm[/tex]
(b)
Wave number,
[tex]K = 0.410 \ rad/cm[/tex]The number of waves per second will be:
→ [tex]f = \frac{1}{T}[/tex]
[tex]= \frac{1}{1.013}[/tex]
[tex]= 0.987 \ Hz[/tex]
(c)
The speed in which the wave crest travel will be:
→ [tex]v = f \lambda[/tex]
[tex]= 15.3\times 0.987[/tex]
[tex]= 15.1 \ cm/s \ or \ 0.151 \ m/s[/tex]
and,
The maximum speed of the cork floater will be:
→ [tex]v_1 = AW[/tex]
[tex]=2.75\times 6.20[/tex]
[tex]= 0.1750 \ m/s[/tex]
Thus the above answers are correct.
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https://brainly.com/question/14313611
The total volume in milliliters of a glucose-water solution is given by the equation below: V = 1001.93 + 111.5282m + 0.64698m2 where m is the molality of the solution. The partial molar volume of glucose ?glucose, is the slope of a V versus m curve, (?V/?m). Find the partial molar volume of glucose in a 0.100m solution of glucose in water
Answer:
111.657596
Explanation:
The expression of volume is given by
[tex]V=1001.93+111.5282+0.64698m^2[/tex]
Partially differentiating the term we get
[tex]\dfrac{\partial V}{\partial x}=\dfrac{\partial (1001.93+111.5282+0.64698m^2)}{\partial x}\\\Rightarrow \dfrac{\partial V}{\partial x}=111.5282+2\times 0.64698m\\\Rightarrow \dfrac{\partial V}{\partial x}=111.5282+1.29396m[/tex]
m = 0.100
[tex]\dfrac{\partial V}{\partial x}=111.5282+1.29396\times 0.100\\\Rightarrow \dfrac{\partial V}{\partial x}=111.657596[/tex]
The partial molar volume of glucose is 111.657596
In the fabrication of an electrical extension cord, the manufacturer wants to reduce the overall resistance of the wires in the extension cord.
Which of the following changes would result in the lowest resistance?
A) decrease the diameter of the wiresB) increase the diameter of the wiresC) choose a metal wire with a larger value of resistivityD) increase the length of the extension cord
Answer:
Option B is correct.
Explanation:
The resistance of a material or wire is represented by the formula below.
R=(ρL)/A
where
R is resistance of wireρ is the resistivity of the materialA is the cross sectional area of wireL is length of wireResistance increases with increase in length and resistivity which means options C and D are wrong.
A=(π[tex]d^{2}[/tex])/4
An increase in diameter will result in a proportional increase in area
From the resistance formula, an increase in area will cause a reduction in the resistance of the material.Thus increased diameter of wire will lower the resistance of wire. Option B is correct
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